Question

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

• A. $\frac{a+b}{a}$
• B. $\frac{ab}{a+b}$
• C. $\frac{1}{a}+\frac{1}{b}$
• D. none of these

Answer 1

Answer: (D)

none of these

Given, $y=\frac{b^2}{a-x}+\frac{a^2}{x}$

$\therefore \frac{dy}{dx}=\frac{-b^2}{{(a-x)}^2}(-1)-\frac{a^2}{x^2}=0$

$\Rightarrow x=\frac{a^2}{a\pm b}$

$\frac{d^{2}y}{d{x}^2}=\frac{2a^2}{x^3}+\frac{2b^2}{{(a-x)}^3}$

so $\frac{d^{2}y}{d{x}^2}{|}_{x=\frac{a^2}{a+b}}>0$

Therefore $y$ is minimum at $x=\frac{a^2}{a+b}$ and

$y_{min}=\frac{b^2(a+b)}{ab}+\frac{a^2(a+b)}{a^2}=(a+b)[\frac{b}{a}+1]=\frac{{(a+b)}^2}{a}$

Ans: D