Question

If the area bounded by the line $y=mx+2$ , $m>0$ and $x=2y-y^{2}$ is minimum and equals to $A$ , then the value of $3A$ is

Answer 1

Given $y=m x+2 \ldots(i)$
$\&\, x=2 y-y^{2} \ldots(i i)$
Solving simultaneously​ for point of intersection
$\Rightarrow \frac{y-2}{m}=2 y-y^{2}$
$\Rightarrow \frac{y-2}{m}=(y-2)(-y)$
$\Rightarrow y-2=0$ or $\frac{1}{m}=-y$
$\Rightarrow y=2,-\frac{1}{m}$

$\therefore$ area bounded by $x=f(y), x=g(y)$ line $y=a$ and $y=b$ is $\int\limits_{a}^{b}|f(y)-g(y)| \cdot d y$
Required area $=\int\limits_{-\frac{1}{m}}^{2}\left[\left(2 y-y^{2}\right)-\left(\frac{y-2}{m}\right)\right] d y$
$=\left(\left|y^{2}-\frac{y^{3}}{3}-\frac{1}{m}\left(\frac{y^{2}}{2}\right)+\frac{2}{m} y\right|\right)_{-\frac{1}{m}}^{2}$
$=\left[4-\frac{8}{3}-\frac{2}{m}+\frac{4}{m}\right]-\left[\frac{1}{m^{2}}+\frac{1}{3 m^{3}}-\frac{1}{2 m^{3}}-\frac{2}{m^{2}}\right]=f(m)$
Now $f(m)=\frac{4}{3}+\frac{2}{m}+\frac{1}{m^{2}}+\frac{1}{6 m^{3}}$ will be minimum when $m \rightarrow \infty$
So, minimum value of $f(m)=\frac{4}{3}$
$A=\frac{4}{3}$
$3 A=4$