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

MATH100 A December 2023

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Question 25
Let $f(x,y)=2x-2x^{2}-xy^{2}$. Find all the critical points. Your answer(s) should include both $x$- and $y$- values, and be written in the form $(-,-)$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Critical points here are pairs $(a,b)$ that satisfy both ${\frac {\partial f}{\partial x}}(a,b)=0$ and ${\frac {\partial f}{\partial y}}(a,b)=0$.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $y=0$. 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 $f$. We shouldn't forget the second option, $y=0$. In this case, the first equation becomes $2-4x=0$, for which the solution is $x=2$. Thus, another critical point of $f$ is $(2,0)$. The final list of critical points is $(0,-{\sqrt {2}}),(0,{\sqrt {2}}),(2,0)$.