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

Question

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

If we take derivative, then I am not getting anything, please guide how to proceed in such problems will be of great help. Thanks.

Answer 1

Differentiation + some work: $$ \begin{align} y' &= -2\left(x+\sqrt{x^2+b^2}\right) +2(a-x)\left(1+\frac{x}{\sqrt{x^2+b^2}}\right) \\ &= -2\sqrt{x^2+b^2}\left(\frac{x+\sqrt{x^2+b^2}}{\sqrt{x^2+b^2}}\right) +2(a-x)\left(\frac{\sqrt{x^2+b^2}+x}{\sqrt{x^2+b^2}}\right) \\ &=0 \end{align} $$ this implies: $$ \sqrt{x^2+b^2}=(a-x) $$ and you can square both sides and work out the equation.

Answer 2

Let $\sqrt{x^2+b^2}+x=z.$ Then $\displaystyle \sqrt{x^2+b^2}-x=\frac{b^2}{z}.$

So $$y=z \bigg(2a-z+\frac{b^2}{z}\bigg)=2az-z^2+b^2=a^2+b^2-(z-b)^2\leq a^2+b^2$$