# Thread: minimum Dimensions of an open box

1. ## minimum Dimensions of an open box

1. The problem statement, all variables and given/known data
Determine the dimensions of a rectangular box, open at the top, having volume
4 m3, and requiring the least amount of material for its construction. Use the
second partials test. (Hint: Take advantage of the symmetry of the problem.)

2. Relevant equations

3. The attempt at a solution

Volume, V= 4m^3
let x = length
y = width
z = height

4m^3 = xyz
x = 4/yz

because it is an open at the top rectangular box,
the Surface, S = 2xz + 2yz + xy

substitute x=4/yz inside the surface equation..
S = 8/y + 2yz + 4/z

to find the critical points, take the partial with respect to y and z.. the equal it to zero..

S'(y)= 2z - 8/y^2
S'(z)= 2y - 4/z^2

solve the equations, i get,
when y = 0, z = 0..
y = 8, z = 1/16

the problem is, what should i do next?

2. Originally Posted by nameck
1. The problem statement, all variables and given/known data
Determine the dimensions of a rectangular box, open at the top, having volume
4 m3, and requiring the least amount of material for its construction. Use the
second partials test. (Hint: Take advantage of the symmetry of the problem.)

2. Relevant equations

3. The attempt at a solution

Volume, V= 4m^3
let x = length
y = width
z = height

4m^3 = xyz
x = 4/yz

because it is an open at the top rectangular box,
the Surface, S = 2xz + 2yz + xy

substitute x=4/yz inside the surface equation..
S = 8/y + 2yz + 4/z

to find the critical points, take the partial with respect to y and z.. the equal it to zero..

S'(y)= 2z - 8/y^2
S'(z)= 2y - 4/z^2

solve the equations, i get,
when y = 0, z = 0..
y = 8, z = 1/16

the problem is, what should i do next?

The solutions are $y=2\,,\,z=1\,\Longrightarrow x=2$ and you're done (just check why you got y = 8...??)

Perhaps you oversaw the fact that you're positioning your box at the origin in the 3-dimensional space, with sides parallel to the three axis. Then, x = 2 is the length of the box's side along the x-axis, y = 2 the length along the y-axis and etc.

If you already studied Lagrange's Multipliers then you can double-check with them that the answer indeed is the above one.

Tonio

3. ok.. i repaired my calculation..
got problem with the simultaneous equation..
i got ur answer where y=2 and z=1..
the i substitute the value into xyz=4..
i got x=2..
however, is this my final answer?
should i use second partial test?

4. Originally Posted by nameck
ok.. i repaired my calculation..
got problem with the simultaneous equation..
i got ur answer where y=2 and z=1..
the i substitute the value into xyz=4..
i got x=2..
however, is this my final answer?
should i use second partial test?

Well...yes, in order to be formal. You'll see that the Hessian is a positive definite matrix and thus you got a minimum (or $f_{yy}>0\,,\,\,f_{yy}f_{zz}-(f_{yz})^2>0$) when evaluated at $(2,2,1)$

Tonio

5. nice! i got it now tonio!!
thanks =)

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