Question

The minimum value of the function $y=\frac{a^2}{x}+\frac{b^2}{a-x},a>0,b>0$ in $(0,a)$ is :

• A. $a+b$
• B. $\frac{1}{a+b}$
• C. $\frac{1}{a}{\left(a+b\right)}^2$
• D. $\frac{1}{a^2}\left(a+b\right)$

Answer 1

Answer: (C)

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 $\frac{d^{2}y}{d{x}^2}=\frac{2a^2}{x^3}+\frac{2b^2}{{\left(a-x\right)}^3}>0$ in $\left(0,a\right)$
$\therefore$ Minimum value attained is
$a+b+\frac{a+b}{ab}{b}^2=\frac{1}{a}{\left(a+b\right)}^2$