Question

Prove that the height of the cylinder of maximum volume inscribed in a right circular cone of semi-vertical angle \(\alpha\) and height \(h\) is one-third of the height of the cone, and the maximum volume of the cylinder is \(\dfrac{4}{27}\pi h^3\tan^2\alpha\).

Hint:

Express r in terms of y via similar triangles, maximise V=πr²y, verify with the second-derivative sign change.

Answer 1

Step 1: Setting up the geometry:
Let the cylinder have height \(y\) (measured from the base of the cone) and radius \(r\). By similar triangles, at height \(y\) from the base, the remaining distance to the apex is \((h-y)\), and the cone's radius there is \((h-y)\tan\alpha\); since the cylinder's top rim touches the cone's slanted surface, \(r=(h-y)\tan\alpha\).

Step 2: Writing the volume as a function of y:
\(V=\pi r^2y=\pi\tan^2\alpha\,(h-y)^2y\).

Step 3: Differentiating and setting to zero:
\(\dfrac{dV}{dy}=\pi\tan^2\alpha\left[(h-y)^2+y\cdot2(h-y)(-1)\right]=\pi\tan^2\alpha(h-y)\left[(h-y)-2y\right]=\pi\tan^2\alpha(h-y)(h-3y)\). Setting \(\dfrac{dV}{dy}=0\) gives \(y=h\) (trivial, gives \(r=0\)) or \(y=\dfrac{h}{3}\).

Step 4: Confirming it's a maximum:
For \(y<h/3\), \(h-3y>0\) so \(dV/dy>0\); for \(y>h/3\), \(h-3y<0\) so \(dV/dy<0\) — the derivative changes from positive to negative, confirming a maximum at \(y=h/3\).

Step 5: Computing the maximum volume:
At \(y=\dfrac h3\): \(r=\left(h-\dfrac h3\right)\tan\alpha=\dfrac{2h}{3}\tan\alpha\); \(V_{\max}=\pi r^2y=\pi\left(\dfrac{2h}{3}\tan\alpha\right)^2\cdot\dfrac h3=\pi\cdot\dfrac{4h^2}{9}\tan^2\alpha\cdot\dfrac h3=\dfrac{4\pi h^3}{27}\tan^2\alpha\).

Final Answer:
So the cylinder's height at maximum volume is \(\dfrac h3\) (one-third of the cone's height), and \(V_{\max}=\dfrac{4}{27}\pi h^3\tan^2\alpha\).\[ \boxed{y=\dfrac h3,\ V_{\max}=\dfrac{4}{27}\pi h^3\tan^2\alpha} \]