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If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :

Option 1)
$6\sqrt{2}\: \pi$

Option 2)
$6\sqrt{3}\: \pi$

Option 3)
$8\sqrt{2}\: \pi$

Option 4)
$8\sqrt{3}\: \pi$

Answers (2)

best_answer

As we learned

Method for maxima or minima -

First and second derivative method :

$Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0$

$Step\:2.\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if\:f'(x)>0$ $and\:local\:minimum\:if\;f'(x)<0.$

$Step\:\:3.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:miximum\:if$ $f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0$

- wherein

$Where\:\:y=f(x)$

$\frac{dy}{dx}=f'(x)$

$Sphere's\: radius =3cm$

r,h be the radius and height of sphere

$Volume \: of\: cone =\frac{1}{3}\pi b^{2}h$

in $\bigtriangleup ABC$ ,using pythagorus theorem

$\left ( h-r \right )^{2}+b^{2}=r^{2}$

$b^{2}=r^{2}-\left ( h-r \right )^{2}-2hr-r^{2}$

$volume=\frac{1}{3}\left [ 2h^{2}r-h^{3} \right ]$

$\frac{\mathrm{dv} }{\mathrm{d} h}=h\left ( 4r-3h \right )=0\: \: \frac{d^{2}v}{dh^{2}}=\frac{1}{3}\left [ 4r-6h \right ]$

So at $h=\frac{4r}{3}$ ; we get max r=3 h = 4

thus $b=2\sqrt{2}$

$CSA = \pi bl=\pi 2\sqrt{2}\sqrt{4^{2}+\left ( 2\sqrt{2} \right )^{2}}=\pi 2\sqrt{2}\sqrt{24}=\pi 2\sqrt{2}\cdot 2\sqrt{3}\sqrt{2}$

$=8\sqrt{3}\pi$

Option 1)

$6\sqrt{2}\: \pi$

Option 2)

$6\sqrt{3}\: \pi$

Option 3)

$8\sqrt{2}\: \pi$

Option 4)

$8\sqrt{3}\: \pi$

Posted by

Himanshu

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