What dimensions minimize the surface area of a box with a fixed volume?

[UMich1344]

[SOLVED] Minimizing the Surface Area

Homework Statement

A box has a bottom with one edge 8 times as long as the other. If the box has no top and the volume is fixed at V, what dimensions minimize the surface area?

Homework Equations

V = lwh

SA (with no top) = lw + 2lh + 2wh

The Attempt at a Solution

l = x
w = 8x
h = V/(8x^2)

Finding an equation for the surface area.

SA = lw + 2lh + 2wh
SA = x(8x) + 2x(V/(8x^2)) + 2(8x)(V/(8x^2))
SA = 8x^2 + V/(4x) + 2V/x

Finding the derivative of the equation in order to set it equal to zero to find the critical points, so the minimum can be found.

(d SA)/(d x) = 16x - V/(4x^2) - 2V/(x^2)
(d SA)/(d x) = (64x^3 - 9V) / (4x^2)

(d SA)/(d x) = 0
(64x^3 - 9V) / (4x^2) = 0
(64x^3 - 9V) = 0
64x^3 = 9V
x^3 = (9V)/64
x = ((9V)/64)^(1/3)

Plugging the solution into the equations for the dimensions.

l = x = ((9V)/64)^(1/3)
w = 8x = 8((9V)/64)^(1/3)
h = V/(8(((9V)/64)^(1/3))^2)



I am unsure if I did the right steps in order to find the solution.
Also, I am not very confident in the work I did for each step.

 

[Snazzy]

It looks good to me, except you can simplify 64^(1/3) into 4 so that you can reduce your answers and it looks cleaner.

 

[UMich1344]

Thanks a lot for looking over it. I just put it into the online grader and it is, in fact, correct.

 

[UMich1344]

I appreciate it.