I am tackling this problem below:
A triangle is formed by the three lines $$\begin{align} y &=10-2x \\ y &= mx \\ y &=-\frac{x}{m} \end{align}$$ where $m>\frac{1}{2}$. Find the value of $m$ for which the area of the triangle is a minimum.
My thoughts are these:
I tried to draw the three graphs nothe rectangular coordinate, and since $m> \frac{1}{2}$, $-\frac{1}{m}>2$, but after that how can I decide the area of triangle formed?
On
Although an answer is given, I want to add my answer too, which is less tedious.
It is clear that the lines $y=mx$ and $y=\frac{-x}{m}$ are perpendicular and their point of intersection is the origin.
The other side is $2x+y-10=0$
Since we need to calculate the area of the triangle, we can always use $\frac{h×b}{2}$ instead of coordinates. Perpendicular distance on the line from origin is $\frac{10}{\sqrt{5}}=2\sqrt{5}$.
Now we use that our triangle is right triangle. It is trivial from here. The minimum area is $(2\sqrt{5})^2=20$.
Let $\ell_1$ be the line with equation $y = 10-2x$, let $\ell_2$ be the line with equation $y = mx$, and let $\ell_3$ be the line with equation $y = \frac{-x}{m}$. Then the lines $\ell_1$ and $\ell_2$ intersect at $$P = \left( \frac{10}{m+2}, \frac{10m}{m+2} \right),$$ and the lines the lines $\ell_1$ and $\ell_3$ intersect at $$ Q = \left( \frac{-10m}{1-2m}, \frac{10}{1-2m} \right). $$ If $O = (0,0)$, then the line segments $\overline{OP}$ and $\overline{OQ}$ are orthogonal, which means that the area of the triangle is given by $$ A = \frac{1}{2} \left|OP\right| \left| OQ\right| = \frac{1}{2} \left( \frac{10 \sqrt{1+m^2}}{m+2} \right) \left( \frac{10 \sqrt{1+m^2}}{1-2m} \right) = \frac{50(1+m^2)}{(m+2)(1-2m)}. $$