A straight line $L$ with negative slope passes through the point $(8,2)$ and cuts the positive axes at points $P$ and $Q$, then find minimum area of $\triangle{OPQ}$ ($O$ is origin).
Book hint:
They wrote area is minimum when $(8,2)$ is the mid point of $PQ$.
I don't understand this, please help.
On
the line through $P(8;2)$ is given by $$y=m(x-8)+2$$ and thus the Point on the $x$ axes is $$x=\frac{8m-2}{m}$$ and on the $y$-axes $y=-8m+2$ and the area of the triangle $A=\frac{1}{2}\frac{(2-8m)(8m-2)}{m}$ differentiating with respect to $m$ we get $$A'(m)=\frac{-2(4m-1)(4m+1)}{m^2}$$ solving the equation $A'(m)=0$ we get $m=\frac{-1}{4}$ and $A''(-\frac{1}{4})=256>0$ so we get a Minimum.
Hint: fold it!
The area of the $POQ$ triangle equals $16$ (i.e. the area of the rectangle with vertices at $O,(8,0),(8,2),(0,2)$) plus the area of the region given by the light orange and rose triangles.
This region always encloses the previous rectangle, hence has an area $\geq 16$. Equality is attained only when $(8,2)$ is the midpoint of $PQ$, leading to $[POQ]=32$.