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

Answer: 4

Given $y=mx+2\dots (i)$
$\&x=2y-y^2\dots (ii)$
Solving simultaneously​ for point of intersection
$\Rightarrow \frac{y-2}{m}=2y-y^2$
$\Rightarrow \frac{y-2}{m}=(y-2)(-y)$
$\Rightarrow y-2=0$ 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 $\underset{a}{\overset{b}{\int }}|f(y)-g(y)|\cdot dy$
Required area $=\underset{-\frac{1}{m}}{\overset{2}{\int }}\left[\left(2y-y^2\right)-\left(\frac{y-2}{m}\right)\right]dy$
$={\left(\left|y^2-\frac{y^3}{3}-\frac{1}{m}\left(\frac{y^2}{2}\right)+\frac{2}{m}y\right|\right)}_{-\frac{1}{m}}^2$
$=\left[4-\frac{8}{3}-\frac{2}{m}+\frac{4}{m}\right]-\left[\frac{1}{m^2}+\frac{1}{3m^3}-\frac{1}{2m^3}-\frac{2}{m^2}\right]=f(m)$
Now $f(m)=\frac{4}{3}+\frac{2}{m}+\frac{1}{m^2}+\frac{1}{6m^3}$ will be minimum when $m\to \infty$
So, minimum value of $f(m)=\frac{4}{3}$
$A=\frac{4}{3}$
$3A=4$