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

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

Answers (1)

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 .

 

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