Question

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3?

Answer 1

First, notice that, by the Pythagorean Theorem,

`x^2+y^2=3^2`

meaning that:

`x^2=9-y^2`

Also, since the volume of a cone with radius r and height h is
`(1)/(3) \pi r^2h` we know that the volume of the cone is:

`(1)/(3) \pi x^2 (3+y) = (1)/(3) \pi (9-y^2)(3+y) = (1)/(3) \pi [27+9y-3y^2-y^3]`

Therefore, we want to maximize the function
`V(y) = (1)/(3) \pi [27+9y-3y^2-y^3]` subject to the constraint
`0 \leq y \leq 3`.

To find the critical points, we differentiate:

`V'(y)= (1)/(3) \pi [9-6y-3y^2] = \pi [3-2y-y^2] = \pi (3+y)(1-y).`

Therefore,
`V'(y) = 0` when

`\pi (3+y)(1-y)=0`

meaning that
`y = -3` or
`y=1`. Only
`y=1` is in the interval
`[0,3]` so that’s the only critical point we need to concern ourselves with.

Now we evaluate
`V` at the critical point and the endpoints:

`V(0) = (1)/(3) \pi [27+9(0) - 3(0)^2] = 9 \pi`

`V(1) = (1)/(3) \pi [27+9(1)-3(1)^2-1^3] = (32 \pi )/(3)`

`V(3) = (1)/(3) \pi [27+9(3) - 3(3)^2-3^2] = 0`

Therefore, the volume of the largest cone that can be inscribed in a sphere of radius 3 is
`(32 \pi )/(3)`