First, notice that, by the Pythagorean Theorem,
[tex]x^2+y^2=3^2 [/tex]
meaning that:
[tex]x^2=9-y^2[/tex]
Also, since the volume of a cone with radius r and height h is [tex] \frac{1}{3} \pi r^2h[/tex] we know that the volume of the cone is:
[tex] \frac{1}{3} \pi x^2 (3+y) = \frac{1}{3} \pi (9-y^2)(3+y) = \frac{1}{3} \pi [27+9y-3y^2-y^3][/tex]
Therefore, we want to maximize the function [tex]V(y) = \frac{1}{3} \pi [27+9y-3y^2-y^3][/tex] subject to the constraint [tex]0 \leq y \leq 3[/tex].
To find the critical points, we differentiate:
[tex]V'(y)= \frac{1}{3} \pi [9-6y-3y^2] = \pi [3-2y-y^2] = \pi (3+y)(1-y). [/tex]
Therefore, [tex]V'(y) = 0[/tex] when
[tex] \pi (3+y)(1-y)=0[/tex]
meaning that [tex]y = -3 [/tex] or [tex]y=1[/tex]. Only [tex]y=1[/tex] is in the interval [tex][0,3][/tex] so that’s the only critical point we need to concern ourselves with.
Now we evaluate [tex]V[/tex] at the critical point and the endpoints:
[tex]V(0) = \frac{1}{3} \pi [27+9(0) - 3(0)^2] = 9 \pi [/tex]
[tex]V(1) = \frac{1}{3} \pi [27+9(1)-3(1)^2-1^3] = \frac{32 \pi }{3} [/tex]
[tex]V(3) = \frac{1}{3} \pi [27+9(3) - 3(3)^2-3^2] = 0[/tex]
Therefore, the volume of the largest cone that can be inscribed in a sphere of radius 3 is [tex] \frac{32 \pi }{3} [/tex]
