1. ## Maxima/Minima

A can of beans has internal radius r and height h. Given that Surface area is S = 2 pi r ( r + h ) and this is equal to 120 pi. Determine r and h such that the volume of beans in the can is a maximum.

2. Hello, Jen1603!

A can of beans has internal radius $\displaystyle r$ and height $\displaystyle h.$
Given that surface area is $\displaystyle S \:=\:2\pi r(r + h)$ and this is equal to $\displaystyle 120\pi$,
determine $\displaystyle r$ and $\displaystyle h$ such that the volume of the can is a maximum.
We are given: .$\displaystyle 2\pi(r^2+rh) \:=\:120\pi \quad\Rightarrow\quad h \:=\:\frac{60-r^2}{r}$ .[1]

The volume of a cylindrical can is: .$\displaystyle V \:=\:\pi r^2h$ .[2]

Substitute [1] into [2]: .$\displaystyle V \;=\;\pi r^2\left(\frac{60-r^2}{r}\right) \quad\Rightarrow\quad V \;=\;\pi(60r - r^3)$

Differentiate and equate to zero: .$\displaystyle V' \;=\;\pi(60-3r^2) \:=\:0$

. . and we have: .$\displaystyle 3r^2\:=\:60\quad\Rightarrow\quad r^2 \:=\:20 \quad\Rightarrow\quad\boxed{ r \:=\:2\sqrt{5}}$

Substitute into [1]: .$\displaystyle h \;=\;\frac{60-20}{2\sqrt{5}} \quad\Rightarrow\quad\boxed{ h \:=\:4\sqrt{5}}$