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1. The problem statement, all variables and given/known data

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 atV, what dimensions minimize the surface area?

2. Relevant equations

V=lwh

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

3. 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.

(dSA)/(dx) = 16x-V/(4x^2) - 2V/(x^2)

(dSA)/(dx) = (64x^3 - 9V) / (4x^2)

(dSA)/(dx) = 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(((9^2)V)/64)^(1/3))

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.

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# Homework Help: Minimizing the Surface Area

**Physics Forums | Science Articles, Homework Help, Discussion**