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Q.
The minimum value of the function $y=\frac{a^{2}}{x}+\frac{b^{2}}{a-x}, a > 0, b > 0$ in $(0, a)$ is :
Application of Derivatives
Solution:
Given, $y=\frac{a^{2}}{x}+\frac{b^{2}}{a-x}$
$\Rightarrow \frac{dy}{dx}=-\frac{a^{2}}{x^{2}}+\frac{b^{2}}{\left(a-x\right)^{2}}=0 \Rightarrow x=\frac{a^{2}}{a\pm b},$
out of which only one $x=\frac{a^{2}}{a+b}$ is in $\left(0, a\right).$
Also $\frac{d^{2}y}{dx^{2}}=\frac{2a^{2}}{x^{3}}+\frac{2b^{2}}{\left(a-x\right)^{3}} > 0$ in $\left(0, a\right)$
$\therefore $ Minimum value attained is
$a+b+\frac{a+b}{ab}b^{2}=\frac{1}{a}\left(a+b\right)^{2}$