Question

A straight line $L$ with negative slope passes through the point $(8,4)$ and cuts the positive co-ordinate axes at points $A$ and $B$. The minimum value of $OA+OB$, as $L$ varies, is (Here $O$ is the origin)

• A. $12+8\sqrt{2}$
• B. $12-8\sqrt{2}$
• C. $12+4\sqrt{2}$
• D. $12-4\sqrt{2}$

Answer 1

The correct option is A $12+8\sqrt{2}$

$m$ is the slope of line $L$, $m<0$
Equation of line $L$ is $(y-4)=m(x-8)$
Therefore, intercepts are $OA=8-\frac{4}{m}$ and $OB=4-8m$

Let $f=OA+OB$
$f\left(m\right)=12-\frac{4}{m}-8m$
$f^{\prime }\left(m\right)=\frac{4}{m^2}-8$
$f^{\prime }\left(m\right)=0\Rightarrow m=\frac{-1}{\sqrt{2}}[\because m<0]$
$f^{\prime \prime }\left(\frac{-1}{\sqrt{2}}\right)=\frac{-8}{m^3}>0$ as $m<0$
So minima will lie at $m=\frac{-1}{\sqrt{2}}$
$f\left(\frac{-1}{\sqrt{2}}\right)=12+4\sqrt{2}+\frac{8}{\sqrt{2}}$
Minimum value $=12+8\sqrt{2}$