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.
 A: 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.
I don't know why you would need to express your absolute extrema to the nearest hundredths (until this problem is part of a group in a textbook for which some of the answers are irrational numbers).  This is a pretty simple function; since it increases from "left-to-right" and "top-to-bottom" across the given region, it isn't surprising that the absolute extrema lie at the lower left and upper right vertices.  You're sure this is the correct function and region description?  If it is, I think your results are already correct.  (I even tried graphing this.)
