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QuarkCharmer

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## 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:

[tex]2W^{2}H=10[/tex]

I let [itex]H=\frac{10}{2W^{2}}[/itex]

Then I claim that the cost of the base is given by:

[tex](2W^{2})10 = 20W^{2}[/tex]

The cost of the sides are given by:

[tex](2WH + 4WH)6[/tex]

So the total cost for the box could be written as:

[tex]20W^{2} + (2WH + 4WH)6[/tex]

Substituting in [itex]H=\frac{10}{2W^{2}}[/itex], I get cost as a function of width?

[tex]C(W) =(2W(\frac{10}{2W^{2}})+4W(\frac{10}{2W^{2}}))6 + 20W^{2}[/tex]

[tex]C(W) = \frac{180+20W^{3}}{W}[/tex]

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?