Question

If $xϵR$, then prove that the maximum value of $2(a-x)(x+\sqrt{x^2+b^2})$ is $a^2+b^2$.

Answer 1

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

$\Rightarrow$ Let $t=(x+\sqrt{x^2+b^2})$

$\frac{1}{t}=\frac{1}{(x+\sqrt{x^2+b^2})}=\frac{\sqrt{x^2+b^2-x}}{x^2+b^2-x}$

$=\frac{\sqrt{x^2+b^2-x}}{(x^2+b^2)-x^2}$

$\frac{1}{t}=\frac{\sqrt{x^2+b^2-x}}{b^2}$

$b^2+\sqrt{x^2+b^2}-tx$

from eq (A) & eq (1)

$y=(2a+2x)t$

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

$\frac{y}{t}=2a-2t+2\sqrt{x^2+b^2}...\left(B\right)$

putting the value of $\sqrt{x^2+b^2}$ from eq (i) into above

$b^2=t(t-x)-tx$

$=t^2-tx-tx$

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

$t-\frac{b^2}{t}=2x$ Now again replace the value of x from eq (1)

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

$t+\frac{b^2}{t}=2\sqrt{x^2+b^2}...\left(ii\right)$

From eq (B) & eq (ii)

$\frac{y}{t}=2a-2t+t+\frac{b^2}{t}.$

$y\Rightarrow 2at-t^2+b^2$

$\Rightarrow (a^2+b^2)-(a^2-2at+t^2)$

$\Rightarrow (a^2+b^2)-(a-t{)}^2$

when $a-t=0;$ the equation get if maximum value.

therefore $y=2(a-x)(x+\sqrt{x^2+b^2})⩽a^2+b^2$ proved