If $a, b0$, then minimum value of $y=\frac{b^2}{a-x}+\frac{a^2}{x}, 0

Question

If $a, b>0$, then minimum value of $y=\frac{b^2}{a-x}+\frac{a^2}{x}, 0 < x < a$ is

Aicharya · Accepted Answer

Given, $y=\frac{b^2}{a-x}+\frac{a^2}{x}$ $	herefore \quad \frac{d y}{d x}=\frac{-b^2(-1)}{(a-x)^2}+\left(\frac{-a^2}{x^2}ight) \Rightarrow \frac{d y}{d x}=\frac{b^2}{(a-x)^2}-\frac{a^2}{x^2}$ $\because \quad \frac{d y}{d x}=0, x=\frac{a^2}{a \pm b}$ $\frac{d^2 y}{d x^2}=\frac{2 a^2}{x^3}+\frac{2 b^2}{(a-x)^3}$ So, $\left.\frac{d^2 y}{d x^2}ight|_{x=\frac{a^2}{a+b}}>0$ Therefore, $y$ is minimum at $x=\frac{a^2}{a+b}$ $y_{\min }=\frac{b^2}{a-\left(\frac{a^2}{a+b}ight)}+\frac{a^2}{(a)^2}(a+b)$ $=\frac{b^2(a+b)}{a^2+a b-a^2}+(a+b)=(a+b)\left[\frac{b}{a}+1ight]$ $=\frac{(a+b)^2}{a}$

If $a, b>0$, then minimum value of $y=\frac{b^2}{a-x}+\frac{a^2}{x}, 0 < x < a$ is

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