Optimization

**millerst** · December 8th 2009, 01:23 PM

A rectangle storage container has a volume of 10 meters cubed and no top. The length is twice the width, the base material costs $10 per meter cubed and the side material costs $6 per meter cubed. Calculate the dimensions of the box that minimize total cost of materials.
$ V = lwh = 10 $
$2l = w$
$V = 2l^2h$

$SA = (10)(2l^2) + 4[6(lh)]$
$SA = 20l^2 + 24lh$
$ H = 5/l^2 $

$SA = 20l^2 + 120/l$
$SA' = 40l - 120/l^2 = 0$
l = cubed root of 3?

is this much correct?

**skeeter** · December 8th 2009, 02:09 PM

Originally Posted by millerst

A rectangle storage container has a volume of 10 meters cubed and no top. The length is twice the width, the base material costs $10 per meter cubed and the side material costs $6 per meter cubed. Calculate the dimensions of the box that minimize total cost of materials.

$V = LWH$

$L = 2W$

$V = 2W^2H = 10$

$H = \frac{5}{W^2}$

$C = 2W^2(10) + 2WH(6) + 4WH(6) $

$C = 20W^2 + 36WH$

$C = 20W^2 + \frac{180}{W}$

$\frac{dC}{dW} = 40W - \frac{180}{W^2} = 0$

$W^3 = \frac{9}{2}$

$W = \sqrt[3]{\frac{9}{2}}$

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