#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 9.

x =2r

Hint: For maximum or minimum value of z must have $\frac{dz}{dx}=0$

Given: Let the length of the side of square and radius of the circle x and y respectively. It's given that the sum of the perimeters of square and circle is constant

Solution: $4x + 2\pi r=k$ (k is constant)

$x= \frac{k-2\pi r}{4}$            .......(1)

Now,

$A=x^2+4\pi r^2$

$A=\frac{(k-2\pi r)^2}{16}+\pi r ^2$             .....rom equation (1)

\begin{aligned} &\frac{d A}{d x}=\frac{(k-2 \pi r)^{2}}{16}+\pi r^{2} \\ &=\frac{2(k-2 \pi r)-2 \pi}{16}+2 \pi r \\ &=\frac{(k-2 \pi r)-\pi}{4}+2 \pi r \end{aligned}

\begin{aligned} &=\frac{(k-2 \pi r)-\pi}{4}+2 \pi r=0 \\ &\frac{(k-2 \pi r) \pi}{4}=2 \pi r \\ &k-2 \pi r=8 r \\ &\frac{d^{2} A}{d x^{2}}=\frac{\pi^{2}}{2}+2 \pi>0 \end{aligned}

So the sum of the areas, A is least when $k-2\pi r=8r$

From (1) and (2) we get

$x=\frac{k-2\pi r}{4}$

$x=\frac{8r}{4}=2r$

$\therefore$ Side of square = Diameter of circle

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle .