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  • Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 15.

Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 15.

Answers (1)

\frac{20}{\pi +4},\frac{10 }{\pi +4}

Hint: For maxima and minima value of A, we must have \frac{dA}{dx}=0

Given: (x+2y)+\pi \left (\frac{x}{2} \right )=10

Solution:

Let the dimensions of the rectangular part be x,y

Radius of the semicircle =\frac{x}{2}

Total perimeter = 10

(x+2y)+\pi \left (\frac{x}{2} \right )=10

2 y=\left[10-x-\pi\left(\frac{x}{2}\right)\right]

y=\frac{1}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right]               (1)

Area A=\frac{\pi}{2}\left(\frac{x}{2}\right)^{2}+x y

=\frac{\pi x^{2}}{8}+\frac{x}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right]       ..... from (1)

\begin{aligned} &A=\frac{\pi x^{2}}{8}+\frac{10 x}{2}-\frac{x^{2}}{2}\left(1+\frac{\pi}{2}\right) \\ &\frac{d A}{d x}=\frac{\pi x}{4}+\frac{10}{2}-\frac{2 x}{2}\left(1+\frac{\pi}{2}\right) \\ &\frac{d A}{d x}=0 \\ &\frac{\pi x}{4}+\frac{10}{2}-\frac{2 x}{2}\left(1+\frac{\pi}{2}\right)=0 \end{aligned}

\begin{aligned} &x\left[\frac{\pi}{4}-1-\frac{\pi}{2}\right]=-5 \\ &x=\frac{-5}{\left(\frac{-4-\pi}{4}\right)}=\frac{20}{\pi+4} \end{aligned}

Substitute value of x in eqn (1)

\begin{aligned} &y=\frac{1}{2}\left[10-\left(\frac{20}{\pi+4}\right)\left(1+\frac{\pi}{2}\right)\right] \\ &=5-\frac{10(\pi+2)}{\pi+4}=\frac{10}{\pi+4} \end{aligned}

\begin{aligned} &\frac{d^{2} A}{d x^{2}}=\frac{\pi}{4}-\frac{\pi}{2}-1=\frac{\pi-2 \pi-4}{4} \\ &=\frac{-\pi-4}{4}<0 \end{aligned}

Thus the area is max when xx=\frac{20}{\pi +4}, y=\frac{10}{\pi +4}

So, l=\frac{20}{\pi +4}m, b=\frac{10}{\pi +4}m

Posted by

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