Prove that the radius of the base of right circular cylinder of greatest curved surface area which can be inscribed - Maths - Application of Derivatives

Question

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

Gursheen kaur. · Accepted Answer

Let VAB be the cone of base of radius , r = OA and height , h = VO.

Let a cylinder of base of radius OC = x be inscribed in the cone.

From the figure , it is clear that ΔVOB∼ΔB'DB

$∴ \frac{V O}{B' D} = \frac{O B}{D B} \Rightarrow \frac{h}{B' D} = \frac{r}{r - x} \Rightarrow B' D = \frac{h (r - x)}{r}$

Let S be the curved surface area of cylinder.

Then, S= 2 $π$ (OC) (B'D)

$\Rightarrow S = 2 π \frac{h (r - x)}{r} = \frac{2 π h}{r} (r x - x^{2}) \Rightarrow \frac{d S}{d x} = \frac{2 π h}{r} (r - 2 x) a n d \Rightarrow \frac{d S^{2}}{d x^{2}} = - \frac{4 π h}{r}$

For maximum or minimum values of S, we must have $\frac{d S}{d x} = 0$

$\Rightarrow \frac{d S}{d x} = 0 \Rightarrow \frac{2 π h}{r} (r - 2 x) = 0 \Rightarrow x = \frac{r}{2} N o w, \frac{d S^{2}}{d x^{2}} = - \frac{4 π h}{r} < 0 f o r a l l x . H e n c e, S i s \max i m u m w h e n x = \frac{r}{2}, i . e . r a d i u s o f c y l i n d e r i s h a l f o f r a d i u s t h e c o n e .$

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