Get Answers to all your Questions

header-bg qa

Need solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Excercise Case Study Based Questions question 6 subquestion (iii)

Answers (1)

Answer:   \frac{28 \pi}{4+\pi}

Hint: use the concept of maxima and minima.


Side of the Square   =\frac{28-x}{4}

Area of the square   =\left[\frac{28-x}{4}\right]^{2}

Thus Total Area   =\frac{x^{2}}{4 \pi}\left[\frac{28-x}{4}\right]^{2}

\begin{aligned} &\frac{d d}{d x}=\frac{2 x}{4 \pi}+\frac{2}{16}(28-x)(-1) \\ \end{aligned}

\frac{d d}{d x}=\frac{x}{2 \pi}-\frac{28-x}{8}

\begin{aligned} &\frac{\mathrm{dA}}{\mathrm{dx}}=0 \\ & \end{aligned}

\frac{\mathrm{x}}{2 \pi}-\frac{28-\mathrm{x}}{8}=0 \\

4 \mathrm{x}=28 \pi-\pi \mathrm{x} \\

4 \mathrm{x}+\pi \mathrm{x}=28 \pi

\begin{aligned} &\mathrm{x}[4+\pi]=28 \pi \\ & \end{aligned}

\mathrm{x}=\frac{28 \pi}{4+\pi}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support