whats the minium cost of this is it 166 or 312 Give me the answer for this, dont copy the answers already published, i have the minimum cost already $166, i want to find the derivatives of the function,  the critical points,  prove that the critical points represent maximum or minimum points and add (eg: with an interval table) and  find the extreme points on your function
- take the derivative of this function
- find the critical points in order to find the maximum and minimum values for your function. 
- prove that the critical points represent maximum or minimum points (eg: with an interval table) 
- find the extreme points on your function
Using this info given already, and can you please number or highlight these 4 answers so its easier, thank you!

Problem:

An open-top cylindrical tank with a volume of 2000 cubic meters is to be constructed using steel. The steel used for the top and bottom costs 0.5 dollars per square meter, while the steel used for the sides costs 0.3 dollars per square meter. Determine the dimensions of the tank that will minimize the cost of steel used.

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Step 3: State the variables of the problem

Variables:

r: radius of the base of the cylinder in meters
h: height of the cylinder in meters
Function: The surface area of the cylindrical tank is given by: A = 2πrh + 2πr2

The cost of steel is given by: C = 0.5(2πr 2) + 0.3(2πrh)

Objective: Minimize the cost of steel C.

Domain:

r and h must be positive numbers. Additionally, since the tank has a volume of 2000 cubic meters, we can use the volume formula for a cylinder to set up an equation relating r and h:

V = πr2h = 2000

Therefore, the domain for r is (0, ∞) and the domain for h is (0, 2000/(πr2)).

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Step 4: Solve the above stated problem

Solution:

Using the equation V = πr2h = 2000, we can solve for h in terms of r: h = 2000/(πr2 )

Substituting h into the cost equation, we get:

C = 0.5(2πr 2) + 0.3(2πr(2000/(πr 2)))

C =  πr 2 + 1200/r

Taking the derivative of C with respect to r, we get:

C' = 2πr - 1200/r 2

Setting C' to zero, we get: 2πr = 1200/r 2

Solving for r, we get: r = (600/π)(1/3)

Substituting r back into the equation for h, we get:

h = 2000/(π((600/π)(2/3)))

Therefore, the dimensions of the cylindrical tank that minimize the cost of steel used are approximately:

r = 5.7588 meters h =  19.1961 meters

Thus, the minimum cost of steel used is:

C = π(5.7588) 2 + 1200/19.1961

C ≈ $166.6998

 

 

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Solution

The minimum cost of the steel required is

$ 166. 70