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March 27, 2015

March 27, 2015

Posted by **Amy** on Saturday, February 12, 2011 at 9:35am.

- Calculus - dimensions of a box? -
**Reiny**, Saturday, February 12, 2011 at 9:46amlet each side of the base be x ft

let the height of the box be y ft

V= (x^2)(y) = 256

y = 256/x^2

form the cost equation ....

C = 2(.10)x^2 + 4(.05)xy

= .2x^2 + 2x(256/x^2)

= .2x^2 + 512/x

C' = .4x - 512/x^2 = 0 for min of C

I get x^3 = 1280

x = 10.86 and y = 2.17

check my arithmetic

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