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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height \(h\) and semi vertical angle $\alpha$ is one-third that of the cone and the greatest volume of cylinder is $\large \frac{4}{27}$$\pi h^3 \: tan^2 \: \alpha $

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Let $VAB$ be the cone of height h,semi vertical angle $\alpha$ and let $x$ be the radius of the base of the cylinder $A'B'DC$ which is inscribed in the cone $VAB$.Then
$OO'$ height of the cylinder$=VO-VO'$
$\qquad\qquad\qquad\qquad\quad=(h-x\cot \alpha)\pi x^2$
Differentiating with respect to $x$
$\large\frac{dV}{dx}$$=2\pi xh-3\pi x^2\cot \alpha$
For maxima or minima $V,\large\frac{dV}{dx}$$=0$
$2\pi xh-3\pi x^2\cot \alpha=0$
$2\pi xh=3\pi x^2\cot \alpha$
$x=\large\frac{3\pi x^2\cot \alpha}{2\pi h}$
$2hx=3\pi x^2\cot \alpha$
$2h=3\pi x\cot\alpha$
$\large\frac{2h}{3\pi \cot \alpha}$$=x$
$\large\frac{1}{\cot\alpha}$$=\tan\alpha$
$x=\large\frac{2h}{3}$$\tan\alpha$
Step 2:
Now $\large\frac{d^2V}{dx^2}$$=2\pi h-6\pi x\cot\alpha$
When $x=\large\frac{2h}{3}$$\tan\alpha$ we have
$\large\frac{d^2V}{dx^2}=$$\pi(2h-4h)=-2\pi h <0$
$\Rightarrow V$ is maximum when $x=\large\frac{2h}{3}$$\tan\alpha$
$OO'=h-x\cot \alpha$
$\qquad=h-\large\frac{2h}{3}$
$\qquad=\large\frac{h}{3}$
Step 3:
The maximum volume of the cylinder is $V=\pi\big(\large\frac{2h}{3}$$\tan\alpha\big)^2(h-\large\frac{2h}{3})$
$V=\large\frac{4}{27}$$\pi h^3\tan^2\alpha$
answered Aug 13, 2013 by sreemathi.v
 

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