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.

  • $\begingroup$ 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. $\endgroup$ – 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.

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.)

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  • $\begingroup$ But your function doesn't have r's in it like it should. Shouldn't it be 2rcosθ+r^2sinθ-2? $\endgroup$ – user73064 Apr 18 '13 at 2:46
  • $\begingroup$ What is the radius of the semicircle? $\endgroup$ – colormegone Apr 18 '13 at 2:58
  • $\begingroup$ Ah, you plug in the 1...thanks for the help but I have one more question: I said the critical point of the function is (1,0) but I've been thinking about it and I don't think that's correct; the first partial wrt y is 2y, so y has to equal 0, and the first partial wrt x is 2, so does that make the critical point (2,0)? $\endgroup$ – user73064 Apr 18 '13 at 4:24
  • $\begingroup$ 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. $\endgroup$ – colormegone Apr 18 '13 at 4:39
  • $\begingroup$ Okay, so by factoring the cosines, I got the critical points of the circle occur when 2(cosθ-1)sinθ=0, so when cosθ=1 or when sinθ=0, which is pi and 0 for both, so the only points I test on the circle are (1,0) and (-1,0)? If so then I guess the min and max of the boundary are +/- 8 (not on the circle), which I sort of find strange because the question asked for the answer rounded to the nearest hundredths. $\endgroup$ – user73064 Apr 18 '13 at 5:01

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