Question

Optimization for Max Volume of a Cone?

A right triangle has legs of length h and r, and a hypotenuse of length 4 (see figure). It is revolved about the leg of length h to sweep out a right circular cone. What values of hand r maximize the volume of the cone?

Answer 1

I got: `r=+sqrt(32/3)=+3.26`

Explanation:

Consider our cone:

The volume of the cone will be:

`V=1/3pir^2h`

let us express `h` as function of the othe sides using Pythagoras Theorem to write:
`h^2+r^2=4^2`
`h^2=16-r^2`
`h=sqrt(16-r^2)`

substitute in the volume:

`V=1/3pir^2sqrt(16-r^2)`

let us now derive this expression with respect to `r` and set the derivative equal to zero to find the Maximum:

`(dV)/(dr)=1/3pi[2rsqrt(16-r^2)+r^2*(-cancel(2)r)/(cancel(2)sqrt(16-r^2))]=1/3pi[(2r(16-r^2)-r^3)/(sqrt(16-r^2))]=1/3pi[(32r-3r^3)/(sqrt(16-r^2))]`

set this derivative equal to zero:

`1/3pi[(32r-3r^3)/(sqrt(16-r^2))]=0`
`32r-3r^3=0`
`r(32-3r^2)=0`

giving:

`r_1=0`...not good;
`r_(2,3)=+-sqrt(32/3)=+-3.26` I can use only the positive one.