Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 25/Solution 1

We compute the partial derivatives and set them equal to 0:

$\begin{array}{rl}\frac{\partial f}{\partial x}(x,y)& =2-4x-y^2=0\\ \frac{\partial f}{\partial y}(x,y)& =0-0-2xy=0.\end{array}$

We now solve for $x$ and $y$. Let's start with the second equation. If the point $(x,y)$ satisfies $-2xy=0$, then either $x=0$ or $y=0$.

1. Let us take the first option and suppose $x=0$. Then the first equation then becomes $2-y^2=0$, for which the solutions are $y=\pm \sqrt{2}$. Thus, the points $(0,-\sqrt{2})$ and $(0,\sqrt{2})$ are critical points of $f$.
2. We shouldn't forget the second option, $y=0$. In this case, the first equation becomes $2-4x=0$, for which the solution is $x=\frac{1}{2}$. Thus, another critical point of $f$ is $(\frac{1}{2},0)$.

The final list of critical points is $(0,-\sqrt{2}),(0,\sqrt{2}),(\frac{1}{2},0)$.