Question

The value of $k>0$ so that the area of the bounded region enclosed between the parabolas $y=x-k{x}^2$ and $y=\frac{x^2}{k}$ is maximum, is

Answer 1

Clearly, the given curves intersect at
$x=0,\frac{k}{k^2+1}$
$\therefore$ Required area $=\underset{0}{\overset{\frac{k}{k^2+1}}{\int }}(x-k{x}^2-\frac{x^2}{k})dx=\frac{1}{6{(k+\frac{1}{k})}^2}$

Now, for maximum area, ${(k+\frac{1}{k})}^2$
is to be minimum.
$\Rightarrow k+\frac{1}{k}=2$
$\Rightarrow k=1$