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.

 

 

 

 
 
 
 
 

Answers (1)

consider the given figure

Let H and R be the height and radius of the base of the cone ABC respectively.
suppose the radius and height of the cylinder inserted in the cone be r and h resp.
Now, DF= r\; \; \; AD= H-h
AS \triangle ADF\sim \triangle AOC
so, \frac{AD}{AO}= \frac{DF}{OC}
\Rightarrow \frac{H-h}{H}= \frac{r}{R}\; \; \; ie\; \; h= \left ( 1-\frac{r}{R} \right )H
Let s be the curved surface area of the cylinder
so, s= 2\pi rh
  \Rightarrow s= 2\pi r\left ( 1-\frac{r}{R} \right )H
\Rightarrow s= 2\pi H\left ( r-\frac{r^{2}}{R} \right )
Now, differentiating w.r.t to r, we get : \frac{ds}{dr}= 2\pi H\left ( 1-\frac{2r}{R} \right )
Again differentiating w.r.t to r, we get : \frac{d^{2}s}{dr^{2}}= 2\pi H\left ( \frac{-2}{R} \right )
                                                                       = \frac{4\pi H}{R}
for  \frac{ds}{dr}= 0,ie, 2\pi H\left ( 1-\frac{2r}{R} \right )= 0
                                  r= \frac{R}{2}
Now \frac{d^{2}s}{dr^{2}}]_{@r= \frac{R}{2}}\; \; = \frac{4\pi H}{R}< 0
so, curved surface area s, of the cylinder is maximum at r= \frac{R}{2}
hence, the radius of the right circular cylinder of greatest curved surface area which can be insulated in a given cone is half of that of the cone.

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