Find the global maximum and global minimum of $f : \mathbb{R^2} \rightarrow \mathbb{R}$ given by $f(x,y) = x^2 + (y-1)^2$ on the set $E = \{(x,y) \in \mathbb{R^2} : x^2+y^2\leq4 \text{ and } y\geq 0\}$
My attempt:
$f_x(x,y) = 2x$
$f_y(x,y) = 2(y-1)$
By setting the above partial derivatives equal to $0$ we obtain the critical point at $(0,1)$.
Now, this is where I am stuck. I know there is (from plotting) a global minimum at $(0,1)$ and global maximum at $(-2,0)$ and $(2,0)$, but I don't know how to find them mathematically. Could someone please help/show how to solve?
$\endgroup$ 02 Answers
$\begingroup$The domain is compact, so the function has global maxima and minima, that are not necessarily taken at critical points, but could also be taken at the frontier.
Over $x^2+y^2=4$, the function becomes $g(y)=4-y^2+(y-1)^2=5-2y$, which is decreasing over $[0,2]$. At $(2,0)$ and $(-2,0)$ we have $f(2,0)=f(-2,0)=5$; at $(0,2)$ we have $f(0,2)=1$.
At the interior critical point we have $f(0,1)=0$.
So the global minimum is $0$ and the global maximum is $5$.
$\endgroup$ $\begingroup$Remember no square of real number is negative$$ \begin{aligned} x^{2}+(y-1)^{2}&\geq0 \end{aligned} $$
Quick check confirmed that $(0,1)$ satisfy the constraints
$$ \begin{aligned} x^{2}+(y-1)^{2}&=x^{2}+y^{2}-2y+1\\ &\leq4-2y+1\\ &\leq5 \end{aligned} $$
Quick check confirmed $(\pm 2,0)$ satisfy the constraints
$\endgroup$
