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  • Confused! kindly explain, The maximum volume (in cu.m) of the right circular cone having slant height 3 m is:

 

The maximum volume (in cu.m) of the right circular cone having slant height 3 m is:

  • Option 1)

     

    6\pi

  • Option 2)

    3\sqrt{3}\pi

  • Option 3)

    \frac{4}{3}\pi

  • Option 4)

    2\sqrt3\pi

Answers (1)

best_answer

 

Method for maxima or minima -

By 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

- wherein

Where\:\:y=f(x)

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

l=3m\left ( slant\: \: height \right )

h=3\cos \left ( \theta \right )

r=3\sin \left ( \theta \right )

Volume of right circular cone 

V=\frac{1}{3}\pi r^{2}h

    =\frac{1}{3}\pi \left ( 3\sin \left ( \theta \right ) \right )^{2}\left ( 3\cos \left ( \theta \right ) \right )

V=\frac{\pi }{3}9\sin ^{2}\theta \cdot 3\cos \left ( \theta \right )=9\pi \sin ^{2}\theta \cdot \cos \theta

for maximum volume 

\frac{\mathrm{dv} }{\mathrm{d} \theta }=0,\; \; \; \; \frac{\mathrm{dv} }{\mathrm{d} \theta }=-9\pi \sin \left ( \theta \right )\left [ \sin ^{2}\left ( \theta \right )-2\cos ^{2} \left ( \theta \right )\right ]=0

\sin \left ( \theta \right )=\sqrt{\frac{2}{3}}

\frac{\mathrm{d}^{2}v }{\mathrm{d} \theta ^{2}}=-ve\; at\; \sin \left ( \theta \right )=\sqrt{\frac{2}{3}}

Volume is maximum when \sin \left ( \theta \right )=\sqrt{\frac{2}{3}}

\therefore V_{max}=2\sqrt{3}\pi \left ( cu.m \right )

 


Option 1)

 

6\pi

Option 2)

3\sqrt{3}\pi

Option 3)

\frac{4}{3}\pi

Option 4)

2\sqrt3\pi
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