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Provide Solution for RD Sharma Class 12 Chapter 17 Maxima and Minima Exercise Case Study Based Questions question 8 subquestion (iii)

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Answer:  \frac{8\left(200-x^{2}\right)}{\sqrt{400-x^{2}}}

Hint: use the concept of maxima and minima.


\begin{aligned} \mathrm{x}^{2}+\mathrm{y}^{2} &=400 \\ \ \end{aligned}

{y}^{2} =400-\mathrm{x}^{2} \\

\mathrm{~A}=1 \times \mathrm{b} \\

\mathrm{A}=2 \mathrm{x} \times 2 \sqrt{400-\mathrm{x}^{2}} \\

= 4 \mathrm{x} \sqrt{400-\mathrm{x}^{2}}

\begin{aligned} &\frac{d A}{d x}=4 \sqrt{400-4 x^{2}}+4\left(\frac{\left(-2 x^{2}\right)}{2 \sqrt{400-x^{2}}}\right) \\ \end{aligned}

\frac{d A}{d x}=\frac{8\left(200-x^{2}\right)}{\sqrt{40 x-x^{2}}}

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