Question

1. the volume of a cylindrical tin can with a top and a bottom is to be $16\pi$ cubic inches. if a minimum amount of tin is to be used to construct the can, what must be the height, in inches, of the can?

a $2\sqrt3{2}$
b $2\sqrt{2}$
c $2\sqrt3{4}$
d 4
e 8

Explanation:

Step1: Define variables and volume formula

Let \(r\) = radius (in), \(h\) = height (in). Volume of cylinder:
$$V = \pi r^2 h = 16\pi$$
Simplify to get \(h = \frac{16}{r^2}\)

Step2: Write surface area formula

Total surface area (top + bottom + side):
$$S = 2\pi r^2 + 2\pi r h$$
Substitute \(h = \frac{16}{r^2}\):
$$S(r) = 2\pi r^2 + 2\pi r \cdot \frac{16}{r^2} = 2\pi r^2 + \frac{32\pi}{r}$$

Step3: Find critical points via derivative

Take derivative of \(S(r)\) with respect to \(r\):
$$S'(r) = 4\pi r - \frac{32\pi}{r^2}$$
Set \(S'(r) = 0\):
$$4\pi r - \frac{32\pi}{r^2} = 0$$
Divide by \(4\pi\):
$$r - \frac{8}{r^2} = 0 \implies r^3 = 8 \implies r = 2$$

Step4: Calculate height from \(r\)

Substitute \(r=2\) into \(h = \frac{16}{r^2}\):
$$h = \frac{16}{2^2} = \frac{16}{4} = 4$$

Answer:

D. 4