Expert: Scotto - 11/16/2011

Question
My problem is as follows...

A storage box with a square base must have a volume of 80 cubic
centimeters. The top and bottom cost $0.20 per square centimeters and the sides costs $0.10 per square centimeter. Find the dimensions that will minimize cost.

Thank You !!

Answer
The box has a square base that is x cm wide.
The volume is the width • depth • height, and this is 80 cm³.
The width and the depth are both x cm, and the height is h cm, so we have x²h = 80 cm³.
Later on I will use this to say that h cm = [80/x²] cm.

The area of the bottom is x² cm², and that is the same as the area of the top.
Since this is two sides, and the cost is 20¢ per cc, the equation for the
cost of the top and bottom is 2(20¢/cm²)(x²cm²) = 40x²¢.

The area of one side is xh cm², so the area of 4 sides is 4xh cm².
Since the cost of the sides is 10¢/cm², that gives 40xh¢.
Now h was [80/x²]cm, so putting that in 40xh¢ for h gives [40x•80/x²]¢ = [3200/x]¢.

The overall cost is then 40x² + 3200/x.  If we let f(x) = 40x² + 3200/x,
it can be seen that f'(x) = 80x - 3200/x².

To find the minimum, which is an extreme point on f, set f'(x) = 0 and solve for x.
This gives 80x - 3200/x² = 0.  Adding 3200/x² to both sides gives 80x = 3200/x².
Multiplying both sides by x² gives 80x³ = 3200.  Dividing both sides by 80 gives
x³ = 40.  Since 40 = 2*2*2*5, taking the cuberoot of both sides means that x = 2[³√5].

If I varied the length of the base up or down by 0.1,
the cost went up by a penny, so this is truly the minimum of this curve.