Science:Math Exam Resources/Courses/MATH105/April 2015/Question 03 (b)/Solution 1

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 $(x,y)=(1,-1)$.