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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is \(\tan^{-1} \sqrt {2}\).

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  • $V=\large\frac{1}{3}$$\pi r^2h$
  • $\large\frac{d}{d\theta}$$(\sin \theta)=\cos \theta$
  • $\large\frac{d}{d\theta}$$(\cos\theta)=-\sin \theta$
Step 1:
Let $V$ be the volume $l$ be the slant height and $\theta$ be the semi-vertical angle of a cone.
Volume of the cone $V=\large\frac{1}{3}$$\pi r^2h$
Vertical height $h=l\cos \theta$
Radius $r=l\sin \theta$
By Substituting the value of $h$ and $r$ we get
$V=\large\frac{1}{3}$$\pi r^2h$
$\;\;=\large\frac{1}{3}$$\pi(l\sin \theta)^2.l\cos\theta$
$\;\;=\large\frac{1}{3}$$\pi(l^2\sin ^2\theta).l\cos\theta$
$\;\;=\large\frac{1}{3}$$\pi l^3\sin ^2\theta.\cos\theta$
Step 2:
By differentiating with respect to $\theta$ we get
$\large\frac{dV}{d\theta}=\frac{1}{3}$$\pi l^3[2\sin \theta.\cos \theta.\cos \theta-\sin^2\theta.\sin\theta]$
$\qquad=\large\frac{1}{3}$$\pi l^3[2\sin \theta.\cos^2\theta-\sin^3\theta)$
$\qquad=\large\frac{1}{3}$$\pi l^3\sin\theta[2\cos^2\theta-\sin^2\theta]$
$\tan\theta=\sqrt 2$
Step 3:
We can write $2\cos^2\theta-\sin^2\theta$=$(\sqrt 2\cos \theta+\sin\theta) (\sqrt 2\cos\theta-\sin\theta)$
Further $\large\frac{dV}{d\theta}=\frac{1}{3}$$\pi l^3\sin\theta(\sqrt 2\cos \theta+\sin\theta)\times (\sqrt 2\cos\theta-\sin\theta)$
$\qquad\;\qquad=\large\frac{1}{3}$$\pi l^3\sin\theta\cos^2\theta(\sqrt 2+\tan\theta)(\sqrt 2-\tan \theta)$
$\qquad\;\qquad=\large\frac{1}{3}$$\pi l^3\sin\theta\cos^2\theta(\tan\theta-\sqrt 2)(\tan \theta+\sqrt 2)$
Step 4:
When $\theta$ is slightly < $\tan^{-1}\sqrt 2$
$\sin \theta\cos^2\theta=+ve$
$\tan\theta-\sqrt 2=-ve$
$\tan\theta+\sqrt 2=+ve$
When $\theta$ is slightly > $\tan^{-1}\sqrt 2$
$\sin \theta\cos^2\theta=+ve$
$\tan\theta-\sqrt 2=+ve$
$\tan\theta+\sqrt 2=+ve$
$\large\frac{dV}{d\theta}$ changes from +ve to -ve
$\therefore$ V is maximum at $\theta=\tan^{-1}\sqrt 2$
answered Aug 9, 2013 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1

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