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October 21st, 2013, 05:21 AM   #1
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Find the absolute minimum and absolute maximum.

Find the absolute minimum and absolute maximum of
f(x,y) = x^2+y^2-2(x-y) in the region defined by the
x^2+y^2?4 ?
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October 21st, 2013, 06:03 AM   #2
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Re: Find the absolute minimum and absolute maximum.

When you're looking for minima and maxima, the first thing to try is to differentiate. In this case, as f is afunction of x and y, you need to take partial derivatives wrt x and y.



So, the partial derivatives of f are 0 when x = 1 and y = -1.

And, as both second derivatives are positive, these represent local minima, hence an absolute minimum for f at (1, -1).

The maxima must then occur on the boundary, as there are no local maxima for f. On the boundary, we have:



You can you can use more calculus to show that a maximum occurs at:

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October 21st, 2013, 05:53 PM   #3
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Re: Find the absolute minimum and absolute maximum.

Quote:
Originally Posted by Pero
When you're looking for minima and maxima, the first thing to try is to differentiate. In this case, as f is afunction of x and y, you need to take partial derivatives wrt x and y.



So, the partial derivatives of f are 0 when x = 1 and y = -1.

And, as both second derivatives are positive, these represent local minima, hence an absolute minimum for f at (1, -1).

The maxima must then occur on the boundary, as there are no local maxima for f. On the boundary, we have:

I had to think about that for a second! Pero replaced the first "" in with 4. Now there are several ways to go to "use more calculus" as Pero says below:
We could replace y with . That would be the hard way!

We could switch to polar coordinates with r= 2 (since is a circle with center at (0, 0) and radius 2) to have and so we are finding min and max of for . That's fairly straight forward.

We could use "Lagrange Multipliers" to minimize f(x,y)= 4- 2x+ 2y subject to the condition that . and . We must have for some number so we must have and , together with , three equations to solve for the three unknowns, x, y, and . Since a specific value of is not necessary for a solution, I find it is often best to eliminate first by dividing one equation by another. Here, that gives which reduces to so y= -x. Together with , we have so that . Together with y= -x, we have two solutions, and .
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October 21st, 2013, 11:38 PM   #4
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Re: Find the absolute minimum and absolute maximum.

It might be useful to remember that on a circle x+y has a maximum when .

In an exam, you'll have to work this out every time(and using polar coordinates is the best way to go), but it's the sort of thing I just remember.

In this case we had to maximise y-x, so that would be at

While is the minimum on the boundary.

Also:

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October 22nd, 2013, 12:29 AM   #5
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Re: Find the absolute minimum and absolute maximum.

When they say 'absolute minimum' are they referring to 'global minimum' or 'minimizer of the absolute value'? I believe they're referring to the second.
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October 22nd, 2013, 12:31 AM   #6
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Re: Find the absolute minimum and absolute maximum.

My bad, they're referring to global extrema, which is the first. I just checked.
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