Get Answers to all your Questions

header-bg qa

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


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


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support