Question

The maximum value of the expression $y=2(a-x)(x+\sqrt{x^2+b^2})$ is

• A. $2a^2+b^2$
• B. $a^2+2b^2$
• C. $a^2+b^2$
• D. $|2a^2-b^2|$

Answer 1

The correct option is C

$a^2+b^2$

Let $t=x+\sqrt{x^2+b^2}$
$\Rightarrow (\sqrt{x^2+b^2}-x)(\sqrt{x^2+b^2}+x)=b^2\Rightarrow \sqrt{x^2+b^2}-x=\frac{b^2}{t}\Rightarrow 2x=t-\frac{b^2}{t}\Rightarrow x=\frac{1}{2}\left(\frac{t^2-b^2}{t}\right)$
Now let y = 2(a - x)t
$=2(a-\left(\frac{t^2-b^2}{2t}\right))t=2at-t^2+b^2=b^2-t^2+2at+a^2-a^2=a^2+b^2-(t-a{)}^2\therefore y\le a^2+b^2$