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Explain solution RD Sharma class 12 Chapter 17 Maxmima and Minima Exercise Case Study Questions question 7 subquestion (iv)

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Answer:   \mathrm{x}=\frac{320}{\sqrt{3}}

Hint: use the concept of maxima and minima.


\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\left[\frac{25600}{\left(x^{2}-25600\right)^{5 / 2}}\right] \\ & \end{aligned}

\frac{d y}{d x}=2-\frac{1}{2 \sqrt{\left(x^{2}-100\right)^{2}}} \cdot 2 x=0

For max or min

\begin{aligned} &\frac{2}{x}=\frac{1}{\sqrt{x^{2}-(160)^{2}}} \rightarrow(1) \\ & \end{aligned}

\frac{d^{2} y}{d x^{2}}=-\left[\frac{2}{x}-x^{2} \cdot \frac{8}{x^{3}}\right] \text { by }(1)=\frac{6}{x}

y is minimum when   \mathrm{x}=\frac{320}{\sqrt{3}}


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