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: Substituting r = R/2 back and comparing to nearby values instead of the second-derivative test:
At \(r=R/2\): \(h=\dfrac{H(R-R/2)}{R}=\dfrac{H}{2}\), and \(S=2\pi\cdot\dfrac{R}{2}\cdot\dfrac{H}{2}=\dfrac{\pi RH}{2}\).

Step 2: Testing a slightly smaller r:
At \(r=R/4\): \(h=\dfrac{H(R-R/4)}{R}=\dfrac{3H}{4}\), \(S=2\pi\cdot\dfrac{R}{4}\cdot\dfrac{3H}{4}=\dfrac{3\pi RH}{8}=0.375\,\pi RH\), which is less than \(\dfrac{\pi RH}{2}=0.5\,\pi RH\).

Step 3: Testing a slightly larger r:
At \(r=3R/4\): \(h=\dfrac{H(R-3R/4)}{R}=\dfrac{H}{4}\), \(S=2\pi\cdot\dfrac{3R}{4}\cdot\dfrac{H}{4}=\dfrac{3\pi RH}{8}=0.375\,\pi RH\), again less than at \(r=R/2\).

Final Answer:
Both neighbouring values give a smaller \(S\) than at \(r=R/2\), confirming it's the maximum: \(\boxed{r=\dfrac{R}{2}}\), i.e. half the cone's radius.