I am asking this question because my professor tells me my procedure is not enough to prove my answer. I disagree. That being said, consider the following:
The temperature of a rectangular plate bounded by the lines $$x = ±1,y = ±1$$ is given by $$T =2x^2−3y^2−2x+10$$ Find the hottest and coldest points of the plate.
My solution was very simple: $$t_x=4x-2=0$$$$\Rightarrow x=\frac12$$
$$t_y=-6y=0$$$$\Rightarrow y=0$$ Now, since $t_{xx}>0, x=\frac12$ is a minimum, and $t_{yy}<0, y=0$ is a maximum. Therefore, $P_1(\frac12,0)$ will not be a maximum nor a minimum. Hence, I only had to check the correct combinations. In other words, for example, since $x=\frac12$ is a minimum $y$ has to be as large as possible if we want to find the minimum. Likewise, at $y=0$, $x$ has to be as large as possible to maximize the temperature. Intuitively, these points have to occur at the boundary (because of the shape of the graph of $T$). Therefore, $P_2(\frac12,±1)$ is the coldest point, and $P_3(-1,0)$ is the hottest.
My question is, is my procedure enough to prove those points are the hottest and coldest, in other words, is it true in general for quadratic equations?
I know this graph has no saddle points because it lacks $xy$ term.
