Question

A right circular cone is inscribed in a sphere of radius r. Find the dimensions of the cone if its volume is to be a maximum. ​

Answer 1

Answer:

h=2r

r=r

volumeof cone 1/3×pi×r^2×h

1/3×22/7×r^2×2r

Answer 2

Basic Concept Used :-

HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A FUNCTION

• Differentiate the given function f(x)

• Put f'(x) = 0 and find critical points.

• Find the second derivative, i.e. f''(x).

• Apply the critical points in the second derivative.

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.

$\large\underline{\sf{Solution-}}$

Let us suppose that

• Radius of cone be 'y' units

and

• height of cone be 'h' units.

So,

Dimensions of cone,

• AB = height of cone = h units

and

• BC = radius of cone = 'y' units.

Now,

Given that,

• A cone is inscribed in a sphere of radius 'r' units.

Let

• AB be the axis of cone and O be the centre of sphere.

• OB = x units

So,

• Height of cone, AB = AO + OB = r + x

Now, In right-angle BOC, we have,

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = {r}^{2}$

$\bf\implies \: {y}^{2} = {r}^{2} - {x}^{2}$

Now, we know

Volume of cone is

$\rm :\longmapsto\:V = \dfrac{1}{3}\pi \: {y}^{2}h$

On substituting the values of y and h, we get

$\rm :\longmapsto\:V = \dfrac{1}{3}\pi \: ({r}^{2} - {x}^{2} )(r + x)$

$\rm :\longmapsto\:V = \dfrac{1}{3}\pi \: ({r} - {x})(r + x)(r + x)$

$\rm :\longmapsto\:V = \dfrac{1}{3}\pi \: (r - x) {(r + x)}^{2}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} V =\dfrac{d}{dx} \dfrac{1}{3}\pi \: (r - x) {(r + x)}^{2}$

$\rm :\longmapsto\:\dfrac{d}{dx} V =\dfrac{1}{3}\pi \:\dfrac{d}{dx}(r - x) {(r + x)}^{2}$

$\rm :\longmapsto\:\dfrac{d}{dx} V =\dfrac{1}{3}\pi \:((r - x)\dfrac{d}{dx}(\: {(r + x)}^{2} + {(r + x)}^{2}\dfrac{d}{dx} (r - x))$

$\rm :\longmapsto\:\dfrac{dV}{dx} = \dfrac{1}{3}\pi \:\bigg((r - x)2(r + x) + {(r + x)}^{2}( - 1) \bigg)$

$\rm :\longmapsto\:\dfrac{d}{dx} V =\dfrac{1}{3}\pi \:(r + x)(2r - 2x - r - x)$

$\rm :\longmapsto\:\dfrac{d}{dx} V =\dfrac{1}{3}\pi \:(r + x)(r - 3x) - - (1)$

For maximum or minimum value,

$\rm :\longmapsto\:\dfrac{d}{dx} V =0$

$\rm :\longmapsto\: \dfrac{1}{3}\pi \:(r + x)(r - 3x) = 0$

$\bf\implies \:x = \dfrac{r}{3} - - - (2)$

Now, on differentiating equation (1) w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \dfrac{1}{3}\pi \:\bigg(( r + x)\dfrac{d}{dx} (r - 3x) + (r - 3x)\dfrac{d}{dx} (r + x)\bigg)$

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \dfrac{1}{3}\pi \: ((r + x)( - 3) + (r - 3x))$

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \dfrac{1}{3}\pi \: ( - 3r - 3x \: + r - 3x)$

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = \dfrac{1}{3}\pi \: ( - 2r - 6x )$

$\rm :\longmapsto\:\dfrac{ {d}^{2} V}{ {dx}^{2} } = - \dfrac{1}{3}\pi \: (2r + 6x )$

$\rm :\implies\:\dfrac{ {d}^{2} V}{ {dx}^{2} } < 0$

$\bf\implies \:Volume \: of \: cone \: is \: maximum.$

As

$\rm :\longmapsto\: {y}^{2} = {r}^{2} - {x}^{2}$

$\rm :\longmapsto\: {y}^{2} = {r}^{2} - {\bigg( \dfrac{r}{3} \bigg) }^{2}$

$\rm :\longmapsto\: {y}^{2} = {r}^{2} - \dfrac{ {r}^{2} }{9}$

$\rm :\longmapsto\: {y}^{2} = \dfrac{9{r}^{2} - {r}^{2} }{9}$

$\rm :\longmapsto\: {y}^{2} = \dfrac{8{r}^{2}}{9}$

$\bf\implies \:y = \dfrac{2 \sqrt{2} }{3} r$

Now,

Height of cone,

$\rm :\longmapsto\:h = r + x$

$\rm :\longmapsto\:h = r + \dfrac{r}{3}$

$\rm :\longmapsto\:h = \dfrac{3r + r}{3}$

$\bf:\implies\:h = \dfrac{4}{3}r$

Hence,

$\begin{gathered}\begin{gathered}\bf\: Dimensions_{(cone)}-\begin{cases} &\sf{radius = \dfrac{2 \sqrt{2} }{3}r} \\ &\sf{height = \dfrac{4}{3} r} \end{cases}\end{gathered}\end{gathered}$