1. ## Optimization

A rectangular storage container with an open top is to have a volume of 10 meters cubed. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs$6 per square meter. Find the cost of materials for the cheapest such container.

My work:

The area of the base is given by:

$A_b = 2w^2$

The area of the sides is given by:

$A_s = 4wh$

I know the cost per square meter so:

$C_b = 20w^2$

$C_s = 24wh$

So the total cost is represented by:

$C = 20w^2 + 24wh$

And I know that the volume is equal to 10 cubic meters so:

$V = l\cdot w\cdot h$

$l\cdot w\cdot h = 10$

$2w^2\cdot h = 10$

$w^2\cdot h = 5$

$h = \frac {5}{w^2}$

I plug that into my cost formula:

$C(w) = 20w^2 + \frac {120}{w}$

Take the derivative:

$C'(w) = 40w - \frac {120}{w^2}$

I find the critical points to be 0, 1.44.

Then I find that the absolute minimum is 1.44 but when I use this number I get the incorrect answer of 124 (ish).

Any suggestions?

2. Area of the sides should be 6wh, you idiot.

3. Even when I use 6wh as the area of the sides, I get an incorrect answer of 81.18

GRRRR, this question is frustrating me.

4. Hmmm, if I use the formula:

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

And I plug in the cubed root of 4.5 for w, then I get an answer of 163.54 but if I do it in two steps and use the knowledge that h = 5/w^2 to solve it I get an answer of 81.18....oh well, 163.54 appears to be correct.