- #1

WMDhamnekar

MHB

- 359

- 28

**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:**

Let

**r**and

**h**be the radius and height of the right circular cylinder inscribed in a given cone of radius

**R**and height

**H**. Let S be the curved surface area of cylinder.

**S = 2πr*h**

h = H*(R – r)/R(

h = H*(R – r)/R

**Would any Math help board member provide me the detailed explanation of the computation of height of right circular cylinder of greatest curved surface inscribed in a given cone with a figure (as far as possible)**?

So

**S = 2πr*H(R – r)/R**

= $\frac{2πH}{R}(r*R – r^2)$

Differentiate w.r.t.r

$\frac{dS}{dr} = \frac{2πH}{R}(R – 2r)$

For maxima or minima

$\frac{dS}{dr} =0$

=> $\frac{2πH}{R}(R – 2r) = 0$

=> R – 2r = 0

=> R = 2r

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

$\frac{d^2S}{dr^2} = \frac{2πH}{R}*(0 – 2)= \frac{-4πH}{R }$(negative)

So for $r = \frac{R}{2},$ S is maximum.