Absolute Extrema Multivariable Function

Find the absolute extrema of the function f(x,y)= x^2 + y^2 -6y on a closed region D bounded by the graphs of y=4-x^2 and y=x+2.

First, find the critical points inside D:
f_x = 2x and f_y = 2y - 6.

Setting these equal to 0 yields the critical point (x, y) = (0, 3), which is inside D.
Note that f(0, 3) = -9.
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Next, check the boundaries of D.
Note that y = 4 - x^2 and y = x + 2 intersect at x = -2 and x = 1 (easy algebra).

Plot:
http://www.wolframalpha.com/input/?i=plo…

(i) y = 4 - x^2 with x in [-2, 1]:
Rewrite as x^2 = 4 - y with y in [0, 4] (by the graph of D).

So, g(y) := f(±√(4 - y)) = (4 - y) + y^2 - 6y = y^2 - 7y + 4.

Setting g' = 0:
g' = 2y - 7 ==> y = 7/2. (so x = ±1/√2)
(Also remember the endpoints at (x, y) = (-2, 0) and (1, 3).)

(ii) y = x + 2 with x in [-2, 1]:
==> g(x) = f(x, x+2) = 2x^2 - 2x - 8.

Setting g' = 0:
g' = 4x - 2 ==> x = 1/2.
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Testing all of the critical points (endpoints included):
f(0, 3) = -9 <---Minimum
f(-2, 0) = 4 <---Maximum
f(1, 3) = -8
f(±1/√2, 7/2) = -33/4
f(1/2, 5/2) = -17/2

I hope this helps!