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  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17 5 question 5 maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17 5 question 5 maths textbook solution.

Answers (1)

2\left ( \frac{50}{\pi} \right )^{\frac{1}{3}}

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

Given: Let r and h be in the radius and height of the cylinder respectively. Then , 

Volume (V) of cylinder = \pi r^2h

100=\pi r^2h

h=\frac{100}{\pi r^2}

Solution: h=\frac{100}{\pi r^2}

Surface area (S) of the cylinder = 2\pi r^2+2\pi rh

\begin{aligned} &=2 \pi r^{2}+2 \pi r \times \frac{100}{\pi r^{2}} \\ &S=2 \pi r^{2}+\frac{200}{r} \\ &\frac{d s}{d r}=4 \pi r-\frac{200}{r^{2}} \end{aligned}

For maximum or minimum,

\frac{ds}{dr}=0

4\pi r-\frac{200}{r^2}=0

4\pi r^3=200

\begin{aligned} &r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \\ &\frac{d^{2} s}{d r^{2}}=4 \pi+\frac{400}{r^{3}} \\ &\frac{d^{2} s}{d r^{2}}>0, \text { when } r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \end{aligned}

Thus surface area is minimum when  r=\left (\frac{50}{\pi} \right )^{\frac{1}{3}}

\begin{aligned} h &=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{1}{3}}} \\ h &=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}} \end{aligned}

Cylinder with radius =\left (\frac{50}{\pi} \right )^{\frac{1}{3}} has min surface area.

 

 

Posted by

infoexpert24

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