# Absolute max and min of multivariable function

• java22
In summary: You should label each point you find as to which side of the triangle it is on.In summary, the function f(x,y) = x^2-4xy+5y^2-8y, enclosed by the triangular region (0,0), (3,0), and (3,3), has an absolute maximum of 9 and an absolute minimum of -8. These values are obtained by evaluating the function along the three sides of the triangle and selecting the maximum and minimum values from those points.
java22

## Homework Statement

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

N/A

## 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.

java22 said:

## Homework Statement

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

N/A

## 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.
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).

java22 said:

## Homework Statement

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

N/A

## 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.

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 (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:
Mark44 said:
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).

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?

java22 said:
Is this whole process correct? Do I now just pick the highest and lowest of all of my max and mins and that will gives me my absolute max and mins? So the absolute max would be 9 and the absolute min would be -8?

## What is the concept of absolute max and min in a multivariable function?

The absolute maximum of a multivariable function is the highest value that the function can attain within a given domain, while the absolute minimum is the lowest value that the function can attain within the same domain. These values represent the extreme points of the function.

## How do you find the absolute max and min of a multivariable function?

To find the absolute max and min of a multivariable function, you need to first find the critical points of the function by taking partial derivatives with respect to each variable and setting them equal to zero. Then, evaluate the function at these critical points as well as at the boundaries of the domain to determine the absolute max and min values.

## Can a multivariable function have more than one absolute max or min?

Yes, a multivariable function can have multiple absolute max or min values. This can occur when the function has multiple critical points or when the boundaries of the domain have multiple extreme points.

## Can the absolute max or min of a multivariable function occur at a point where the function is not defined?

Yes, the absolute max or min of a multivariable function can occur at a point where the function is not defined. This can happen if the function has a discontinuity at that point and the values on either side of the discontinuity are the absolute max or min.

## How is the concept of absolute max and min of a multivariable function used in real-world applications?

The concept of absolute max and min is often used in optimization problems, where the goal is to find the maximum or minimum value of a function within a given domain. This is commonly seen in fields such as economics, engineering, and physics, where finding the optimal solution is crucial.

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