Optimization of box, varied material cost.

[QuarkCharmer]

Homework Statement

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.

Homework Equations

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?

 

[ideasrule]

Nope, that's right. :)