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

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

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

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 $x={\tfrac {1}{2}}$. Thus, another critical point of $f$ is $({\tfrac {1}{2}},0)$.

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