# 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.

• 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. – colormegone 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. – colormegone Apr 18 '13 at 4:39