Posted by **kieran** on Monday, October 1, 2012 at 10:31pm.

A closed box is to be rectangular solid with a square base. If the volume is 32in^3, determine the dimensions for which the surface area is minimum.

- Calculus -
**Reiny**, Monday, October 1, 2012 at 11:04pm
make a sketch

let the base be x by x, and the height be y

so x^2 y = 32

y = 32/x^2

SA = 2x^2 + 4xy

= 2x^2 + 4x(32/x^2)

= 2x^2 + 128/x

d(SA)/dx = 4x - 128/x^2 = 0 for a min of SA

4x = 128/x^2

4x^3 = 128

x^3 = 32

x = 32^(1/3) or appr 3.175

y = 32/3.175^2

Well , what do you know, it happens to be a perfect cube.

- Calculus -
**Anonymous**, Friday, December 5, 2014 at 1:17am
60.48

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