1. ## Optimization

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.
$\displaystyle V = lwh = 10$
$\displaystyle 2l = w$
$\displaystyle V = 2l^2h$

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

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

is this much correct?

2. 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.
$\displaystyle V = LWH$

$\displaystyle L = 2W$

$\displaystyle V = 2W^2H = 10$

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

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

$\displaystyle C = 20W^2 + 36WH$

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

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

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

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