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Please Solve RD Sharma Class 12 Chapter 17 Maxima and Minima Exercise Case Study Based Questions question 1 subquestion (iv) Maths Textbook Solution.

Answers (1)

Answer: $x= 0$

Hint: use concept of maxima and minima.

Solution:

Total Area = =Area of rectangle+2× Area of semicircle.

$=A+2 \times \frac{1}{2} \pi r^{2} \\$

$\begin{aligned} & &=A+\pi r^{2} \end{aligned}$

Putting

$\mathrm{r}=\frac{\mathrm{y}}{2} \\$

$=\mathrm{A}+\pi(\mathrm{y} / 2)^{2} \\$

$\begin{aligned} & &=\mathrm{A}+\frac{\pi \mathrm{y}^{2}}{4} \end{aligned}$

Putting A value,

$=\frac{2}{\pi}\left(100 \mathrm{x}-\mathrm{x}^{2}\right)+\frac{\pi \mathrm{y}^{2}}{4}$

Putting

$\mathrm{y}=\frac{1}{\pi}(200-2 \mathrm{x}) \\$

$\begin{aligned} & &=\frac{2}{\pi}\left(100 \mathrm{x}-\mathrm{x}^{2}\right)+\frac{\pi}{4}\left(\frac{1}{\pi}(200-2 \mathrm{x})\right)^{2} \end{aligned}$

$=\frac{2}{\pi}\left(100 \mathrm{x}-\mathrm{x}^{2}\right)+\frac{\pi}{4} \times \frac{1}{\pi^{2}}(200-2 \mathrm{x})^{2} \\$

$\begin{aligned} & &=\frac{2}{\pi}\left(100 \mathrm{x}-\mathrm{x}^{2}\right)+\frac{\pi}{4} \times \frac{1}{\pi^{2}} 2^{2} \times(100-\mathrm{x})^{2} \end{aligned}$

$=\frac{200 \mathrm{x}}{\pi}-\frac{2 \mathrm{x}^{2}}{\pi}+\frac{100^{2}}{\pi}+\frac{\mathrm{x}^{2}}{\pi}-\frac{200 \mathrm{x}}{\pi} \\$

$\begin{aligned} & &=\frac{-\mathrm{x}^{2}}{\pi}+\frac{100^{2}}{\pi} \end{aligned}$

Total Area $=\frac{-x^{2}}{\pi}+\frac{100^{2}}{\pi}$

Let Total Area =Z

$z=\frac{-x^{2}}{\pi}+\frac{10000}{\pi} \$

$\begin{aligned} &\ &\frac{d z}{d x}=\frac{-2 x}{\pi} \end{aligned}$

$\frac{d z}{d x}=0 \\$

$\frac{-2 x}{\pi}=0 \\$

$\begin{aligned} & &x=0 \end{aligned}$

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