Question

Find the dimensions of the right circular cone of the greatest lateral area that can be inscribed in a sphere of radius α

Answer 1

Solution:

Given, a cone inscribed in a sphere of radius a.

Let the radius, height and slant height of cone be $r,h,l$ respectively.

$OB=OA=a,BC=r,AC=h,OC=h-a,AB=l$

Using pythagoras theorem in triangle BOC,

$BC^2+CO^2=OB^2 \\\Rightarrow r^2+(h-a)^2=a^2 \\\Rightarrow r^2+h^2+a^2-2ha=a^2 \\\Rightarrow r^2+h^2=2ha \ ...(i)$

Using pythagoras theorem in triangle BAC,

$BC^2+CA^2=AB^2 \\\Rightarrow r^2+h^2=l^2\ ...(ii)$

From (i) and (ii)

$2ha=l^2$

Now, lateral area of cone, $L=\pi rl$

$L^2=\pi^2 r^2 l^2 \\=\pi^2 (l^2-h^2)l^2\ [From\ (ii)] \\=\pi^2(2ha-h^2)(2ha) \\=\pi^2(4h^2a^2-2h^3a)$

On differentiating w.r.t. h

$2L\dfrac{dL}{dh}=\pi^2(8ha^2-6h^2a) ...(iii) \\ \Rightarrow\dfrac{dL}{dh}=\dfrac{\pi^2 (2ha\pi^2 (4a-3h))}{2L}$

Now, put $\dfrac{dL}{dh}=0$

$\\ \Rightarrow\dfrac{\pi^2 (2ha\pi^2 (4a-3h))}{2L}=0 \\ \Rightarrow 4a-3h=0 \\ \Rightarrow h=\dfrac{4a}3\ ...(iv)$

Again, differentiating (iii) w.r.t h,

$2L\dfrac{d^2L}{dh^2}+2(\dfrac{dL}{dh})^2=\pi^2(8a^2-12ha) \\=\pi^2(8a^2-12a(\dfrac{4a}3))\ [Using\ (iv)] \\=-8\pi^2a^2<0 \\\Rightarrow\ maxima$

Put (iv) in (i)

$\\r^2+(\dfrac{4a}3)^2=2(\dfrac{4a}3)a \\\Rightarrow r^2=\dfrac{8a^2}3-\dfrac{16a^2}9 \\\Rightarrow r^2=\dfrac{8a^2}9 \\\Rightarrow r=\dfrac{2\sqrt2a}3$

Thus, the dimensions of cone are $r=\dfrac{2\sqrt2a}3;h=\dfrac{4a}3$