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  • Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 25 maths.

Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 25 maths.

Answers (1)

h = 16cm

Hint: For maxima or minima f'(x) must be zero f'(x) = 0

Given: Let r,l,v be radius, height and volume respectively.

Solution: 

h = R + \sqrt{R^2-r^2}

Squaring on both the sides

h^2+R^2-2hR=R^2-r^2

r^2=2hR-h^2                    ........(1)

Now,

V=\frac{1}{3}\pi r^2h

V=\frac{\pi }{3}(2h^2R-h^3)                      ....... from (1)

\begin{aligned} &\frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right) \\ &\frac{d V}{d h}=0 \\ &\frac{\pi}{3}\left(4 h R-3 h^{2}\right)=0 \\ &4 h R=3 h^{2} \\ &h=\frac{4 R}{3} \end{aligned}

Now,

\begin{aligned} &\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h) \\ &\frac{\pi}{3}(4 R-8 R)=0 \\ &\frac{-4 \pi R}{3}<0 \end{aligned}

So, the volume is maximum when h = \frac{4R}{3}

h = \frac{4\times 12}{3}=16cm

 

Posted by

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