Question

Find all possible value of $b$ so that area bounded between $y=x-b{x}^2$ and $y=\frac{x^2}{b}$ is maximum.

Answer 1

Given:

First equation: $y=x-b{x}^2$

Second equation: $y=\frac{x^2}{b}$

$\therefore$ First finding the intersecting point:

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

$\Rightarrow x-(b{x}^2+\frac{x^2}{b})=0$

$\Rightarrow x[1-x\left(\frac{1+b^2}{b}\right)]=0$

$\therefore x=0$ or $\frac{b}{1+b^2}$

$\therefore y=0$ or $\frac{b}{(1+b^2{)}^2}$

$\therefore$ Finding the shaded Area, A, with respect to x-axis:

$\therefore A={\int }_0^{\frac{b}{1+b^2}}(x-b{x}^2-\frac{x^2}{b})dx$

$={\int }_0^{\frac{b}{1+b^2}}(x-x^2\left(\frac{1+b^2}{b}\right))dx$

$={[\frac{x^2}{2}-\frac{x^3}{3}(\frac{1+b^2}{b}\left)\right]}_0^{\frac{b}{1+b^2}}$

$=\frac{b^2}{6(1+b^2{)}^2}$

Given: Area, A is maximum

$\therefore \frac{dA}{db}=0$

$\Rightarrow \frac{1}{6}\left[\frac{2b(1+b^2{)}^2-4b^3(1+b^2)}{(1+b^2{)}^4}\right]=0$

$\Rightarrow 2b(1+b^2{)}^2-4b^3(1+b^2)=0$

$\Rightarrow 2b(1+b^2)[1+b^2-2b^2]=0$

$\Rightarrow 2b(1+b^2)(1-b^2)=0$

Only $(1-b^2)=0$

$\therefore b=\pm 1$