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## How to Calculate and Solve for the Length of a Side and Perimeter of a Hexagon | The Calculator Encyclopedia

The image above is a hexagon.

To compute the perimeter of a hexagon, one essential parameter is needed and this parameter is length of side (a).

The formula for calculating the perimeter of a pentagon:

P = 6a

Where;
P = Perimeter of the hexagon
a = length of side

Let’s solve an example;
Find the perimeter of a hexagon when the length of side is 15 cm.

This implies that;
a = length of side = 15 cm

P = 6a
P = 6 x 15
P = 90

Therefore, the perimeter of the hexagon is 90 cm.

Calculating the length of side (a) using the Perimeter of the hexagon.

a = P / 6

Where;
a = length of side
P = Perimeter of the hexagon

Let’s solve an example;
Find the length of side when the perimeter of the hexagon is 120 cm.

This implies that;
P = Perimeter of the hexagon = 120 cm

a = P / 6
a = 120 / 6
a = 20

Therefore, the length of side (a) is 20 cm.

## How to Calculate and Solve for the Length of a Side and Area of a Hexagon | The Calculator Encyclopedia

The image above is a hexagon.

To compute the area of a hexagon, one essential parameter is needed and this parameter is length of side (a).

The formula for calculating the area of a hexagon:

A = (a2)3√32

Where;
A = Area of the hexagon
a = Length of side

Let’s solve an example;
Find the area of a hexagon when the length of side is 35 cm.

A = (a2)3√32
A = (352)3√32
A = 1225 x 3(1.73)2
A = 1225 x 5.1962
A = 6365.2872
A = 3182.64

Therefore, the area of the hexagon is 3182.64 cm2.

Calculating the length of side (a) using the area of the hexagon.

a = √(2A3√3)

Where;
a = length of side
A = Area of the hexagon

Let’s solve an example;
Given that the area of the hexagon is 120 cm2. Find the length of side?

This implies that;
A = Area of the hexagon = 120 cm2

a = √(2A3√3)
a = √(2 x 1205.196)
a = √(2405.196)
a = √46.189
a = 6.796

Therefore, the length of side (a) is 6.796 cm.

## How to Calculate and Solve for the Length of a Side and Perimeter of a Pentagon | Nickzom Calculator

The image above is a pentagon.

To compute the perimeter of a pentagon, one essential parameter is needed and this parameter is length of side (a).

The formula for calculating the perimeter of a pentagon:

P = 5a

Where;
P = Perimeter of the pentagon
a = length of side

Let’s solve an example;
Find the perimeter of a pentagon when the length of side (a) is 25 cm.

This implies that;
a = length of side = 25 cm

P = 5a
P = 5 x 25
P = 125

Therefore, the perimeter of the pentagon is 125 cm.

Calculating the length of side using the perimeter of the pentagon.

a = P / 5

Where;
a = Length of side
P = Perimeter of the pentagon

Let’s solve an example;
Given that the area of the pentagon is 250 cm. Find the length of side?

This implies that;
P = Perimeter of the pentagon = 250 cm

a = P / 5
a = 250 / 5
a = 50

Therefore, the length of side (a) is 50 cm.

## How to Calculate and Solve for the Length and Area of a Pentagon | Nickzom Calculator

The image above is a pentagon.

To compute the area of a pentagon, one essential parameter is needed and this parameter is length of side (a).

The formula for calculating the area of a pentagon:

A = a2(5(5 + 2√5) / 4

Where;
A = Area of the pentagon
a = length of side

Let’s solve an example;
Find the area of the pentagon when the length of side is 30 cm.

This implies that;
a = length of side = 30 cm

A = a2(5(5 + 2√5) / 4
A = 302(5(5 + 4.47) / 4
A = 900√(5(9.47) / 4
A = 900√47.36 / 4
A = 900√11.84
A = 900 x 3.44
A = 3096

Therefore, the area of the pentagon is 3096 cm2.

Calculating the length of side using the area of the pentagon.

a = √(4A / √5(5 + 2√5))

Where;
a = length of side
A = Area of the pentagon

Let’s solve an example;
Find the length of side with an area of 150 cm2.

This implies that;
A = Area of the pentagon = 150 cm2

a = √(4A / √5(5 + 2√5))
a = √(4 x 150 / √5(5 + 1.148))
a = √(600 / √5(6.148))
a = √(600 / √30.74)
a = √(600 / 5.54)
a = √108.30
a = 10.41

Therefore, the length of side is 10.41 cm.

## How to Calculate the Volume, Length, Width and Height of a Cuboid | Nickzom Calculator

The image above represents a cuboid.
To compute the volume of a cuboid requires three essential parameters which are the length, width and height of the cuboid.

The formula for computing the volume of a cuboid is:

V = lwh

Where:
V = Volume of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example
Find the volume of a cuboid with a length of 6 cm, width of 2 cm and a height of 10 cm.

This implies that:
l = length of the cuboid = 6
w = width of the cuboid = 2
h= = height of the cuboid = 10

V = lwh
V = 6 x 2 x 10
V = 120

Therefore, the volume of the cuboid is 120 cm3.

Calculating the Length of a cuboid when Volume, Width and Height are Given

The formula is l = V / (w)(h)

Where;
V = Volume of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example:
Find the length of a cuboid with a volume of 440 cm3 , a width of 5 cm and a height of 11 cm.

This implies that;
V = Volume of the cuboid = 440 cm3
w = width of the cuboid = 5 cm
h  = height of the cuboid = 11 cm

l =  V / (w)(h)
l = 440 / (5)(11)
l = 440 / 55
l = 8 cm

Therefore, the length of the cuboid is 8 cm.

Calculating the Width of a cuboid when Volume, Length and Height are Given

The formula is w = V / (l)(h)

Where;
V = Volume of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example:
Find the width of a cuboid with a volume of  180 cm3 , a length of 6 cm and a height of 10 cm

This implies that;
V = Volume of the cuboid = 180 cm3
l = length of the cuboid = 6 cm
h  = height of the cuboid = 10 cm

w =  V / (l)(h)
w = 180 / (6)(10)
w = 180 / 60
w = 3 cm

Therefore, the width of the cuboid is 3 cm.

Calculating the Height of a cuboid when Volume, Length and Width are Given

The formula is h = V / (l)(w)

Where;
V = Volume of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example:
Find the height of a cuboid with a volume of 195 cm3 , a length of 5 cm and a width of 3 cm

This implies that;
V = Volume of the cuboid = 195 cm3
l = length of the cuboid = 5 cm
w  = width of the cuboid = 3 cm

h =  V / (l)(w)
h = 195 / (5)(3)
h = 195 / 15
h = 13 cm

Therefore, the height of the cuboid is 13 cm.

## How to Calculate and Solve for the Length and Perimeter of a Rhombus | Nickzom Calculator

The image above is a rhombus.

To compute the area of a rhombus, one essential parameter is needed and this parameter is length of side (a).

The formula for calculating the perimeter of a rhombus:

P = 4a

Where;
P = Perimeter of the rhombus
a = length of side

Let’s solve an example;
Find the perimeter of a rhombus when the length of side is 30 cm.

This implies that;
a = Length of side = 30 cm

P = 4a
P = 4 x 30
P = 120

Therefore, the perimeter of the rhombus is 120 cm.

Calculating the Length of Side using the Perimeter of the Rhombus.

a = P / 4

Where;
a = Length of Side
P = Perimeter of the rhombus

Let’s solve an example;
Given that the perimeter of the rhombus is 180 cm. Find the length of side?

a = P / 4

This implies that;
P = perimeter of the rhombus = 180 cm

a = P / 4
a = 180 / 4
a = 45

Therefore, the length of side is 45 cm.

## How to Calculate and Solve for the Area, Length, Width and Height of a Cuboid | Nickzom Calculator

The image above represents a cuboid.
To compute the area of a cuboid requires three essential parameters which are the length, width and height of the cuboid.

The formula for computing the area of a cuboid is:

A = 2(lw + lh + wh)

Where:
A = Area of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example
Find the area of a cuboid with a length of 5 cm, width of 3 cm and a height of 9 cm.

This implies that:
l = length of the cuboid = 5
w = width of the cuboid = 3
h= = height of the cuboid = 9

A = 2(lw + lh + wh)
A = 2(5 x 3 + 5 x 9 + 3 x 9)
A = 2(87)
A = 174

Therefore, the area of the cuboid is 174 cm2.

Calculating the Length of a cuboid when Area, Width and Height are Given

The formula is l = A – 2(w)(h) / 2(w) + 2(h)

Where;
A = Area of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example:
Find the length of a cuboid with an area of 140 cm2 , a width of 4 cm and a height of 12 cm

This implies that;
A = Area of the cuboid = 140 cm2
w = width of the cuboid = 4 cm
h  = height of the cuboid = 12 cm

l =  A – 2(w)(h) / 2(w) + 2(h)
l = 140 – 2(4)(12) / 2(4) + 2(12)
l = 140 – 96 / 8 + 24
l = 44 / 32
l = 1.375 cm

Therefore, the length of the cuboid is 1.375 cm.

Calculating the Width of a cuboid when Area, Length and Height are Given

The formula is w = A – 2(l)(h) / 2(l) + 2(h)

Where;
A = Area of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example:
Find the width of a cuboid with an area of 200 cm2 , a length of 5 cm and a height of 12 cm

This implies that;
A = Area of the cuboid = 200 cm2
l = length of the cuboid = 5 cm
h  = height of the cuboid = 12 cm

w =  A – 2(l)(h) / 2(l) + 2(h)
w = 200 – 2(5)(12) / 2(5) + 2(12)
w = 200 – 120 / 10 + 24
w = 80 / 34
w = 2.353 cm

Therefore, the width of the cuboid is 2.353 cm.

Calculating the Height of a cuboid when Area, Length and Width are Given

The formula is h = A – 2(l)(w) / 2(l) + 2(w)

Where;
A = Area of the Cuboid
l = Length of the Cuboid
w = Width of the Cuboid
h = Height of the Cuboid

Let’s solve an example:
Find the height of a cuboid with an area of 300 cm2 , a length of 6 cm and a width of 2 cm

This implies that;
A = Area of the cuboid = 300 cm2
l = length of the cuboid = 6 cm
w  = width of the cuboid = 2 cm

h =  A – 2(l)(w) / 2(l) + 2(w)
h = 300 – 2(6)(2) / 2(6) + 2(2)
h = 300 – 24 / 12 + 4
h = 276 / 16
h = 17.25 cm

Therefore, the height of the cuboid is 17.25 cm.

## How to Calculate and Solve for the Length of the Diagonal and Area of a Rhombus | The Calculator Encyclopedia

The above image is a rhombus.

To compute the area of a rhombus, two essential parameters are needed and this parameters are length of the diagonal (p) and length of the diagonal (q).

The formula for calculating the area of a rhombus:

A = pq ⁄ 2

Where;
A = Area of the rhombus
p = length of the diagonal
q = length of the diagonal

Let’s solve an example;
Find the area of a rhombus when the length of the diagonal (p) is 10 cm and the length of the diagonal (q) is 18 cm.

This implies that;
p = length of the diagonal = 10 cm
q = length of the diagonal = 18 cm

A = pq2
A = 10 x 18 / 2
A = 180 / 2
A = 90

Therefore, the area of the rhombus is 90 cm2.

Calculating the length of the diagonal (p) using Area of the Rhombus and length of the diagonal (q).

p = 2A / q

Where;
A = Area of the rhombus
q = length of the diagonal

Let’s solve an example;
Given that the length of the diagonal (q) is 20 cm with an area of 60 cm2. Find the length of the diagonal (p)?

This implies that;
A = Area of the rhombus = 60 cm2
q = length of the diagonal = 20 cm

p = 2A / q
p = 2 x 60 / 20
p = 120 / 20
p = 6

Therefore, the length of the diagonal is 6 cm.

## How to Calculate and Solve for the Perimeter of a Trapezium | Nickzom Calculator

The image above is a trapezium.

To compute the perimeter of a trapezium, four essential parameters is needed and they are the length of the side (a), length of the side (b), length of the side (c) and length of the side (d).

The formula for calculating the perimeter of a trapezium is;

P = a + b + c + d

Where;

P = Perimeter of the trapezium
a = length of the side
b = length of the side
c = length of the side
d = length of the side

Let’s solve an example;
Find the perimeter of a trapezium with length of side (a) 4 cm, length of side (b) 8 cm, length of side (c) 12, length of side (d) 17 cm.

This implies that;
a = length of side = 4 cm
b = length of side = 8 cm
c = length of side = 12 cm
d = length of side = 17 cm

P = a + b + c + d
P = 4 + 8 + 12 + 17
P = 41

Therefore, the perimeter of the trapezium is 41 cm.

## How to Calculate and Solve for the Volume and Radius of a Sphere | The Calculator Encyclopedia

The image above is a sphere.

To compute the volume of a sphere, one essential parameter is needed and this parameter is the radius of the sphere (r). You can also use diameter of the sphere (d).

The formula for calculating the volume of a sphere:

V = (4/3)πr³

Where;

V = Volume of the sphere
r = Radius of the sphere

Let’s solve an example:
Find the volume of a sphere when the radius of the sphere is 10 cm.

This implies that;

r = Radius of the sphere = 10 cm

V = (4/3)πr³
V = (4/3)π x (10)³
V = (4/3)π x 1000
V = (4/3)3.142 x 1000
V = (4/3)3142
V = 1.33 x 3142
V = 4188.7

Therefore, the volume of the sphere is 4188.7 cm³.

Calculating the Volume of a Sphere using Diameter of the Sphere.

V = 4πd³ / 24

Where;

V = Volume of the sphere
d = Diameter of the sphere

Let’s solve an example:
Find the volume of a sphere when the diameter of the sphere is 20 cm.

This implies that;

d = Diameter of the sphere = 20 cm

V = 4πd³ / 24
V = 4 x 3.142 x (20)³ / 24
V = 4 x 3.142 x 8000 / 24
V = 100544 / 24
V = 4189.3

Therefore, the volume of the sphere with diameter is 4189.3 cm3.

Calculating the Radius of a Sphere using Volume of the Sphere.

r = 3√(3V / )

Where;

V = Volume of the sphere
r = Radius of the sphere

Let’s solve an example:
Find the radius of a sphere when the volume of the sphere is 250 cm3.

This implies that;

V = Volume of the sphere = 250 cm3

r = 3√(3V / )
r = 3√(3 x 250 / 4 x 3.142)
r = 3√(750 / 12.568)
r = 3√59.675
r = 3.907

Therefore, the radius of the sphere is 3.907 cm.

Calculating the Diameter of a Sphere using Volume of the Sphere.

d = 3√(24V / )

Where;

A = Area of the sphere
d = Diameter of the sphere

Let’s solve an example:
Find the diameter of a sphere when the volume of the sphere is 40 cm3.

This implies that;

V = Volume of the sphere = 40 cm3

d = 3√(24V / )
d = 3√(24 x 40 / 4 x 3.142)
d = 3√(960 / 12.568)
d = 3√76.38
d = 4.24

Therefore, the diameter of the sphere is 4.24 cm.


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HomeEnglishClass 12MathsChapterConic Sections

Find the area of the quadrilat...

# Find the area of the quadrilateral formed by common tangents drawn from a point P to the circle x^(2) + y^(2) = 8 and the parabola y^(2) =16x, chord of contact of tangents to the circle and chord of contact of tangents to the parabolas.

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Text Solution

Solution :
Let any tangent to the parabola y^(2) = 8x be y = mx + (4)/(m). It also touches the circle x^(2) + y^(2) = 8
 therefore Length of perpendicular from centre of the circle to the line = radius of the circle.
 rArr|(0-0+(4)/(m))/(sqrt(1+m^(2)))|=2sqrt(2)

 ((4)/(m))^(2) = 8 + 8m^(2)
 m^(2) + m^(2) -2 = 0
 (m^(2) + 2) (m^(2) -1) =0
 m = pm 1
 therefore The common tangents are
 y = x + 4 and y = - x - 4
 Point of intersection of these tangents is P(-4, 0) Equation of chord of contact to the circle from P(-4, 0) is
 -4x + 0 = 8 rArr x + 2 = 0
 Equation of chord of contact of tangents to the parabola is
 0 = 4(x -4)
 rArr x - 4 = 0
 Redrawing the figure formed by four lines, we require the area of the shaded region i.e., trapezium
 Required area  =(1)/(2) (AD + BC) xx MN
 =(1)/(2) (4+16)xx6
 =60 sq. units

Transcript

TimeTranscript
00:00 - 00:59hello Siri question says find the area of the quadrilateral drawn to the circle x 8 and contact of tangent to the parabola ok we know that a standard form is look something like this is equal to 16 that he will be equal to for this the tangent is equal to Y = MX + c then the X and Y Y is equal why
01:00 - 01:59is MX + for a circle the distance of the centre of the circle from the equal to radius circle as it 2000 and it is 22 Centre little bcs-040 and radius is equal to root therefore I can say that Mod of why will be 0 - 10 - 2 root 2 upon under root of coefficient of Y + Y square + x square by under root 1 + sin square is equal to I am so sorry I made a mistake dishoom equal to minus 4 by m
02:00 - 02:59m is equal to 22 so I can solve square and square is equal to 8 x + sin square is equal to MC square into X square + Y square will be equal to is equal to 1 and the other possibility is equal to minus 2 but since MS real this is not possible if I am still be equal to an equal to Angle B equal to plus minus one I can get the now the equation of the tangent will be by = S minus x + 4
03:00 - 03:59OK so I will get that the point of the point P - 404 and tigers 2 - 3 x + 4 equal to now I can find the contact with inner circle using the on people's 20 to see the circle is equal to zero is equal to zero record of music portal contact replacing x square with a square with xx1 y square with yy1 X with X + 1 by 2 Y + Y + 1 by 2 and she will remain cox1 and Y1 presence what is the external point to a circle
04:00 - 04:59are so as to from this we can say that people will represent the chord of contact of circle which will be in this case - 4 x + 20 equal to C is it that implies X equal to minus 2 upon x square x y z will be zero because I can write 16 X 16 Into X + 3 x minus 4 upon x is equal to s a
05:00 - 05:59x + 4 and Y equal to minus y equal to X + 4 and Y + 1 is equal to plot a graph say it will look something like it will do something like this and the points will be this this this and this is a 4 and the sin is equal to minus 2 is the exact and this will be the why so now we can see that it is a trapezium and the area of trapezium is equal to sum of parallel sides that is let it be L1 and L2 So album plus L2 and the height BH so into edge at this distance BH area so I can say that
06:00 - 06:59I have to find first L1 and L2 decided by cos x + 4 Allah to solve it with X and get this is this will be a comma Sunny -2 kam 2 and minus minus and this will be and aur Seth 1602 into the 16 - 24 - 641 finance
07:00 - 07:59options are not given that the answer is 240 thank you

format: plain


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## Articulo

• Similares en SciELO

## versión On-line ISSN 2448-6698versión impresa ISSN 2007-1124

### Rev. mex. de cienc. pecuarias vol.10 no.4 Mérida oct./dic. 2019  Epub 30-Abr-2020

#### https://doi.org/10.22319/rmcp.v10i4.4927

Articles

Technical optimum milk and meat production levels in dual-purpose cattle systems in tropical Mexico

a Universidad Autónoma de Tamaulipas. Facultad de Medicina Veterinaria y Zootecnia. Carretera Mante Km 5. 87000. Ciudad Victoria Tamaulipas, México.

b Instituto Nacional de Investigaciones Forestales Agrícolas y Pecuarias (INIFAP). Centro Nacional de Investigación Disciplinaria en Fisiología y Mejoramiento Animal, Ajuchitlan Querétaro, México.

e Campo Experimental General Terán, INIFAP. General Terán. Nuevo León, México.

Abstract

Inputs directly affect profitability in livestock production, although what effects they have vary in response to production system and input type. An analysis was done of the results from a milk and meat production function using data from dual-purpose system (DP) production units in three locations in tropical Mexico. Data were collected through monthly surveys and covered milk production, meat production, income and financial costs over a twelve-month period. The functions were estimated by the indirect linear regression method with transformed data for a Cobb-Douglas function. The milk function showed the feed and cows inputs to explain 91 % of production. Elasticity coefficients were 0.34 for feed and 0.5 for cows. Marginal products were 0.75 for milk and 892.2 for cows, with values of $4.03 L for milk and$ 4,800.20 per cow. Both inputs are in stage II of production with diminishing marginal returns. For meat production both the feed and cows’ inputs explained 72 % of production, with elasticities of production coefficients of -0.20 for feed and 1.11 for cows. Feed was in stage III of production with negative marginal returns, but the cows input was in stage I with increasing marginal returns. The sum of the coefficients was less than one for both functions (0.92 for feed, 0.91 for cows), indicating decreasing returns to scale. The optimum technical production levels were 488.97 L milk per day and 10 calves per year. In the studied producers the inputs for milk production were being used rationally, although in meat production feed appears to be overused and should be evaluated.

Key words Cobb-Douglas, Elasticity, Marginal product, Returns to scale; technical optimum

Resumen

El objetivo de este estudio fue estimar y analizar los resultados obtenidos de una función de producción de leche y carne en unidades de producción del sistema de doble propósito (DP). Los datos se obtuvieron a través de encuestas mensuales, donde se registró información de producción de leche, carne, ingresos y egresos económicos, durante 12 meses. Las funciones se estimaron por el método indirecto de regresión lineal con datos transformados para una función Cobb-Douglas. La función para leche mostró que los insumos alimento y vacas explican el 91 % de la producción, con coeficientes de elasticidad de 0.34 y 0.5, productos marginales de 0.75 y 892.2 con valores de $4.03 y$4,800.2 respectivamente. Los insumos utilizados para la producción de leche se encuentran en la etapa II de producción con rendimientos marginales decrecientes, mientras que para la producción de carne los insumos alimento y vacas explican el 72 % de la producción, con coeficientes de elasticidades de producción de -0.20 y 1.11 respectivamente; el alimento se ubicó en la etapa III de producción con rendimientos marginales negativos, y el insumo vacas se encontró en la etapa I con rendimientos marginales crecientes. La suma de los coeficientes de ambas funciones que fueron 0.92 y 0.91 por ser menores a uno, tienen rendimientos decrecientes a escala. El nivel óptimo técnico de producción fue de 488.97 L diarios y 10 becerros al año. Los insumos para producción de leche se están utilizando de forma racional, pero se debe evaluar la cantidad de alimento utilizada para producción de carne, ya que refleja que es sobre utilizado.

Palabras clave Cobb-Douglas; Elasticidad; Producto marginal; Rendimientos a escala y Óptimo técnico

Introduction

A total of 11.8 billion liters of milk were produced domestically and 3.7 million liters imported to meet domestic demand in Mexico in 2017. For the same period domestic beef production was 1.85 million tons with 136,000 t imported to meet demand1. Dairy and beef producers in Mexico clearly do not generate enough product to meet domestic demand, highlighting the need for quantitative analysis of the efficiency of specialized, semi-specialized, dual-purpose and family dairy and beef production systems to optimize resource use. Implemented mainly in tropical regions2, dual-purpose cattle systems (DP) are characterized by milk production coupled with sale of weaned calves for beef3. One of its main advantages is that feed costs are reduced since most systems are based on grazing with supplementation for lactating cows2.

A vital aspect of DPs is the need for efficiency analysis of production units to maximize appropriate use of inputs for production4. Parametric methodologies have been developed based on estimation of production functions to study the functioning of these systems and manifest cause-and-effect relationships in them. These methodologies identify the relationship between the amounts of different inputs and the quantities of resulting products, as well as associating each input with the maximum production level per period. This data can then be used to formulate productive development strategies for a specific region.

Physiological and non-physiological factors such as feed quantity and quality, fodders, herd size, season, and lactation number and stage, among others5, influence milk and beef production. It is therefore important to understand the factors that best explain production to facilitate selection of the inputs to be used and make optimal use of them6. The Cobb-Douglas function is widely used to identify production functions in livestock systems, and has been applied to estimate milk and beef production in different systems and regions in Mexico7,8,9. Indicators can be calculated using the properties of Cobb-Douglas type functions and the theory of production. Principal among these is elasticity of production, which is the percentage change in the amount produced relative to the percentage change in input levels10. Another indicator is marginal return, which describes production decreases or increases in response to addition of an input, and, depending on its behavior (i.e. increase, decrease, zero or negative), can indicate whether the input analyzed is in stage I, II or III of a classic production function. This indicator also allows identification of return types at the livestock production unit level, which helps to explain how production behaves in response to proportional and simultaneous variation of all inputs, which can be increasing, constant or decreasing. Input sales price is used to estimate the indicator marginal product, which is the variation in the quantity produced in response to unit increases in any production input (ceteris paribus) as well as the marginal product value, which is the additional income earned by a livestock company for each additional input unit11. These data are useful for economic advisers in the livestock sector, extension representatives advising producers and producers themselves. They help in making decisions on rational resource use, and appropriate increases or decreases in inputs for the production process, all aimed at augmenting profits. The present study objective was to use production functions to analyze data from representative dual-purpose (DP) system milk and beef producers in the Mexican tropics to estimate those inputs that have the greatest influence on production, and calculate the technical optimum levels subject to input price and milk and meat sale price to determine if they are being used rationally.

Material and methods

Study area

The study was done in production units (PU) in three states representative of the tropics in Mexico, and where the DP system predominates. Production units in the state of Tabasco (17°51’ N; 93°23’ W) were at 2 m asl, in an area with a warm humid climate and abundant summer rains, a 26.4 °C average annual temperature and 190.85 mm mean monthly rainfall. The units in Chiapas (15°41’12” N; 93°12’33” W) were at 57 m asl in an area with a warm subhumid climate, 28 °C average annual temperature, and 80 mm mean monthly rainfall. In Sinaloa (23°14’29” N; 106°24’35” W) the units were at 10 m asl, in an area with 26.0 °C average annual temperature, and 63 mm mean monthly rainfall12. All PU used Bos indicus x Bos taurus animals. Feeding was based on extensive grazing using supplementation with commercial balanced diets based on net lactation energy and 17 % protein during lactation for tall and medium-height cows. The average number of producing cows among the PU was 39. These were milked once a day using the calf to stimulate milk flow, extracting three-quarters of the udder for sale and leaving a quarter to feed the calf. Meat production in all units consisted of the sale of calves weaned at 160 kg average weight.

Input classification and productive variables

Data were collected via monthly producer surveys from June 2012 to July 2013, in 30 PU, 10 per state. Production unit (PU) selection was done by unrestricted random sampling from the PU registered in cooperating local cattle associations. The surveys consisted of forms with sections on herd structure, land use, income from sale of milk and meat, and expenses from supply purchases. The variables used in the production function were based on recommendations for estimating problems when a producer generates multiple products such as livestock and agricultural crops, and when these can change substantially from one region to another; for example, the quantities of concentrate feed, animals, labor and fuel13. A total of ten variables were recorded: total annual milk production (liters); total annual calf production; total amount of concentrate feed used in the PU per year (kg); number of producing cows; grazing area (hectares); preserved fodder used in PU per year (kg); full-time and seasonal labor (days); number of sires; operating supply costs (electricity, gasoline, diesel).

Data analysis

Milk and meat production functions were estimated using the indirect method to generate a Cobb-Douglas type function, which consists of a linear regression with the original data transformed to Neperian logarithms of the dependent and independent variables6.

After each of the variables was converted, the model that best explains milk and meat production was selected using the STEPWISE procedure in the SAS program. This procedure begins by calculating the simple correlation matrix, based on the correlation values; the independent variable (Xi) with the highest correlation to the response variable (Yi) is included in the model. Selection of the variable to include in the model was done using the partial correlation coefficients (R2). At each step the contribution of each variable to the model is examined by applying the partial F test as a criterion; therefore, at each stage all variables are examined for their unique contribution to the model, and those that do not meet a previously established criterion are eliminated.

The estimated specific model for milk is:

lnY1= β0+β1lnX1+ β2lnX2+β3lnX3+β4lnX4+β5lnX5+β6lnX6+β7lnX7+ε

The estimated specific model for meat is:

lnY2= β0+β1lnX1+ β2lnX2+β3lnX3+β4lnX4+β5lnX5+β6lnX6+β7lnX7+ε

Where:

Y1 = milk production;

Y2 = annual calf production;

X1 = concentrated feed used in PU in kg yr-1;

X2 = lactating cows;

X3 = grazed area in hectares;

X4 = preserved forage used in PU in kg yr-1;

X5 = labor;

X6 = sires;

X7 = operating supply costs (electricity, gasoline, diesel);

βi = parameters to be estimated (i = 0, 1, …,7);

ε= residual term.

After estimating the Cobb-Douglas function with the variables that best explain meat and milk production, the input coefficient values were used to calculate the elasticity of production, marginal returns and the production stage of each input. In a Cobb-Douglas function, each input’s coefficient value is equal to its elasticity of production. If this is greater than 1 the input has increasing marginal returns and if it is less than 1 it has diminishing marginal returns. In addition, each input’s elasticity of production indicates the production stage in which it is located: a value βi> 1 indicates stage I; βi <1 is stage II; and βi <0 is stage III11. The type of returns to scale of the studied livestock producers was identified using the sum of the input coefficients of the milk and meat production functions. Calculations were also done of the marginal product (PMgXi) and the value of the marginal product (VPMgXi) of the inputs derived from the elasticity formula using the means of total milk production and of the inputs, with the following formulas14.

Epb1=Y/YXi/Xi=XiYYXi=PMgXiPPXI

PMgXi= Epb1* PPXi

VPMgXi =PMgXi*PYI

Where,

PMgXi= Marginal product of inputXi

Epb1= Elasticity ofYi

Yi= Mean of annual milk or calf production

Xi= Mean of input used

VPMgXi= Value of marginal productXi

PYI= Unit price ofYi

PPXi= Average product of inputXiused

The average product of each input was the ratio between mean production (milk, calves) and input average (feed, cows).

Technical optimum milk and meat production levels were estimated by the Lagrange multiplier method, optimizing the milk and meat production functions (objective functions), subject to the prices of the inputs used and product sale price (one liter of milk and one calf).

L=fX1,X2-λ (PX1+PX2+M)

Where,

L = Lagrange Function

λ = Lagrange Multiplier

F (X1, X2) = Cobb-Douglas production function for milk and meat

PX1, PX2 = Price of variable inputs

M = Product unit price

The algebraic procedure consisted of subtracting the constraint from the objective function, and L (first order condition) was partially derived from X1, X2, and λ. Using the maximization rule, the ratio of partial derivatives was equalized to X1 and X2, which were limited to the input price ratio. The solution provided the values of X1 and X2, which were substituted in the Cobb-Douglas function, thus estimating the technical optimum levels for milk (in liters) and calf production.

Average sale price for a liter of milk was $5.38 and that for calves was$ 6,020. The average cost of one kilo feed was $4.00. The cost of a producing cow cost was estimated using the capital recovery formula15, where purchase cost of a replacement heifer was$ 18,000.00, assumed use life was 8 yr in a DP system, annual return rate was 12.5 %, and estimated cost of one cow per year was $500.00. Calculation of the technical optimum level in the milk production function was done considering the price of one cow per day; that is the quotient of the price of one cow per year / 365 d, which was$ 1.36.

Results and discussion

Herd structure in the studied PU varied with production intensity and the area avai/lable for livestock within the PU (Table 1). Constant movement also occurred due to cow physiological condition (heifer, dry, lactating) or animal purchase and sale16.

Table 1 Herd structure in double-purpose production units

Variable n Mean y- SD (S) CV
Producing cows 30 38.9 15.19 38.81
Dry cows 30 18.97 9.91 52.24
Heifers 30 19.80 11,71 59.14
Bull calf 30 14.22 8.54 60.05
Cow calf 30 12.40 6.73 54.27
Sires 30 2.49 1.49 59.83

n= number of production units; SD= standard deviation; CV= coefficient of variation.

In the milk and meat production function model, the coefficient of determination (R2) indicates that most of the variability in production is explained by the independent variables Feed (91.9 %) and cows (72.4 %) (Table 2). The percentage of unexplained variation in both models can be attributed to differences between PU such as herd management practices or environmental conditions. In milk production systems the feed input explains a greater percentage of variation in milk production than other inputs15,17. As a result, fresh fodder, preserved fodder and concentrate feed can be used strategically to increase milk production5,18. Feed handling and quality is clearly important in dairy production systems since it is directly related to production.

Table 2 Regression models selected for milk and meat production

Estimated parameter Standard error Pr>F
Milk
Intercept 6.099 0.310 <.0001
InX1 0.346 0.034 <.0001
In X2 0.542 0.095 <.0001
R2 0.919
bi= 0.888
Meat
Intercept 0.853 0.430 0.0579
InX1 -0.205 0.047 0.0002
In X2 1.118 0.133 <.0001
R2 0.724
bi= 0.913

InX1= Neperian logarithm of kilograms feed concentrate; In X2= Neperian logarithm of producing cows; R2 = coefficient of determination; bi= sum of bi coefficients.

Average herd size was 39 producing cows, average annual milk production was 93,678.5 L, and average annual calf production was 14 (Table 3).

Table 3 Means for annual milk and calf production, inputs used in double-purpose system production units

Function MPL MPB MIA MIV PPA PPV
Milk 93,678.5 ----------- 45,678.9 38.9 2.05 2,408.18
Meat ---------- 14.22 45,678.9 38.9 3.11X10-04 0.365

MPL= mean milk production kg PU yr-1; MPB= mean calf production heads PU yr-1; MIA= mean feed input kg PU yr-1; MIV= mean cow input head PU yr-1; PPA= average product of feed input; PPV= average product of cow input.

Cobb-Douglas production function for milk

InY1=6.099+0.346 InX1+0.542 InX2 (Equation 1)

After transformation with antilogarithms:

Milk=e6.099X10.346X20.542 (Equation 2)

Milk=445.69 X1 0.346X2 0.542 (Equation 3)

Elasticities of production for milk

In the present results a 1% increase in the X1 input (feed) raised total milk production by 0.34 % (ceteris paribus). A 1% increase in the X2 input (cows) raised total milk production by 0.54 %. Increases in either input positively affected milk production, although increasing the number of cows provided a better response in production than increasing feed. This coincides with reports from other dairy production systems in tropical climates similar to those of the study area in which the cow input provided the most elasticity in the studied inputs, with values ranging from 0.40 to 0.60 %, whereas the feed input produced elasticities of 0.15 to 0.30 %14. Understanding to what degree inputs impact production is vital when analyzing livestock producers since under certain circumstances increases in inputs can decrease production, resulting in negative elasticities10. For example, dairies in the tropics of India report a 2.4 % decrease in milk production as feed intake increases19, resulting in financial losses to producers.

Marginal returns and input production stage for milk production

The elasticity coefficients for the milk production function were 0.34 for feed and 0.54 for cows (Equation 1). According to the law of marginal returns both inputs have diminishing marginal returns because their values are less than 1. Also, they are in stage II of a classic production function, meaning that increases in these inputs will increase milk production. These production increases will become progressively less as input levels increase, until production becomes constant or begins to decrease, beginning stage III of production10. Increasing feed availability to cows early on will increase milk production, but when the animals reach maximum feed efficiency (i.e., the amount of feed in kilograms required by the animal to produce a liter of milk)20 their metabolism will be unable to absorb all the nutrients and translate them into greater milk production. These are then expelled in the urine and feces, representing financial losses for producers in the form of costs for excess feed. Increasing the number of cows (ceteris paribus) would reduce the availability of resources such as feed, causing a decrease in total milk production.

Returns to scale for milk production

The milk production homogeneous function exhibited decreasing returns to scale since the sum of the coefficients β1 and β2 was 0.888. Therefore a similar percentage increase in all inputs will cause a percentage increase of smaller magnitude in the product11. Similar results have been reported in milk production systems in the state of Sinaloa7 in which this effect is attributed to overuse of producer resources and absence of technology use in the system. In this scenario large livestock producers experience increasing returns to scale due to specialization in capital and labor8. Presence of this type of return to scale in PU using DP requires evaluation of input use because increasing feed and cow inputs will not raise income from greater production21, rather it will cause financial losses due to unnecessary input costs.

Marginal product value for milk production

Milk production marginal product results for the feed input indicated that adding 1 kg of feed would increase milk production by 0.75 L, generating additional income of $4.03 per unit of added input (ceteris paribus). Raising the number of cows in a herd would increase milk production by 892.2 L per year, generating additional income of$ 4,800.20. These results are similar to a study of a DP system in Sinaloa in which marginal product values greater than zero for the cow input caused diminishing marginal returns7. This means that increasing herd size to increase milk production is not the best option for improving efficiency in DP systems. Rather, a better approach is to make optimal use of the inputs that have the greatest impact on production. This is supported by the present marginal product values for the feed and cow inputs: both indicate positive economic benefits in producers, but at values less than 1. The law of marginal returns would classify these as diminishing marginal returns, placing them in stage II of a classic production function. Continued increases in these inputs will therefore cause the marginal product to continue decreasing until reaching zero, eventually becoming negative and leading to financial losses. Examples of this dynamic include production units in the eastern portion of the state of Yucatan, Mexico5, and tropical dairy production systems in India9, both of which have negative marginal returns for the feed input, and, even though marginal product values remain positive, milk production no longer increases.

Technical optimum milk production levels

Y1=445.69 X10.346X20.542    subject to 4.0 X1+1.36 X2=5.38

Using the Lagrange method:

L=445.69 X10.346X20.542-λ (4.0X1+1.36 X2-5.38)

The partial derivative of L for X1 and X2, under a first order condition:

4.0 λ=154.257X1-0.653 X20.542

1.36 λ=241.64 X10.346 X2-0.457

Equalizing the partial derivatives for X1 and X2 , and substituting X2 in the constraining equation generates the optimum amount for this input.

154.257X1-0.653 X20.542241.64 X10.346 X2-0.457=4.01.36

X2=4.0X10.229=17.41X1

X1=5.3827.677=0.19

X2=17.41 0.19=3.38

Substituting the X1 and X2 in the Cobb-Douglas function generates the optimum amount of milk produced.

Y1=445.69 (0.19) 0.3463.380.542=488.97 l

Livestock producers in the study area attained an optimal milk production of 488.97 L per day, which is equivalent to producing 12.53 L per day per cow, since on average they had 39 cows in production. In semi-intensive systems, cows must produce 35.38 L per day to achieve optimum milk production using a combination of concentrate feed and fodder inputs22. The main difference between the DP and semi-intensive systems is that the latter use a larger amount of concentrate.

Cobb-Douglas production function for meat

InY2=0.85366-0.20523InX1+1.11829InX2 (Equation 4)

Transformation via antilogarithms results in:

Y2=e0.85366 X1-0.20523X2 1.11829 (Equation 5)

Y2=2.348 X1 -0.20523X21.11829 (Equation 6)

Elasticities of production for meat

The meat production function (Equation 6) shows that a 1% change in the number of cows would increase calf production by 1.11 %, while the same change in the feed input would decrease it by 0.20 % (ceteris paribus). In DP systems, calf feeding is based on controlled lactation in the form of one quarter of the milk in the udder at milking and the quantity and quality of forage consumed when grazing23. Nutritional supplementation of calves with good quality diets does improve weaning weight in DP systems, but this does not increase the prices paid to the producer for calves. Producers therefore search for alternatives to reduce the weaning period through supplementation with alternative forages (e.g. forage trees and bushes) that improve pre- and post-weaning calf development24. The present results indicate that the amount of concentrate feed included in calf diets should be reduced because it does not improve overall production performance. Use of concentrate feed generally increases productive variables in livestock production systems25, although any improvements will depend on feed quantity and quality, since lack of data on the appropriate amount of feed can generate unnecessary costs and cause financial losses for producers.

Marginal returns and input production stage for meat production

The elasticity coefficient for the meat production function is negative for the feed input (X1), but greater than one for the cow input (X2) (Equation 4). Values greater than one indicate increasing marginal returns and that the input is in stage I of a classical production function10. Therefore, increasing the cows input would increase milk production at this stage, making it unadvisable for the producer to lower this input and consequently slow or stop production. Similar behavior has been reported in grazing systems in the State of Mexico and Yucatan8,9. In other words, increasing herd size within the resources available to the studied PU would raise yield as represented by production variables. In contrast, the feed input exhibited an elasticity of less than zero, representing negative marginal returns and placing it in stage III10. That is, increasing the amount of feed does not benefit calf production, and indeed could decrease it. A similar effect has been reported for beef production Yucatan, where increases in the amount of concentrate feed did not improve production9. Rather, a more effective way of increasing calf weaning weight was to properly manage existing pastures by using high quality forages that meet animal nutritional requirements. If the feed input is in stage III of production, the livestock producer is not economically viable because it is spending money on an input that does not increase income from increased production.

Returns to scale for meat production

The meat production functions exhibited diminishing returns to scale because b1  is less than one. That is, increasing all inputs in the same proportion would not increase total production, an effect similar to that reported in small- and medium-sized producers in the State of Mexico8. Large producers, in contrast, attain increasing returns to scale through genetic improvement (capital) and management efficiency strategies (labor)8. An alternative for improving returns to scale for small and medium producers using DP would be to increase adoption of technology to improve efficiencies.

Marginal product and marginal product value for meat production

The marginal product of the feed input for meat production was less than zero (Table 4), indicating that total calf production would no longer increase, and that maximum production was attained with a smaller amount of feed. In the studied DP systems this input was used excessively, generating losses of 0.38 cents for each additional unit of feed. In contrast, increasing the number of cows in production by one unit generates $2,460.93 incomes, and because it is in stage I of the production function, the marginal product would continue to increase, as would its value. The same trend has been reported in PU in the state of Yucatan in which increases of one animal unit raised meat production to 980.7 kg, which is attributed to their production being less than maximum due to the limited use of breeding programs and genetic improvement9. Considering this, the producers studied here need not stagnate calf production by maintaining the cows input unchanged since by increasing this input they could enter stage II of production. Table 4 Marginal product and marginal product value of inputs used in a double-purpose milk and meat production system Function Unit Price$ Feed Cows
Milk Calves PMg VPMg PMg VPMg
Milk 5.38 ---------- 0.75 4.03 892.2 4800.2
Meat ------- 6020 -6.38exp-05 -0.38 0.408 2460.93

PMg= marginal product; VPMg= value of marginal product.

Technical optimum meat production level

Y2=2.34 X1-0.205X21.118    subject to 4.0 X1+500 X2=6,020

By the Lagrange method:

L= X1-0.205X21.118-λ (4.0X1+500 X2-6,020)

Partial derivative of L for X1 and X2, under first order condition:

4.0 λ=0.481X1-1.205 X21.188

500 λ=2.625 X1-0.205 X2-0.188

Equalizing partial derivatives for X1 and X2

0.481X1-1.205 X21.1882.625 X1-0.205 X2-0.188=4.0500

X2=4.0X191.61=0.043 X1

Substituting X2 in the constraining equation produces the optimum level for this input, and substituting the X1 and X2 values in the Cobb-Douglas function produces the optimum amount of milk produced.

X1=6,02025.5=236.07

X2=0.043 233.07=10.151

Y2=2.348 (236.07) -0.205(10.151)1.118=10.22 calves

The technical optimum meat production level in the studied PU was 10.22 calves annually. Combining the X1=236.07 and X2=10.15 inputs maximizes the calf production isoquant.

Conclusions and implications

Feed and cows are the inputs that best explained milk and meat production in dual-purpose livestock producers in the studied areas. Producers need to place more emphasis on their use of these inputs since irrational use can decrease production variables. The elasticities of production indicated that, ceteris paribus, increasing these inputs in milk and meat production raises total production, except for the feed input in meat production. In milk production both inputs exhibited diminishing returns to scale, and were in stage II of production, the stage during which emphasis is needed on production. In meat production, the feed input had a negative marginal product value, placing it in stage III of a production function, and generating financial losses; operating in this stage will cause losses. Although in milk production both inputs had positive marginal product values they should not be increased since they are operating within the law of diminishing marginal returns. The studied livestock producers have generally diminishing returns to scale. One alternative for improving these returns is to increase the use of technology in different areas (feed, pasture management, forage conservation, reproduction, technical training) with the purpose of specializing capital and labor in these systems.

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Received: June 07, 2018; Accepted: September 29, 2018

Este es un artículo publicado en acceso abierto bajo una licencia Creative Commons


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# Math Help - Showing the closest point in a set to a point outside the set is orthogonal to

1. ## Showing the closest point in a set to a point outside the set is orthogonal to

Hi everyone,

I have a bit of an algebra/convexity type problem. I've been asked to show that if V is a linear subspace in R^n and y is not in V, then show that x* in V is the closest point in V to y if and only if y-x* is orthogonal to V. I'm a bit stuck on this. I know any subspace must be a hyperplane, so intuitively, this makes sense, but I'm having trouble bridging the gap to make it rigorous. Any ideas?

Thanks for any help!

Best,

Alex

2. Originally Posted by cutmeout
Hi everyone,

I have a bit of an algebra/convexity type problem. I've been asked to show that if V is a linear subspace in R^n and y is not in V, then show that x* in V is the closest point in V to y if and only if y-x* is orthogonal to V. I'm a bit stuck on this. I know any subspace must be a hyperplane, so intuitively, this makes sense, but I'm having trouble bridging the gap to make it rigorous. Any ideas?

Thanks for any help!

Best,

Alex
let $V^{\perp}=\{z \in \mathbb{R}^n: \ =0, \ \forall v \in V \}$ be the orthogonal complement of $V$ in $\mathbb{R}^n.$ we know that $\mathbb{R}^n=V \oplus V^{\perp}.$ so: $y=v_0+w_0,$ for some unique $v_0 \in V, \ w_0 \in V^{\perp}.$ now let $v$ be any element
of $V.$ then: $||y-v||^2=||v_0 - v +w_0||^2==||v_0-v||^2 + ||w_0||^2.$ (note that $==0,$ because $v_0-v \in V$ and $w_0 \in V^{\perp}.$)

so we showed that for all $v \in V: \ ||y-v||^2=||v_0-v||^2+||w_0||^2.$ now $v_0,w_0$ are constants. thus $||y-v||$ is minimum if and only if $||v_0-v||$ is minimum, i.e. $v=v_0.$ but then $y-v=w_0 \in V^{\perp}.$

conversely if $y - v_1 =w_1 \in V^{\perp},$ for some $v_1 \in V,$ then $y=v_1+w_1=v_0+w_0.$ thus $v_1=v_0, \ w_1=w_0.$ hence $||y-v_1||=||y-v_0||,$ and as we showed: $||y-v_0||=\min_{v \in V} ||y-v||. \ \ \ \Box$

3. It makes so much more sense now. Thank you very much!


text:
Use a graph to determine where a function is increasing, decreasing, or constant

As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function.

Figure 3. The function $f\left(x\right)={x}^{3}-12x$ is increasing on $\left(-\infty \text{,}-\text{2}\right){{\cup }^{\text{ }}}^{\text{ }}\left(2,\infty \right)$ and is decreasing on $\left(-2\text{,}2\right)$.

This video further explains how to find where a function is increasing or decreasing.

While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is “local minima.” Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is “extremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term local.

Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain.

For the function in Figure 4, the local maximum is 16, and it occurs at $x=-2$. The local minimum is $-16$ and it occurs at $x=2$.

Figure 4

To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 5 illustrates these ideas for a local maximum.

Figure 5. Definition of a local maximum.

These observations lead us to a formal definition of local extrema.

A General Note: Local Minima and Local Maxima

A function $f$ is an increasing function on an open interval if $f\left(b\right)>f\left(a\right)$ for any two input values $a$ and $b$ in the given interval where $b>a$.

A function $f$ is a decreasing function on an open interval if $f\left(b\right)<f\left(a\right)$ for any two input values $a$ and $b$ in the given interval where $b>a$.

A function $f$ has a local maximum at $x=b$ if there exists an interval $\left(a,c\right)$ with $a<b<c$ such that, for any $x$ in the interval $\left(a,c\right)$, $f\left(x\right)\le f\left(b\right)$. Likewise, $f$ has a local minimum at $x=b$ if there exists an interval $\left(a,c\right)$ with $a<b<c$ such that, for any $x$ in the interval $\left(a,c\right)$, $f\left(x\right)\ge f\left(b\right)$.

Example 7: Finding Increasing and Decreasing Intervals on a Graph

Given the function $p\left(t\right)$ in the graph below, identify the intervals on which the function appears to be increasing.

Figure 6

Solution

We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from $t=1$ to $t=3$ and from $t=4$ on.

In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval $\left(4,\infty \right)$.

Analysis of the Solution

Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at $t=1$ , $t=3$ , and $t=4$ . These points are the local extrema (two minima and a maximum).

Example 8: Finding Local Extrema from a Graph

Graph the function $f\left(x\right)=\frac{2}{x}+\frac{x}{3}$. Then use the graph to estimate the local extrema of the function and to determine the intervals on which the function is increasing.

Solution

Using technology, we find that the graph of the function looks like that in Figure 7. It appears there is a low point, or local minimum, between $x=2$ and $x=3$, and a mirror-image high point, or local maximum, somewhere between $x=-3$ and $x=-2$.

Figure 7

Analysis of the Solution

Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 7 provides screen images from two different technologies, showing the estimate for the local maximum and minimum.

Figure 8

Based on these estimates, the function is increasing on the interval $\left(-\infty ,-{2.449}\right)\\$
and $\left(2.449\text{,}\infty \right)\\$. Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at $\pm \sqrt{6}$, but determining this requires calculus.)

Try It 4

Graph the function $f\left(x\right)={x}^{3}-6{x}^{2}-15x+20\\$ to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing.

Solution

Example 9: Finding Local Maxima and Minima from a Graph

For the function $f$ whose graph is shown in Figure 9, find all local maxima and minima.

Figure 9

Solution

Observe the graph of $f$. The graph attains a local maximum at $x=1$ because it is the highest point in an open interval around $x=1$. The local maximum is the $y$ -coordinate at $x=1$, which is $2$.

The graph attains a local minimum at $\text{ }x=-1\text{ }$ because it is the lowest point in an open interval around $x=-1$. The local minimum is the y-coordinate at $x=-1$, which is $-2$.

Analyzing the Toolkit Functions for Increasing or Decreasing Intervals

We will now return to our toolkit functions and discuss their graphical behavior in the table below.

Function Increasing/Decreasing Example

Constant Function

$f\left(x\right)={c}$

Neither increasing nor decreasing

Identity Function

$f\left(x\right)={x}$

Increasing

$f\left(x\right)={x}^{2}$

Increasing on $\left(0,\infty\right)$

Decreasing on $\left(-\infty,0\right)$

Minimum at $x=0$

Cubic Function

$f\left(x\right)={x}^{3}$

Increasing

Reciprocal

$f\left(x\right)=\frac{1}{x}$

Decreasing $\left(-\infty,0\right)\cup\left(0,\infty\right)$

Reciprocal Squared

$f\left(x\right)=\frac{1}{{x}^{2}}$

Increasing on $\left(-\infty,0\right)$

Decreasing on $\left(0,\infty\right)$

Cube Root

$f\left(x\right)=\sqrt[3]{x}$

Increasing

Square Root

$f\left(x\right)=\sqrt{x}$

Increasing on $\left(0,\infty\right)$

Absolute Value

$f\left(x\right)=|x|$

Increasing on $\left(0,\infty\right)$

Decreasing on $\left(-\infty,0\right)$


text:
How to Interpret the Factor Analysis Findings

Zainab

New Member
Kindly

If someone can help interpreting results as attached please

Last edited:


text:
# Доказательство гипотезы Больших Чисел Дирака. Proof of the Hypothesis of Large Dirac Numbers.

## Recommended Posts

1 hour ago, Aeromash said:

F=c4G

Force  in the SI system.

What?

1 hour ago, Aeromash said:

φ=v2=Gmr

The square of speed is the gravitational potential.

A car moving on the surface of the earth can have a wide range of speeds without changing its gravitational potential.

If you square GM/r you get G2M2/rwhich is not the force (GmM/r2)

• Replies 68
• Created

#### Posted Images

40 minutes ago, studiot said:

L-12 does not equal L-6L-5

Thanks! My mistake!

$L^{-12}=L^{-5}L^{-7}=F=qE$

$L^{-5}$

charge.

$L^{-7}$

electric field strength.

$L^{-6}$

Рotential.

Edited by Aeromash
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31 minutes ago, swansont said:
1 hour ago, Aeromash said:

φ=v2=Gmr

The square of speed is the gravitational potential.

The square of the velocity is not the gravitational potential, the gravitational potential is the square of the escape velocity.

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45 minutes ago, swansont said:

A car moving on the surface of the earth can have a wide range of speeds without changing its gravitational potential.

You are talking about the speed of the car. And not about the square of the speed or the gravitational potential of the car. You are confusing physical quantities.

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2 minutes ago, Aeromash said:

You are talking about the speed of the car. And not about the square of the speed or the gravitational potential of the car. You are confusing physical quantities.

You are confusing this. The gravitational potential of the Earth, not the car. If the car moves so fast that the square of its speed becomes equal to the gravitational potential of the Earth not on its surface, then the car will fly into the Solar System.

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49 minutes ago, swansont said:

If you square GM/r you get G2M2/rwhich is not the force (GmM/r2)

$\frac{\varphi^{2}}{G} =\frac{G^{2}M^{2}}{Gr^{2}}=F$

G is the constant of the transition from the LT system to SI.

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1 minute ago, Aeromash said:

φ2G=G2M2Gr2=F

G is the constant of the transition from the LT system to SI.

The force is a vector, not a scalar

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46 minutes ago, Aeromash said:

You are talking about the speed of the car. And not about the square of the speed or the gravitational potential of the car. You are confusing physical quantities.

If the speed of the car is variable then so is the square of the speed. I thought that would be self-evident. I did mention the gravitational potential; it is constant on the surface of the sphere of the earth.

If the square of the speed is variable, then it cannot be equal to a constant.

42 minutes ago, Aeromash said:

φ2G=G2M2Gr2=F

G is the constant of the transition from the LT system to SI.

Sorry, G is taken already, and this is a new fudge factor you have added because your original claim is wrong.

It's still not the force, because the gravitational potential does not include the mass of the object, and the force does. Squaring the potential will not make the mass of the object magically appear.

Using unit analysis gives you proportionalities, at best. Not equalities.

1 hour ago, SergUpstart said:

The square of the velocity is not the gravitational potential, the gravitational potential is the square of the escape velocity.

No, it's not. There's a pesky factor of 2 that need to be included. It has no units, so unit analysis and manipulation would not reveal it.

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14 hours ago, swansont said:

Using unit analysis gives you proportionalities, at best. Not equalities.

Yes! Using unit analysis gives you proportionalities. But using unit analysis gives me equalities in L and LT systems. For the transition to the SI system need the gravitational constant G or, in electrodynamics, epsilon . Newton knew nothing about the constant G. Mass in his era had dimensions:

$\frac{m^{3}}{s^{2}}$

In general, this is terrible, but the SI system is a mixture of the LMT, LT and L systems. Therefore, thinking that we are working in SI, in fact, we use the LT or L system. When we work in SI, then coefficients of the G or epsilon type appear in the formulas right away. When you calculate the volume in SI by multiplying the area by the height, you are actually working in the L system. If you added mass to the LT system, that is, switched to the LMT system, then the constant G immediately appears in the formulas. The larger the dimension of the system, the more different coefficients(constants) you need! This is what we observe in modern physics.

System L does not need coefficients! This is its beauty.

16 hours ago, SergUpstart said:

The force is a vector, not a scalar

Vector if the radius is a vector.

$\vec{r}$

Edited by Aeromash
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17 hours ago, swansont said:

A car moving on the surface of the earth can have a wide range of speeds without changing its gravitational potential.

A car moves at different speeds relative to the Earth's surface, but its speed does not exceed the square root of the gravitational potential at the Earth's surface. In this matter, it is a gravitationally bound object, which is one with the Earth. Unlike Tesla's car, which is part of the solar system.

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4 hours ago, Aeromash said:

Yes! Using unit analysis gives you proportionalities. But using unit analysis gives me equalities in L and LT systems. For the transition to the SI system need the gravitational constant G or, in electrodynamics, epsilon . Newton knew nothing about the constant G. Mass in his era had dimensions:

m3s2

In general, this is terrible, but the SI system is a mixture of the LMT, LT and L systems. Therefore, thinking that we are working in SI, in fact, we use the LT or L system. When we work in SI, then coefficients of the G or epsilon type appear in the formulas right away. When you calculate the volume in SI by multiplying the area by the height, you are actually working in the L system. If you added mass to the LT system, that is, switched to the LMT system, then the constant G immediately appears in the formulas. The larger the dimension of the system, the more different coefficients(constants) you need! This is what we observe in modern physics.

System L does not need coefficients! This is its beauty.

70 posts, some of them have involved a lot of work and I still don't understand the basis of what you are saying about dimensions. Most of what you have posted assumes the reader understands and agrees with your basic statement about dimensions.

I don't understand or agree.

You mention unit analysis and also Tesla.

There is a form of analysis in electrical power engineering called 'per unit' analysis.

I don't know if you are trying to emulate this.

The problem with 'unit analysis' is that the dimensions are hidden in the 'units' employed.

They haven't really gone away.

So can I respectfully suggest you put some of this effort into developing (proving) your underlying claim once and for all?

Perhaps showing one simple example developed from start to fininsh, but involving as many as possible of the seven basic quantities in the LMTAθIN system.

For your information here are two free websites where you build mathematical formulae and copy/paste into ScienceForums.

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3 hours ago, Aeromash said:

A car moves at different speeds relative to the Earth's surface, but its speed does not exceed the square root of the gravitational potential at the Earth's surface.

That was not your claim.  You said "The square of speed is the gravitational potential." with no caveats.

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Since I asked for a clear example it is only fair that I give you one showing what i mean.

On a per unit basis, the molecular weight of water is 18.

Superficially, that apppears to be a pure number.

But that is because the the real dimensions are hidden in the per unit, which is a gramme-molecule.

So a complete statement is the molecular weight of water is 18 grammes per gramme-molecule.

The interesting part of this story is that the full dimensional statement is MN, although the N is usually 'understood' and not stated explicitly.

Edited by studiot
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4 hours ago, studiot said:

Perhaps showing one simple example developed from start to fininsh, but involving as many as possible of the seven basic quantities in the LMTAθIN system.

A good question takes a long time to answer. I am preparing a general table of the transition of the one-dimensional laws of the system L to the SI system.

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1 hour ago, Aeromash said:

A good question takes a long time to answer. I am preparing a general table of the transition of the one-dimensional laws of the system L to the SI system.

If I can understand and agree with it I can then follow all your other long posts that have taken so much effort.

As a matter of interest the MLT system is not unique, nor is it specifically SI, although yes Système International d'Unités has adopted it.

But there are also other systems in use for instance one with mass replaced by force - the FLT system.

The FLT system finds favour in Fluid Mechanics as it make this system easier (at least for Fluid Mechanics) as can be seen from this small table. (after Olson :  Engineering Fluid Mechanics)

Note at the bottom of the table the comment about temperature.

I did ask and you have not said how your L system handles temperature.

Edited by studiot
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17 hours ago, studiot said:

Note at the bottom of the table the comment about temperature.

I did ask and you have not said how your L system handles temperature.

$E \sim T^{\circ}$

Energy is proportional to temperature. Which means that the temperature has the same degree as the energy in the system  L.

$L^{-11}=T^{\circ}$

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50 minutes ago, Aeromash said:

ET

Energy is proportional to temperature. Which means that the temperature has the same degree as the energy in the system  L.

L11=T

Really ?

What is n ?

That leaves me still waiting for some unit free constant that couples length to any one of the other independent physical dimensions I have mentioned.

Do you know what they are?

If not are you not curious about my symbols ?

I certainly don't agree that all forms of 'Energy' in general are proportional to temperature.

Edited by studiot
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35 minutes ago, studiot said:

What is n ?

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27 minutes ago, Aeromash said:

I don't think that answers the question

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text:
setPheno {AlphaSimR} R Documentation

## Set phenotypes

### Description

Sets phenotypes for all traits by adding random error from a multivariate normal distribution.

### Usage

setPheno(
pop,
h2 = NULL,
H2 = NULL,
varE = NULL,
corE = NULL,
reps = 1,
fixEff = 1L,
p = NULL,
onlyPheno = FALSE,
traits = NULL,
simParam = NULL
)

### Arguments

 pop an object of Pop-class or HybridPop-class h2 a vector of desired narrow-sense heritabilities for each trait. See details. H2 a vector of desired broad-sense heritabilities for each trait. See details. varE error (co)variances for traits. See details. corE an optional matrix for correlations between errors. See details. reps number of replications for phenotype. See details. fixEff fixed effect to assign to the population. Used by genomic selection models only. p the p-value for the environmental covariate used by GxE traits. If NULL, a value is sampled at random. onlyPheno should only the phenotype be returned, see return traits an integer vector indicate which traits to set. If NULL, all traits will be set. simParam an object of SimParam

### Details

There are three arguments for setting the error variance of a phenotype: h2, H2, and varE. The user should only use one of these arguments. If the user supplies values for more than one, only one will be used according to order in which they are listed above.

The h2 argument allows the user to specify the error variance according to narrow-sense heritability. This calculation uses the additive genetic variance and total genetic variance in the founder population. Thus, the heritability relates to the founder population and not the current population.

The H2 argument allows the user to specify the error variance according to broad-sense heritability. This calculation uses the total genetic variance in the founder population. Thus, the heritability relates to the founder population and not the current population.

The varE argument allows the user to specify the error variance directly. The user may supply a vector describing the error variance for each trait or supply a matrix that specify the covariance of the errors.

The corE argument allows the user to specify correlations for the error covariance matrix. These correlations are be supplied in addition to the h2, H2, or varE arguments. These correlations will be used to construct a covariance matrix from a vector of variances. If the user supplied a covariance matrix to varE, these correlations will supercede values provided in that matrix.

The reps parameter is for convenient representation of replicated data. It is intended to represent replicated yield trials in plant breeding programs. In this case, varE is set to the plot error and reps is set to the number of plots per entry. The resulting phenotype represents the entry-means.

### Value

Returns an object of Pop-class or HybridPop-class if onlyPheno=FALSE, if onlyPheno=TRUE a matrix is returned

### Examples

#Create founder haplotypes
founderPop = quickHaplo(nInd=10, nChr=1, segSites=10)

#Set simulation parameters
SP = SimParam$new(founderPop) SP$addTraitA(10)

#Create population
pop = newPop(founderPop, simParam=SP)

#Add phenotype with error variance of 1
pop = setPheno(pop, varE=1)

[Package AlphaSimR version 1.3.2 Index]


text:
# Q&A - Ask Doubts and Get Answers

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Q

Q 10.     An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Given, Radius of the circular orbit =   Circumference of the orbit =  The satellite takes 24 hours to revolve around the earth. We know,  Therefore, the speed of the artificial satellite is

Q 9.     State which of the following situations are possible and give an example for each of these:

(c)     an object moving in a certain direction with an acceleration in the perpendicular direction.

(c) The given situation is possible. For an object moving in a circular path with constant speed, the direction of its velocity at any point will be tangential to that point. However, its acceleration will be directed radially inwards. (Constant speed but still having an acceleration - Due to continuous change in direction.)

Q 9.        State which of the following situations are possible and give an example for each of these:

(b)     an object moving with an acceleration but with uniform speed.

(b) The given situation is possible.  An object moving in a circular path with uniform speed, i.e covering equal distance in equal amount of time is still under acceleration. Because, the velocity keeps on changing due to continuous change in the direction of motion. Therefore, circular motion is an example of an object moving with an acceleration but with uniform speed.

Q 9.     State which of the following situations are possible and give an example for each of these:

(a)     an object with a constant acceleration but with zero velocity

(a) The given situation is possible. When an object is thrown upwards (under gravity only), it reaches to a maximum height where its velocity becomes zero. However, it still has an acceleration acting in the downward direction (acceleration due to gravity). Note: This is possible for a given point of time, however, it is not possible for a period of time.

Q 8.     The speed-time graph for a car is shown is Figure:

(b)     Which part of the graph represents uniform motion of the car?

In uniform motion, the speed of car will become constant which is represented by line parallel to the time axis. In the given figure, the straight line graph from  represents the uniform motion of the car.

Q 8.     The speed-time graph for a car is shown is Figure:

(a)     Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

Given is a speed-time graph. The area under the curve will give the distance travelled by the car. In time , the distance travelled by the car will be equal to the area under the curve from  Considering this part of the graph as a quarter of a circle whose radius = 4 unit. Therefore, required area =  Therefore, distance the car travelled in the first 4 seconds is

Q 7.     A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground?  After what time will it strike the ground?

Considering downward direction as positive direction. Given, Height from which ball is dropped,  Acceleration of the ball,  =  Initial velocity,  (i) We know,  (In downward direction) Therefore, the ball will strike the ground with a velocity of  (ii) Now, we know,  Therefore, the ball reaches the ground in . Note:  was rejected because in this case, the negative sign implies the velocity in...

Q 6.     Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

(d)     How far has B travelled by the time it passes C?

Given is a distance-time graph. The graph of B and C intersect at a point whose y-coordinate is 5. Hence, B has travelled by the time it passes C.

Q 6.     Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

(c)     How far has C travelled when B passes A?

Given is a distance-time graph. Any point on the curve will give the distance of object from O. To find how far C has travelled when B passes A, draw a perpendicular from the intersection point of A and B on time axis. The point where it intersects on the C graph, from C draw a perpandicular to y axis . Therefore, distance travelled by C will be (Final distance from O - Initial distance from...

Q 6.     Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

(b)     Are all three ever at the same point on the road?

Given is a distance-time graph. Any point on the curve will give the distance of object from O. Since there is no intersection point of all the three graphs, they never meet at the same point on the road.  (Although any two of them do meet at some point on the road!)

Q 6.     Figure shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions:

(a)     Which of the three is travelling the fastest?

Given is a distance-time graph. The slope of this graph gives us speed. Hence, the graph with the highest slope will have the highest speed. Since B has the highest slope(inclination), it travels the fastest.

Q 5.     A driver of a car travelling at 52 km h-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h-1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

The initial speed   After 5 sec the car stops The graph is represented by the blue line ( x-axis is time and the y-axis is speed)  For the car with 3Kmh-1. Initial speed   .  The graph which is represented by the golden line The area covered by the blue graph is greater than the golden graph so the car with 15 m/s initial velocity travells large distance.

Q 4.     A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of $3.0\ m s^{-2}$ for 8.0 s. How far does the boat travel during this time?

Given, The motorboat starts from rest. Hence, initial speed of the motorboat, u =  Acceleration of the motorboat,  =  Time taken,  (b) We know,  Therefore, the distance travelled by the motorboat is

Q 3.     Abdul, while driving to school, computes the average speed for his trip to be $20\ km h^{-1}$. On his return trip along the same route, there is less traffic and the average speed is $30\ km h^{-1}$. What is the average speed for Abdul’s trip?

Given, Average speed while going to school,  And Average speed while returning back from school,  Let the distance between starting point and school be  And time taken by Abdul during the two trips be  We know,  And,              -(i) Now, Total distance that Abdul covers =  And total time Abdul takes to cover this distance = Therefore, the average speed for Abdul's trip is  (Note: )

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging

(a) from A to B and

(b) from A to C?

Given, (a) Distance between A and B = Time taken to reach from A to B =    Average speed from A to B =  And, Average velocity from A to B =  (In this case, average speed is equal to the average velocity) (b) Distance travelled from A to reach C =   And, Displacement from A to C =  Also, time taken to reach C from A =   Average speed from A to C =  And, Average velocity from A to C =  (In this...

Q 1.     An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given, Diameter of the circular track =  The circumference of the circular track,   The athlete completes one round of a circular track in 40 s.  Speed of the athlete =  In , Distance travelled by the athlete =  Also, number of rounds the athlete will complete in  =  Therefore, the final position of the athlete after  will be diametrically opposite to his initial point. (3 complete rounds and...

Q 5.     A stone is thrown in a vertically upward direction with a velocity of $5\ m s^{-1}$. If the acceleration of the stone during its motion is $10\ m s^{-2}$ in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Taking upward direction as positive (+) direction: Given,          (This is due to gravitational force!) The stone will move up until its velocity becomes zero. We know,  Therefore, the stone reaches to a height of  Now,  We know,  Therefore, the time taken by the stone to reach the maximum height is .

A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?

Given, Initial speed of the racing car, u =  Acceleration of the car,  =  Time taken,  We know,  Therefore, the distance travelled by the racing car in  is

A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest.

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A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start?

Given, The trolley starts from rest. Hence, the initial speed of the trolley, =  Acceleration of the trolley,  =  Time is taken,  (a) We know,  Therefore, the velocity of the trolley after 3 sec is
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# How do you solve x^2-5x-3/4=0 by completing the square?

May 13, 2016

$x = \frac{5}{2} \pm \sqrt{7}$

#### Explanation:

Premultiply by $4$ first to cut down on the fractions involved:

$0 = 4 \left({x}^{2} - 5 x - \frac{3}{4}\right)$

$= 4 {x}^{2} - 20 x - 3$

$= {\left(2 x\right)}^{2} - 2 \left(2 x\right) \left(5\right) - 3$

$= {\left(2 x - 5\right)}^{2} - {5}^{2} - 3$

$= {\left(2 x - 5\right)}^{2} - 28$

$= {\left(2 x - 5\right)}^{2} - {2}^{2} \cdot 7$

$= {\left(2 x - 5\right)}^{2} - {\left(2 \sqrt{7}\right)}^{2}$

$= \left(\left(2 x - 5\right) - 2 \sqrt{7}\right) \left(\left(2 x - 5\right) + 2 \sqrt{7}\right)$

$= \left(2 x - 5 - 2 \sqrt{7}\right) \left(2 x - 5 + 2 \sqrt{7}\right)$

$= 4 \left(x - \frac{5}{2} - \sqrt{7}\right) \left(x - \frac{5}{2} + \sqrt{7}\right)$

So $x = \frac{5}{2} \pm \sqrt{7}$


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• •

### 温度和盐度对中间球海胆（Strongylocentrotus intermedius）早期胚胎发育率的联合效应

1. （大连海洋大学， 农业农村部北方海水增养殖重点实验室， 辽宁大连 116023）
• 出版日期:2021-02-10 发布日期:2021-07-09

### Interactive effects of temperature and salinity on embryonic development rate of Strongylocentrotus intermedius.

ZHEN Hao, WU Yang-lei, WANG Luo*, PEI Hong-lin, TAN Ba-mei, DING Jun, CHANG Ya-qing#br#

1. (Key Laboratory of Mariculture & Stock Enhancement in North China’s Sea, Ministry of Agriculture and Rural Affairs, Dalian Ocean University, Dalian 116023, Liaoning, China).
• Online:2021-02-10 Published:2021-07-09

Abstract: The growth of marine organisms is currently challenged by multiple environmental stressors. The early life stages of sea urchin Strongylocentrotus intermedius are sensitive to environmental changes. Temperature and salinity are important environmental factors affecting the growth and development of marine animals. To understand the temperature and salinity tolerance of the early embryonic development of sea urchin and its optimal environmental conditions, we used the central composite design (CCD) and response surface methodology (RSM) to investigate the interactive effects of temperature (12-26 ℃) and salinity (22-34) on the early development of the sea urchin embryo. The quantitative relationship models between temperature, salinity and the fertilization rate, floating rate and metamorphosis rate of sea urchin embryos were established, and the best combination of temperature and salinity was obtained through statistical optimization methods. The results showed that both high and low temperature and salinity were not conducive to the early embryonic development rate of sea urchins. The effect of temperature on the early embryonic development of S intermedius was greater than that of salinity, with certain antagonism between temperature and salinity. The primary effects of temperature and salinity, and the secondary effects of temperature significantly affected the fertilization rate, floating rate, and metamorphosis rate of the S. intermedius embryos (P<0.01). The interactive effects of temperature and salinity significantly affected the embryo fertilization rate and metamorphosis rate of S. intermedius (P<0.05), but did not affect the floating rate. The secondary effect of salinity significantly (P<0.01) affected the fertilization rate of the S. intermedius embryos, but did not affect the floating rate and metamorphic rate. The determination coefficients of the fertilization rate, floating rate and metamorphosis rate were 0.9736, 0.9946, and 0.9925, respectively, indicating that the models were effective and could be used to predict the fertilization rate, floating rate and metamorphosis rate of S. intermedius. The model optimization and verification tests showed that, at 17.71 ℃ and salinity of 31.85, the fertilization rate, floating rate and metamorphosis rate reached the maximum value, with a satisfaction rate of 0.969. Our findings can provide theoretical basis for the artificial breeding of sea urchin.