# Optimization of box, varied material cost.

1. Jul 4, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data
Stewart Calculus 6E: 4.7 #14
A rectangular storage container with an open top is to have a volume of 10m³. The length of it's base is twice the width. Material for the base costs $10 per square meter. Material for the sides cost$6 per square meter. Find the cost of materials for the cheapest such container.

2. Relevant equations

3. The attempt at a solution
I let the width of the box be W, the length be 2W, and the height be H.

Since:
$$2W^{2}H=10$$
I let $H=\frac{10}{2W^{2}}$

Then I claim that the cost of the base is given by:
$$(2W^{2})10 = 20W^{2}$$

The cost of the sides are given by:
$$(2WH + 4WH)6$$

So the total cost for the box could be written as:
$$20W^{2} + (2WH + 4WH)6$$
Substituting in $H=\frac{10}{2W^{2}}$, I get cost as a function of width?
$$C(W) =(2W(\frac{10}{2W^{2}})+4W(\frac{10}{2W^{2}}))6 + 20W^{2}$$
$$C(W) = \frac{180+20W^{3}}{W}$$

So I can minimize that function, and I find that the minimum is when the width is 1.65, and the cost of the box is about $163.50. So$163.50 is the solution? Does anyone see a problem with this?

2. Jul 4, 2011

### ideasrule

Nope, that's right. :)