Maxima/Minima

**Jen1603** · December 2nd 2008, 04:13 AM

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.

**Soroban** · December 2nd 2008, 07:11 AM

Hello, Jen1603!

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 the can is a maximum.

We are given: . $2\pi(r^2+rh) \:=\:120\pi \quad\Rightarrow\quad h \:=\:\frac{60-r^2}{r}$ .[1]

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

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

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

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

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

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