Maximization: Volume of paraboloid within cone?

Question

Given a right circular cone with the line of symmetry along $x=0$, and the base along $y=0$, how can I find the maximum volume paraboloid (parabola revolved around the y-axis) inscribed within the cone? Maximising the volume of the paraboloid relative to the volume of the right circular cone. In 2-D, the parabola has 2 points of tangency to the triangle, one of each side of the line of symmetry. I have tried using the disk method to find the volume of the cone, and the parabola, both with arbitrary equations such as $y=b-ax$, and $y=c-dx^2$, but I end up with a massive equation for several variables, instead of a simple percentage answer. Any help is appreciated! Thanks in advance.

Answer 1

Hint:

Note that here $a$ and $b$ are given.

The tangency condition is $$a^2-4\,d\,(b-c)=0$$ If you decide to work with $c\,$, then $$d=\frac {a^2}{4\,(b-c)}$$ The volume of the paraboloids, inscribed in the cone, is $$V(c)=\pi \int_0^c \frac {c-y}d\, dy=\pi \int_0^c (c-y)\, \frac {4(b-c)}{a^2}\, dy=\frac {2\pi}{a^2}\,(b\,c^2-c^3)$$ Then solve $$V'(c)=0$$ to find the largest volume for $$c=\frac 23\, b$$ Verify that $$V_{max}=\frac 89\, V_{cone}$$