Question

You are asked to build an open cylindrical can with volume 180.5 cubic inches. You will cut the bottom from a square of metal and form the curved side by bending a rectangular sheet of metal.

Find the exact value of the radius and height of the can that will minimize the total amount of material required for the square and the rectangle.

Answer 1

This was the image included with the problem.

Answer 2

Generate a function for the height of the cyliner as a function of radius. Use this function to generate a function of r for the material used. Take the derivative of that function, set it to zero, and solve for r.

Answer 3

\[V=\pi r^2h\]h=d
h=d=2r
h=2r
\[V=\pi r^2 2r\]

Answer 4

Am I approaching this correctly so far?

Answer 5

\[V(r,h)=180.5=\pi r^2h \implies h=\frac{180.5}{\pi r^2}\]\[\implies A(r)=r(r+h)=r^2+\frac{180.5}{\pi r}\]Differentiate, set the derivative to zero, and solve.

Answer 6

Your assumption that there is a direct relationship between the height and the radius is incorrect. As the radius increases, the height decreases, since the volume is a constant.

Answer 7

Actually, my function for the area of the can has an error. The function should be\[A(r)=2r(2r+h)=4r^2+\frac{361}{\pi r}\]The square the can bottom is cut from is 2r on each side, not r.

Answer 8

So now I differentiate the area function.

Answer 9

Yep. You think you've got this?

Answer 10

\[A'(r)=8r-\frac{ 361 }{ \pi r^2 }\]

Answer 11

I'd say that's not too far off. Hope I was helpful. Do math every day.

Answer 12

Yes, thank you! That was very helpful.
I have one more question. After I set the derivative to zero, which derivative test should I use to make sure it is a minimum? Is it easier to use the first derivative test?

Answer 13

I would take the second derivative, and confirm that it's positive at that value of r. The setup of the problem is such that it has to be a minimum.

Answer 14

Thank you!

Answer 15

No sweat. Do math every day.