3-D Absolute Max/Min over closed&bounded region

Find the absolute max and min values of $f(x,y)=2x+y^2-2$ on the closed and bounded region that lies outside the upper half-circle of $\{(x,y)| x^2+y^2=1\}$, and inside the rectangle given by $[-3,3]\times [0,2]$. This region should look like a solid arch.

I know the critical point of the function is just $f(1,0)=0$, but I am having trouble checking the boundary...I think the max is $f(3,2)=8$ and the min is $f(-3,0)=-8$, but that's without really take the semicircle into account.

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On the semi-circle, you might try polar coordinates to write the function as $2 \cos\theta + \sin^2\theta - 2 = 2 \cos\theta - \cos^2 \theta - 1$ and search for "critical angles" that way. –  RecklessReckoner Apr 17 '13 at 16:29

I would propose that you express $f(x,y)$ in polar coordinates on the semi-circle as I describe in my comment above and then show that the bounds on the function place it between the "corner point extrema" you've already found.
EDIT: Sorry, I must have looked at one of my scribbled notes and thought I was looking at the derivative, instead of the function. Yes, you have on the semi-circle $f(\theta) = 2 \cos\theta - \cos^2\theta - 1$, so $\frac{df}{d\theta} = -2 \sin\theta + 2 \cos\theta \sin\theta = -2 \sin\theta (1 - \cos\theta) = 0$ . The critical points on the semi-circle are thus at $\theta = 0$ and $\theta = \pi$, which give local extrema on the curve (at the points you indicate in your comment), but not for the entire boundary.
You're looking for places where the partial derivatives of the function equal zero. Since $\frac{\partial f}{\partial x} \neq 0$, there isn't any value of x that produces a critical point in the interior of the region. (I guess I should have caught that: there isn't anything special about $( 1 , 0 )$.) That leaves testing the boundary, and then the corner points; that means looking at the semi-circle, the edges at $x = \pm 3$, $y = 0$, $y = 2$, and then the vertices of the rectangle and endpoints of the semi-circle. A lot of this can be done pretty quickly for such a simple function. –  RecklessReckoner Apr 18 '13 at 4:39