Science:Math Exam Resources/Courses/MATH105/April 2015/Question 03 (b)

MATH105 April 2015

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Question 03 (b)
Find the maximum and minimum values of the function $f(x,y)=(x-1)^{2}+(y+1)^{2}$ over the region $R=\left\{(x,y):x^{2}+y^{2}\leq 4\right\}$.

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

The maximum and the minimum values of $f$ can be attained only at critical points (i.e., where all partial derivatives of $f$ vanish) or on the boundary of $R$. Compute the critical points of the function and the value the function takes at those points. Then, compare these to the extrema on the boundary, which were computed in Question 3(a), to determine the global maximum and minimum of $f$ over $R$.

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. The minimum or maximum of $f$ can only occur at its critical points (points where both $f_{x},f_{y}$ are zero) or on the boundary $x^{2}+y^{2}=4$ of $R$. We will find the critical points (the extrema on the boundary were found in Question 3(a)) and then compare all these points to determine the global maximum and minimum of $f$ on $R$. Since $\left(f_{x},f_{y}\right)=\left(2(x-1),2(y+1)\right),$ there is a single critical point in $R$ where $x-1=0,y+1=0$; namely, $(x,y)=(1,-1)$. At this point, $f(x,y)=0$. Using the solution of Question 3(a), the function has a maximum on the boundary at $(x,y)=(-{\sqrt {2}},{\sqrt {2}})$, where $f(x,y)=6+4{\sqrt {2}}$, and a minimum on the boundary at $(x,y)=({\sqrt {2}},-{\sqrt {2}})$, where $f(x,y)=6-4{\sqrt {2}}$. Comparing all three values, we see that $\max _{(x,y)\in R}f(x,y)={\color {blue}6+4{\sqrt {2}}}$, attained at $(x,y)=(-{\sqrt {2}},{\sqrt {2}})$, and $\min _{(x,y)\in R}f(x,y)={\color {blue}0}$, attained at $(x,y)=(1,-1)$.