Question

[25 POINTS] A cylinder has a volume of 400π cm3. Find the dimensions that minimize its surface area.

Answer 1

Answer:

r = 5.848cm, h = 11.696cm

Step-by-step explanation:

Surface area: $SA=2\pi rh+2\pi r^2$

Volume: $V=\pi r^2h$

Therefore, the height of a cylinder given its volume would be $h=\frac{V}{\pi r^2}$, thus:

$SA=2\pi r(\frac{V}{\pi r^2})+2\pi r^2\\\\SA=\frac{2V}{r}+2\pi r^2\\\\SA=\frac{2(400\pi)}{r}+2\pi r^2\\\\SA=\frac{800\pi}{r}+2\pi r^2$

We now find the derivative of the surface area of the cylinder with respect to its radius and set it equal to 0, solving for the radius:

$\frac{d(SA)}{dr}=-\frac{800\pi}{r^2}+4\pi r$

$0=-\frac{800}{r^2}+4\pi r$

$0=800\pi+4\pi r^3$

$-800\pi=4\pi r^3$

$-200=r^3$

$r\approx5.848$

If $0If [tex]r>5.848$, then the surface area of the cylinder increases

Therefore, the surface area is minimized when the radius is $r=5.848cm$, making the minimum height $h=\frac{V}{\pi r^2}=\frac{400\pi}{\pi(5.848)^2}\approx11.696cm$.

In conclusion, the 2nd option is correct.