I'm solving problems on extrema of multivariable functions and Lagrange multipliers. The problem is:
"Through a fixed point M inside a fixed angle draw a line, that cuts a triangle with minimal area from the angle".
I know the way of solving this problem using a function of one variable, i.e. writing equation of the line $y=k(x-a)+b$ and then minimizing area of the triangle this line cuts from the given angle using $k$ as a variable. But we need to minimize a function of at least two variables. Maybe we could denote length of line segment from the given point $M$ to the intersection with one side of the angle as $x$, and to the intersection with another side of the angle as $y$? But then how do we write the area of the triangle as a function of $x,y$? Or maybe there is another way of solving this problem using multivariable function minimization?
We can solve this problem without using multivariable function minimization.
Indeed, let $\angle BAC$ be our angle, where $B$ and $C$ be chose such that $M$ is a midpoint of $BC$.
Easy to see that $\Delta ABC$ has a minimal area.
Indeed, let $B_1$ and $C_1$ be placed on rays $AB$ and $AC$ respectively such that $M\in B_1C_1$.
Now, let $B_1M>C_1M$, $B_2\in B_1M$ such that $B_2M=C_1M$.
Hence, $$\Delta MBB_2\cong\Delta MCC_1,$$ which says that $$S_{\Delta BB_1M}>S_{\Delta CC_1M}$$ and from here $$S_{\Delta AB_1C_1}=S_{ABMC_1}+S_{BB_1M}>S_{ABMC_1}+S_{CC_1M}=S_{\Delta ABC}.$$
The rest is easy:
Let $D$ be a point on $AM$ such that $M$ is a midpoint of $AD$.
Now, we can get the parallelogram $ABDC$ and we are done!