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  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 40.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 40.

Answers (1)

x = 2r

Hint: For maximum or minimum value of V, we must have \frac{dV}{dr}=0

Given:

Let r be the radius of the sphere, x is side of the cube and S be the sum of the surface area of both of them

S= 4 \pi r^2 + 6x^2

x= \left ( \frac{S-4\pi r^2}{6} \right )                    ......(1)

Solution:

Sum of Volumes, V=\frac{4}{3}\pi r^3+x^3

V=\frac{4}{3} \pi r^{3}+\left[\left(\frac{S-4 \pi r^{2}}{6}\right)\right]^{\frac{3}{2}}

From eqn (1)

\frac{dV}{dr}=4 \pi r^{2}-2 \pi r\left[\left(\frac{S-4 \pi r^{2}}{6}\right)\right]^{\frac{1}{2}}

\frac{dV}{dr}=0                    .........(2)

\begin{aligned} &4 \pi r^{2}-2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}=0 \\ &4 \pi r^{2}=2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}} \end{aligned}

4 \pi r^2 -2\pi rx from eqn (1),

\begin{aligned} &\frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}-\frac{2 \pi r}{2}\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{-\frac{1}{2}}\left(\frac{-8 \pi r}{b}\right) \\ &=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}+\frac{4 \pi^{2} r^{2}}{3}\left[\frac{6}{\left(S-4 \pi r^{2}\right)}\right]^{\frac{1}{2}} \\ &=8 \pi r-2 \pi x+\frac{4 \pi^{2} r^{2}}{3} \frac{1}{x}=8 \pi r-4 \pi r+\frac{2}{3} \pi^{2} r \\ &\frac{d^{2} V}{d r^{2}}=4 \pi r+\frac{2}{3} \pi^{2} r>0 \end{aligned}

Volume is minimum when x = 2r

Posted by

infoexpert24

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