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  • Need solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise Very short answer type question 7

Need solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise Very short answer type question  7

Answers (1)

Answer:  2 \sqrt{a b}

Given: f(x)=a x+\frac{b}{x}

Solution:

        \begin{aligned} &f(x)=a x+\frac{b}{x} \\\\ &f^{\prime}(x)=a-\frac{b}{x^{2}} \end{aligned}

For a local maxima or a local minima we must have

        \begin{aligned} &f^{\prime}(x)=0 \\\\ &a-\frac{b}{x^{2}}=0 \\\\ &x^{2}=\frac{b}{a} \end{aligned}

        \begin{aligned} &x=\sqrt{\frac{b}{a}},-\sqrt{\frac{b}{a}} \\\\ &\text { But } x>0 \\\\ &x=\sqrt{\frac{b}{a}} \end{aligned}

Now

        \begin{aligned} &f^{\prime \prime}(x)=\frac{2 b}{x^{3}} \\\\ &\text { At } x=\sqrt{\frac{b}{a}} \end{aligned}

        f^{\prime \prime}\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b}{\left(\sqrt{\frac{b}{a}}\right)^{3}}=\frac{2 a^{\frac{3}{2}}}{b^{\frac{1}{2}}}>0 \ldots \ldots . . .[a>0 \text { and } b>0]

So,  x=\sqrt{\frac{b}{a}} is a point of local minimum.

Hence the least value is

        f\left(\sqrt{\frac{b}{a}}\right)=a \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}=\sqrt{a b}+\sqrt{a b}=2 \sqrt{a b}

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