Question

If $x$ is real then the maximum value of $y=2(a-x)(x+\sqrt{x^2+b^2})$ is _____________.

• A. $a^2-b^2$
• B. $a^2+b^2$
• C. $\sqrt{a^2-b^2}$
• D. $a\sqrt{a^2-b^2}$
• E. $Noneofthese$

Answer 1

The correct option is B $a^2+b^2$

Given $y=2(a-x)(x+\sqrt{x^2+b^2})$ (1)

Put $x+\sqrt{x^2+b^2}=t$

$\therefore \sqrt{x^2+b^2}=t-x$

Squaring both sides, we get,

$x^2+b^2={(t-x)}^2$

$\therefore x^2+b^2=t^2-2tx+x^2$

$\therefore b^2=t^2-2tx$

$\therefore 2tx=t^2-b^2$

$\therefore x=\frac{t^2-b^2}{2t}$

Thus, from equation (1),

$y=2(a-\frac{t^2-b^2}{2t})t$

$\therefore y=2(at-\frac{t^2-b^2}{2})$

$\therefore y=2at-(t^2-b^2)$

$\therefore y=2at-t^2+b^2$ (2)

For y to be maximum, differentiate y w.r.t. x and equate to zero

$\therefore \frac{dy}{dx}=\frac{d}{dx}(2at-t^2+b^2)=0$

$\therefore 2a-2t=0$

$\therefore 2a=2t$

$\therefore t=a$

Put this value in equation (2), we get,

$y=2a\left(a\right)-{\left(a\right)}^2+b^2$

$\therefore y=2a^2-a^2+b^2$

$\therefore y=a^2+b^2$