Question

Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Answer 1

Let ${}^{\prime }{H}^{\prime }$ be the height of the cone, ${}^{\prime }{R}^{\prime }$ be the radius of the cone, ${}^{\prime }{h}^{\prime }$ be the height of the cylinder, ${}^{\prime }{r}^{\prime }$ be the radius of the cylinder and ${}^{\prime }{S}^{\prime }$ be the lateral surface area of the cylinder.

To prove- $r=\frac{R}{2}$.

Radius of cylinder, $r=R(1-\frac{h}{H})$

Now since $S=2\pi rh$, put $r=R(1-\frac{h}{H})$ in $S=2\pi rh$ as shown below:

$S=2\pi \left(R(1-\frac{h}{H})\right)h$

Differentiating with respect to $h$ we get,

$\frac{dS}{dh}=2\pi \left(\frac{R}{H}\right)(H-2h)$

Again differentiating with respect to $h$ that is

$\frac{d^{2}S}{d{h}^2}=-4\pi \left(\frac{R}{H}\right)$

Now put $\frac{dS}{dh}=0$ to find the stationary points:

$H-2h=0$

Implies $h=\frac{H}{2}$

Now consider,

$r=R(1-\frac{h}{H})$

Put $h=\frac{H}{2}$ in $r=R(1-\frac{h}{H})$, we get

$r=\frac{R}{2}$.