Finding Absolute Minima & Maxima of a Function

In summary, when finding the absolute minima and maxima of a function, it is important to not only consider the critical points found through the Hessian Matrix, but also to check the four bounding line segments of the region. This is because the endpoints of these segments may also be absolute minima or maxima, even if they are not critical points. This is similar to how, in a one-variable function, the endpoints and points where the derivative does not exist must also be considered when finding the absolute minima and maxima.
  • #1
Amaelle
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Homework Statement
f(x,y)=x^2 +xy+y^2 R:[(x,y)/ -2<=x<=2 ; -1<=y<=1]
Relevant Equations
Hessian Matrix
Good day,

I have a question regrading how to find the absolute minima maxima of a function , I understand that first we need to calculate the Hessian Matrix to find the relative minima /maxim but after we need to check the borders of the region ( a rectangle in our case)
for example we put x=-2 and y=y
and we tried to find the relative maxima minima that we need to compare with the other critical points we found with the Hessian ,but my confusion is the following: why do you need to check the end points (for instance the point (-2,1) and (-2,-1)? if when put the substitution x=-2 and y=y, those two points are already included?
many thanks in advance!
 
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  • #2
Amaelle said:
Homework Statement:: f(x,y)=x^2 +xy+y^2 R:[(x,y)/ -2<=x<=2 ; -1<=y<=1]
Relevant Equations:: Hessian Matrix

Good day,

I have a question regrading how to find the absolute minima maxima of a function , I understand that first we need to calculate the Hessian Matrix to find the relative minima /maxim but after we need to check the borders of the region ( a rectangle in our case)
for example we put x=-2 and y=y
and we tried to find the relative maxima minima that we need to compare with the other critical points we found with the Hessian ,but my confusion is the following: why do you need to check the end points (for instance the point (-2,1) and (-2,-1)? if when put the substitution x=-2 and y=y, those two points are already included?
many thanks in advance!
What do you mean by "already included"?
 
  • #3
Amaelle said:
Homework Statement:: f(x,y)=x^2 +xy+y^2 R:[(x,y)/ -2<=x<=2 ; -1<=y<=1]
Relevant Equations:: Hessian Matrix

Good day,

I have a question regrading how to find the absolute minima maxima of a function , I understand that first we need to calculate the Hessian Matrix to find the relative minima /maxim but after we need to check the borders of the region ( a rectangle in our case)
for example we put x=-2 and y=y
and we tried to find the relative maxima minima that we need to compare with the other critical points we found with the Hessian ,but my confusion is the following: why do you need to check the end points (for instance the point (-2,1) and (-2,-1)? if when put the substitution x=-2 and y=y, those two points are already included?
many thanks in advance!
Two things:
1) Endpoints -- For this problem, you're not checking endpoints. You need to check the four bounding line segments: x = -2, -1 <= y <= 1; x = 2, -1 <= y <= 1; y = -1, -2 <= x <= 2; y = 1, -2 <= x <= 2.
Maybe this is what you meant when you wrote "for example we put x=-2 and y=y".
2) Where a maximum or minimum will occur --
If you have a continuous function of one variable, that is defined on a closed, bounded interval, the absolute maximum or minimum can occur at any of these places:
  • Points at which the derivative is zero (assuming that the function is also differentiable)
  • Endpoints of the interval
  • Points at which the function is defined but at which the derivative doesn't exist

For the second bullet point above, the graph of ##y = -x^2 + 1## for ##x \in [-1, 2]## has a maximum at x = 0, but its minimum is at x = 2. This point can't be found by using the derivative.
For the third bullet point, the graph of y = |x| has an absolute minimum at x = 0. Although this function is continuous everywhere, its derivative doesn't exist at x = 0, so using the derivative is no help.

The situation is similar for functions of two or more variables, except you don't just have endpoints of an interval.
 
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  • #4
thank you very much, this exactely how the problem is solved
I want to ask you about something in particular
you said :"You need to check the four bounding line"
I totally agree let's take for example the first line x=-2 and y=y
why do we need to check if end points (-2;1) and (-2;-1) are also absolute minima / maxima even though they belong to the same line? why the first derivative of that line can not detect them?
I hope I could explain my doubt?
thank you very much!
 
  • #5
Amaelle said:
thank you very much, this exactely how the problem is solved
I want to ask you about something in particular
you said :"You need to check the four bounding line"
I totally agree let's take for example the first line x=-2 and y=y
You don't need to say "and y = y." The equation x = -2 is a vertical line (the y values are arbitrary). If you want to specify a segment of this line, then you can add -1 <= y <= 1 or whatever the limits are.
Amaelle said:
why do we need to check if end points (-2;1) and (-2;-1) are also absolute minima / maxima even though they belong to the same line? why the first derivative of that line can not detect them?
I hope I could explain my doubt?
Because when you set x = -2 (for the line x = -2), you now have a function of one variable. If the max or min occurs in the interior that line segment, the derivative will find it, but if it occurs at an endpoint, the derivative won't help.

Keep in mind that you will need to check all four bounding line segments, and the corner points.
 
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  • #6
thanks a million!
 
  • #7
PeroK said:
What do you mean by "already included"?
I meant they belong to that line, so there is no need to check them independently.
 
  • #8
Amaelle said:
I meant they belong to that line, so there is no need to check them independently.
You haven't differentiated along the line. In general, you don't have ##\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0## at a minimum or maximum on the boundary.
 
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  • #9
thanks a lot!
 

FAQ: Finding Absolute Minima & Maxima of a Function

1. What is the definition of absolute minima and maxima of a function?

Absolute minima and maxima refer to the lowest and highest values of a function, respectively, over its entire domain. These points are also known as global minima and maxima.

2. How do you find the absolute minima and maxima of a function?

To find the absolute minima and maxima of a function, you need to take the derivative of the function and set it equal to zero. Then, solve for the critical points (points where the derivative is equal to zero or undefined). These critical points will be potential candidates for the absolute minima and maxima. Finally, plug in these critical points and the endpoints of the function's domain into the original function to determine the absolute minima and maxima.

3. What is the difference between absolute and relative minima and maxima?

Absolute minima and maxima are the lowest and highest values of a function over its entire domain. On the other hand, relative minima and maxima refer to the lowest and highest values of a function within a specific interval. Relative minima and maxima can exist within the range of absolute minima and maxima.

4. Can a function have multiple absolute minima and maxima?

Yes, a function can have multiple absolute minima and maxima. This can occur when the function has multiple critical points that are all potential candidates for the absolute minima and maxima. In this case, you would need to plug in each critical point and the endpoints of the function's domain to determine the actual absolute minima and maxima.

5. Why is finding absolute minima and maxima important in mathematics and science?

Finding absolute minima and maxima is important in mathematics and science because it allows us to determine the optimal values of a function. These optimal values can represent the minimum or maximum values of a physical quantity or represent the most efficient solution to a problem. This information is crucial in various fields such as optimization, economics, and physics.

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