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: Parametrising by the cylinder radius instead:
Let the cylinder have radius \(r\); then by similar triangles the height of the cylinder is \(y=h-\dfrac{r}{\tan\alpha}\) (the cylinder's top is at the level where the cone's radius equals \(r\)).

Step 2: Volume as a function of r:
\(V(r)=\pi r^2\left(h-\dfrac{r}{\tan\alpha}\right)=\pi r^2h-\dfrac{\pi r^3}{\tan\alpha}\).

Step 3: Differentiating w.r.t. r:
\(\dfrac{dV}{dr}=2\pi rh-\dfrac{3\pi r^2}{\tan\alpha}\). Set to zero: \(2\pi rh=\dfrac{3\pi r^2}{\tan\alpha}\Rightarrow 2h=\dfrac{3r}{\tan\alpha}\) (dividing by \(\pi r\), \(r\ne0\)) \(\Rightarrow r=\dfrac{2h\tan\alpha}{3}\).

Step 4: Finding the corresponding height:
\(y=h-\dfrac{r}{\tan\alpha}=h-\dfrac{2h}{3}=\dfrac h3\), matching the height found by the other method exactly.

Step 5: Substituting back for maximum volume:
\(V_{\max}=\pi\left(\dfrac{2h\tan\alpha}{3}\right)^2\cdot\dfrac h3=\pi\cdot\dfrac{4h^2\tan^2\alpha}{9}\cdot\dfrac h3=\dfrac{4\pi h^3\tan^2\alpha}{27}\).

Final Answer:
\[ \boxed{y=\dfrac h3,\quad V_{\max}=\dfrac{4}{27}\pi h^3\tan^2\alpha} \]