How can I find the vertices of a triangle by optimization?

Question

Here is the information provided, and the hypotenuse length is minimum. How can I find the vertices of a triangle by optimization? Thanks.

Answer 1

Let's solve a similar but different problem so I'm not doing your homework for you. Let's change the point to $(8,27)$.

The line going through the point $(8,27)$ has equation $y-27=m(x-8)$.

We can find the vertices by solving for when $x=0$ and when $y=0$, this gives $y=27-8m$ and $x=8-\frac{27}m$, respectively.

The square of the hypotenuse is therefore $(27-8m)^2 + (8-\frac{27}m)^2=64m^2-432m+793-\frac{432}m+\frac{729}{m^2}$.

Minimizing the square of the hypotenuse will also minimize the hypotenuse, so we take the derivative $128m-432+\frac{432}{m^2}-\frac{1458}{m^3}$, multiply by $\frac{m^3}2$, and find critical points. $64m^4-216m^3+216m-729=0$.

This group factors as $(8m-27)(8m^3+27)=(8m-27)(2m+3)(4m^2-6m+9)$. The last term has no zeros, and the slope must be negative, so the critical point is at $m=-\frac32$.

The equation of the line is therefore $y-27=-\frac32(x-8)$, and the vertices, as solved for previously, are at $y=39$ and $x=26$.