Question

Prove that the height of the right circular cone of maximum volume inscribed in a sphere of radius \(r\) is \(\dfrac{4r}{3}\).

Hint:

Use R^2 = 2rh - h^2 from the sphere geometry, then maximize V(h) using calculus.

Answer 1

Step 1: Setting up the geometry:
Let a right circular cone be inscribed in a sphere of radius \(r\) with center \(O\).
Let \(h\) be the height of the cone and \(R\) be the radius of its base circle.
The base circle and the apex both lie on the sphere, so the right triangle formed by \(R\), \(r\) and \((h-r)\) gives the relation \(R^2=r^2-(h-r)^2\).

Step 2: Writing volume as a function of one variable:
Simplify the relation to get \(R^2=2rh-h^2\), valid for \(0<h<2r\).
The volume of the cone is \(V=\dfrac{1}{3}\pi R^2 h\).
Substitute \(R^2\) to write V purely in terms of h:
\[ V(h)=\dfrac{1}{3}\pi(2rh^2-h^3), \quad 0<h<2r \]

Step 3: Differentiating and finding the critical point:
Differentiate V with respect to h.
\[ \dfrac{dV}{dh}=\dfrac{1}{3}\pi(4rh-3h^2)=\dfrac{1}{3}\pi h(4r-3h) \]
Set \(\dfrac{dV}{dh}=0\). Since \(h\neq0\) for a real cone, \(4r-3h=0\), so \(h=\dfrac{4r}{3}\).

Step 4: Verifying it is a maximum:
Find the second derivative.
\[ \dfrac{d^2V}{dh^2}=\dfrac{1}{3}\pi(4r-6h) \]
At \(h=\dfrac{4r}{3}\): \(\dfrac{d^2V}{dh^2}=\dfrac{1}{3}\pi(4r-8r)=-\dfrac{4\pi r}{3}<0\), so V is maximum at this h.

Final Answer:
The volume is maximum when the height equals four-thirds of the radius.
\[ \boxed{h=\dfrac{4r}{3}} \]