Maximize the volume of a cone

Question

How do I maximize the volume of a cone which is inscribed inside a sphere of radius $r$. I know that $V=\pi r^2h$. But how do you inscribe that into a sphere with radius $r$.

Answer 1

As per the comment, once you let the radius of the cone's base be a different variable (say $r$) than the radius of the sphere (say $R$), then all you need to do is come up with a formula for $r$ in terms of $R$ and $h$. Hint: think about the relationship between $(h-R)$ and $r$.

Answer 2

$V = \dfrac{π r^2 h}{3}$

And that of a sphere of radio $R$: $V = \dfrac{4π R^3}{3} $

The problem can be reduced to two dimensions considering the case of a circle centered on the origin of coordinates and radius $R$ that has inscribed a triangle of base $2r$ and height $R + a$, where $``a"$ is variable.

When both lines are rotated around the $X$-axis, they describe a sphere with an inscribed cone of radius $R$ and $r$ respectively and cone height $h = R + a$:

It is necessary to maximize the volume of the cone with the condition that it is inscribed in the sphere, that is, to maximize the function: $V (a, r) ​​= \dfrac{πr^2 (R + a)}{ 3}$. Provided that the point $(a, r)$ ​​belongs to the circle $x^2 + y^2 = R^2$, that is, that the relation is fulfilled:

$a^2 + r^2 = R^2$

$r^2 = R^2 - a^2 ........... (1)$

And substituting in the formula of the volume of the cone we have eliminated the variable $"r"$ and it remains:

$V (a) = \dfrac{π (R^2 - a^2) (R + a)}{3} = \dfrac{π (R^3 + aR^2 - a^2 R - a^2)}{3}$

Let's see for what value of "a" said volume is maximum. Deriving and equating to zero:

$V'(a) = \dfrac{π(R^2-2aR-3a^2)} { 3} = 0$

$R^2-2aR-3a^2 = 0\Rightarrow 3a^2+2aR-R^2 = 0$

Quadratic equation in "a", which by the resolvent leaves us

$a = \dfrac{-2R \pm \sqrt{4R^2 + 12 R^2}}{6} $

whose two roots are:

$a_1 = \dfrac{- 2R + 4R}{6} = \dfrac{R}{3}$

$a_2 = \dfrac{- 2R - 4R} {6} = - R$

Taking the only possible value, the positive one, i.e $a_1$, we substitute it in equation (1), we have left

$r^2 = R^2 - \dfrac{R^2}{9} = \dfrac{8R^2}{9}$

That is, the maximum volume of the cone inscribed in the sphere of radius R is obtained for a conical radius:

$R = \dfrac{\sqrt{8R^2}}{3} = \dfrac{2R \sqrt{2}}{3}$

And for a cone height of:

$h = R + a = R + \dfrac{R}{3} = \dfrac{4R}{3}$

The maximum volume of the inscribed cone will therefore be:

$V = π (\dfrac{8R^2}{9})(\dfrac{4R}{3}) = \dfrac{32 πR^3}{81}$