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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 20.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 20.

Answers (1)

h =\frac{2}{3}

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

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

Solution:  Let h, r and R be in the height and radius and base of the cone

\begin{aligned} &h=R+\sqrt{R^{2}-r^{2}} \\ &(h-R)^{2}=R^{2}-r^{2} \\ &h^{2}+R^{2}-2 h r=R^{2}-r^{2} \end{aligned}

r^2=2hR-h^2               .......(1)

Volume of cone = \frac{1}{3}\pi r^2h

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

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

Now,

\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}(4hR-3h^2)=0

4hR=3h^2

h=\frac{4R}{3}

Substituting y value in eqn (1)

\begin{aligned} &r^{2}=4\left(r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}\right) \\ &=4\left(r^{2}-\frac{r^{2}}{2}\right)=4\left(\frac{r^{2}}{2}\right)=2 r^{2} \end{aligned}

x=r\sqrt{2}

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

Volume is maximum when h =\frac{4R}{3}

h =\frac{2}{3}  (Diameter of sphere)

Hence proved

 

Posted by

infoexpert24

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