Determining maximum and minimum of the function

Question

Determine the minimum and maximum of the function $f(x,y) = xy$ in the annulus $A = \{(x,y) \mid 1 \leq x^2 + y^2 \leq 4\}$.

I'm wondering if in cases like this, where we have two conditions for the constrained extremum, we proceed like usual with the constrained extremum and then add these two conditions together like $F(x,y,g_1,g_2) = xy + g_1(x^2+y^2-1) + g_2(x^2+y^2-4)$, or is there an easier way, as this leads to more complicated calculations?

Answer 1

$f(x,y) \leq \frac 1 2 {(x^{2}+y^{2})}=2$ and the value $2$ is attained when $x=\sqrt 2=y$. So the maximum value is $2$. I will let you change the sign of $x$ to conclude that the minimum value is $-2$, attained when $x =-\sqrt 2, y=\sqrt 2$.

Answer 2

Note that for this function, at maximum and minimum, we would want the magnitude to be large, hence at optimal value, we will have $x^2+y^2=4$, or using polar coordinate $r^2=2^2$ and we have $$f(r, \theta)=r^2\cos\theta \sin \theta=\frac{r^2}2\sin2\theta=2\sin2\theta$$

The maximum happens when $\sin 2 \theta=1$ and the minimum happens when $\sin 2 \theta = -1$.