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  • Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 26 maths.

Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 26 maths.

Answers (1)

r = 7cm

Hint: For maxima or minima value of s we must have \frac{ds}{dr}=0

Given: Let the height, radius of base and surface area of cylinder be h,r,v.

Solution:

Volume =\pi r^2 h

2156=\pi r^2 h

2156=\frac{22}{7} r^2 h

h = \frac{2156 \times 7}{22r^2},h = \frac{686}{r^2}                          .........(1)

Surface area = 2\pi rh+\pi r^2h 

=\frac{4132}{r}+\frac{44}{r^2}                 ......from (1)

\begin{aligned} &\frac{d s}{d r}=\frac{4312}{-r^{2}}+\frac{88 r}{7} \\ &\frac{d s}{d r}=0 \\ &-\frac{4312}{r}+\frac{88 r}{7}=0 \\ &\frac{4312}{r}=\frac{88 r}{7} \Rightarrow r^{3}=\frac{4321 \times 7}{88} \\ &r^{3}=343, r=7 \mathrm{~cm} \end{aligned}

Now,

\begin{aligned} &\frac{d^{2} s}{d r^{2}}=\frac{8624}{r^{3}}+\frac{88}{7} \\ &=\frac{8624}{343}+\frac{88}{7}=\frac{176}{7}>0 \end{aligned}

So, the surface area is maximum when r = 7

 

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