Question

Prove that the radius of the largest (maximum surface area) right circular cylinder that can be inscribed in a cone is half of the radius of the cone.

Hint:

Express \(S=2\pi rh\) with \(h\) from similar triangles, then maximize \(S(r)\) via \(dS/dr=0\) and the second-derivative test.

Answer 1

Step 1: Understanding the Concept:
Let the cone have fixed base radius \(R\) and height \(H\). An inscribed cylinder of radius \(r\) and height \(h\) has its top rim touching the slanted surface, so \(r\) and \(h\) are linked by similar triangles; express the cylinder's curved surface area in terms of \(r\) alone and maximize.

Step 2: Relating h and r via similar triangles:
Looking at the cone's axial cross-section, the cylinder's top edge lies on the cone's slant, giving \(\dfrac{h}{H}=\dfrac{R-r}{R}\), so \(h=\dfrac{H(R-r)}{R}\).

Step 3: Writing the curved surface area as a function of r:
\(S=2\pi rh=2\pi r\cdot\dfrac{H(R-r)}{R}=\dfrac{2\pi H}{R}(rR-r^2)\).

Step 4: Differentiating and setting to zero:
\(\dfrac{dS}{dr}=\dfrac{2\pi H}{R}(R-2r)\). Setting \(\dfrac{dS}{dr}=0\) gives \(R-2r=0\), i.e. \(r=\dfrac{R}{2}\).

Step 5: Confirming it's a maximum:
\(\dfrac{d^2S}{dr^2}=\dfrac{2\pi H}{R}(-2)<0\), a negative second derivative confirms \(r=R/2\) gives a maximum (not a minimum) of \(S\).

Final Answer:
The surface area is maximized when \(r=\dfrac{R}{2}\), i.e. \(\boxed{\text{the cylinder's radius is half the cone's radius}}\).