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September 1, 2014

September 1, 2014

Posted by **Megan** on Thursday, May 15, 2008 at 10:05pm.

- Math-Geometry -
**Reiny**, Friday, May 16, 2008 at 8:07amThere is additional information needed.

I am guessing from previous questions like this that the base is a square.

Let the base be x metres by x metres, an let the height be y metres.

then (x^2)y = 200 and y = 200/x^2

Cost = 3(x^2) + 1(4xy + x^2)

= 4x^2 + 4xy

= 4x^2 + 4x(200/x^2)

= 4x^2 + 800/x

Cost' = 8x - 800/x^2 = 0 for a max/min of Cost

8x = 800/x^2

x^3 = 100

x = 100^(1/3) = 4.64 m

then y = 200/4.64^2 = 9.28 m

the box should have a base of 4.64m by 4.64m and a height of 9.28 m

I also have a feeling that the volume was given as 2000, then the answer would have been exactly x=10 and y = 20

which would be in the same height:base length of 2:1

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