Absolute maximum and minimum, Multivariable

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October 25th 2009, 04:14 PM
valleyman

Absolute maximum and minimum, Multivariable

Find the absolute maximum and minimum:

f(x,y) = 3x2 + 2y, on set D = {(x,y) | x2 + y2 <= 2}

Within the boundary, there are no critical points:
Partial derivative with respect to y is 2, which is never zero.

If there are no critical points within the boundary, the extrema must occur on the boundary x2 + y2 = 2.

I'm not quite sure how to calculate the extrema on the circle x2 + y2 = 2. Here's what I've got, but I'm not sure how close I am to an answer:

x2 + y2 = 2, thus x2 = 2 - y2
So, substituting this into the original equation, f(x,y) = 3(2-y2) + 2y = 6-3y2+2y
We need to find the extrema on 6-3y2+2y, so we'll find the critical points:

-6y+2=0 when y = 1/3

I don't know what to do at this point, or if I've even done the right thing up until this point. I know there's another absolute max/min problem that was posted just before this one, but I know how to work such a problem. This one, on the other hand, is confusing me.
October 25th 2009, 04:43 PM
Bruno J.

$y=1/3$ so $x=\pm \sqrt{2-y^2} = \pm \sqrt{17}/3$ . Now we test the two points; we find that $f$ has equal value at both, with $f(\pm\sqrt{17}/3,1/3)=19/3$ .

Another way would have been to set $x=\sqrt{2}\cos \theta,\ y=\sqrt{2}\sin \theta$ . Then the function along the boudary is $F(\theta)=6\cos^2\theta+2\sqrt{2}\sin \theta$ , and $\frac{dF}{d\theta}=-12\cos\theta\sin\theta + 2\sqrt{2}\cos\theta$ $=2\cos\theta(\sqrt{2}-6\sin\theta)=\sqrt{2}x(\sqrt{2}-3\sqrt{2}y)=2x(1-3y)$ , so $F$ has extrema when $x=0,\ y=\pm\sqrt{2}$ or when $y=1/3,\ x=\pm \sqrt{17}/3$ .
October 25th 2009, 04:48 PM
Bruno J.

Welcome to the forum, by the way! (Hi)
October 25th 2009, 05:26 PM
valleyman

Quote:

Originally Posted by Bruno J.

$y=1/3$ so $x=\pm \sqrt{2-y^2} = \pm \sqrt{17}/3$ . Now we test the two points; we find that $f$ has equal value at both, with $f(\pm\sqrt{17}/3,1/3)=19/3$ .

Another way would have been to set $x=\sqrt{2}\cos \theta,\ y=\sqrt{2}\sin \theta$ . Then the function along the boudary is $F(\theta)=6\cos^2\theta+2\sqrt{2}\sin \theta$ , and $\frac{dF}{d\theta}=-12\cos\theta\sin\theta + 2\sqrt{2}\cos\theta$ $=2\cos\theta(\sqrt{2}-6\sin\theta)=\sqrt{2}x(\sqrt{2}-3\sqrt{2}y)=2x(1-3y)$ , so $F$ has extrema when $x=0,\ y=\pm\sqrt{2}$ or when $y=1/3,\ x=\pm \sqrt{17}/3$ .

I understand how you made it to the first conclusion: We have critical points at [plus/minus] sqrt(17)/3, 1/3

And F is 19/3 at these points.

I actually made it to this point before and was confused by the fact that I only have one extrema with a value of 19/3, and I need a minimum and a maximum. I see that you made the trig substitution, but I don't understand why it is necessary. Obviously, it gives us another extrema, but why can't we find both extrema with the other calculation? Aren't we essentially doing the same thing with each calculation?

Thank you for the quick reply!
October 25th 2009, 06:04 PM
Bruno J.

It's because you have not defined the domain of the function properly. You searched for extrema of $6-3y^2+2y$ in all of $\mathbb{R}$ ... but let's see:

When you say $f(x,y)=6-3y^2+2y$ on $D=\{(x,y) \in \mathbb{R}^2\ : \ x^2+y^2=2\}$ , then in fact the only values of $y$ in the domain $D$ are in the range $[-\sqrt{2}, \sqrt{2}]$ . You did not explicitly state that. So you would have needed to test those boundary values of $y$ also to see if they were extrema of $f(x,y)$ on $D$ , which they are.

The trigonometric approach allows you to get rid of those pesky issues; the angular parametrization is a much more natural one for the circle than the two-piece parametrization $(x, \sqrt{2-x^2})$ , $(x, -\sqrt{2-x^2})$ where you need to be much more careful.

Hope that helps!
October 25th 2009, 06:42 PM
valleyman

Thanks much, it makes sense now!
October 25th 2009, 07:50 PM
valleyman

So I'm having some more trouble with another similar problem:

Find absolute max/min of f(x,y) = (x-y)/(x2+y2) on D = {x,y | 1 <= x2+y2 <= 2}

I've found the partial derivatives:

fx = (-x2+y2+2xy)/(x2+y2)2
fy = (-x2+y2-2xy)/(x2+y2)2

In order to find the critical points, we set these equations equal to zero, and thus, to each other. With them equated, we are able to eliminate the denominators, -x2, and y2 to obtain:

2xy = -2xy = 0, so 4xy = 0

At this point, we see that when either x OR y is zero, fx and fy are zero. How can we obtain critical points from this?

And when I solve for the extrema on the inner boundary (the inner circle):
x2 + y2 = 1
y = $\pm$ sqrt(1-x2),
so, substituting back into the original equation, f(x,y) = (x-( $\pm$ sqrt(1-x2))/(x2+1-x2) = x - ( $\pm$ sqrt(1-x2))

And we have two equations (due to the plus-minus):
x-sqrt(1-x2) and x+sqrt(1-x2)

We need to find the extrema on the inner circle, but I'm not sure how to deal with the fact that instead of one equation, we now have two equations. I can find that, on x + sqrt(1-x2), we have a maximum of 1.414 at x = 0.7071. However, if we plug x = 0.7071 into x-sqrt(1-x2), we find that it is equal to 0. These two equations represent the same 'reinterpretation' of the original equation, so how can we get two different values? And how do we deal with this?

I'm also a little freaked out by the fact that, in order to find these values, I'm using a calculator. Our professor doesn't assign work that requires a calculator. In fact, they're not even allowed on the tests.

Utterly confused