# Absolute max and min of multivariable function

1. Mar 6, 2016

### java22

1. The problem statement, all variables and given/known data

f(x,y) = x2-4xy+5y2-8y ; enclosed by the triangular region (0,0), (3,0), and (3,3)

Find the absolute maximum and minimum

2. Relevant equations

N/A

3. The attempt at a solution

First I took the partial derivatives with respect to x and y:

fx = 2x - 4y
fy = -4x + 10y -8

I then set both of these equal to zero and solved them as a system of equations. This gives the critical point of (8,4)

Now, this is where I'm lost. Since the critical point (8,4) doesn't exist in the bounded region given, is the problem simply not solvable, and there are no absolute minima or maxima? I assume I am going about it wrong because calculus is never that easy.

2. Mar 6, 2016

### Staff: Mentor

Now that you have found that the only critical point is outside the region, you need to look along the three boundary line segments. For the segment along the x-axis, all points have a y-value of 0, and the x-value ranges from 0 to 3. So the function simpifies to $f(x, 0) = 5y^2 - 8y$, essentially a function of one variable. What are the maximum and minimum function values along this segment?

Do the same for the vertical segment, and for the segment between (0, 0) and (3, 3).

3. Mar 6, 2016

### Ray Vickson

No: there ARE absolute maxima and minima for any continuous function on a closed and bounded region, which is the case here. That is a theorem in analysis, but you may not have seen it in an introductory calculus course. See, eg.,
http://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx
or
http://math.harvard.edu/~ytzeng/worksheet/1017_sol.pdf [Broken] (especially examples 5 and 6).

When you have inequality restrictions the derivatives need not equal zero. The derivatve = 0 condition applies to an interior point max or min (that is, a point not on the boundary of the region. As you found, that stationary point lies completely outside the allowed region, so that means there are no interior point optima. The solution must lie on the boundary.

So, you can look at each side of the triangle separately and find the absolute max/min on the side. There will be three possible maxima; you need to evaluate all three numerically and take the best one.

Last edited by a moderator: May 7, 2017
4. Mar 6, 2016

### java22

Ok so for the line segments of the triangle I got x=3, y=0, and y=x

Along the segment x=3, I plugged 3 in for x in the original function and got 5y2-20y+9
When evaluating this for y for 0 and 3, I got that the max would be y=0 which equals 9, and the min would be y=3 which is -6

Along the segment y=0, I plugged 0 in for y in the original function and got x2
When evaluating this for x for 0 and 3, I got that the max would be x=3 which equals 9 and the min would be x=0 which equals 0

Along the segment y=x, I plugged in x for all of the y's in the original function and got 2x2-8x
I also took the derivative of this to get 4x-8 which gives x=2 as a critical point.
When evaluating for x as 0, 3, and 2, I got a max when x=0 which gave 0, and a min when x=2 which gives -8.

Is this whole process correct? Do I now just pick the highest and lowest of all of my max and mins and that will give me my absolute max and mins? So the absolute max would be 9 and the absolute min would be -8?

5. Mar 6, 2016

### Staff: Mentor

Yes, that's the idea. I haven't checked your work, so can't confirm your answers.

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