Finding Absolute Minima & Maxima of a Function

[Amaelle]

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!

 

[PeroK]

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"?

 

[Mark44]

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.

 

[Amaelle]

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!

 

[Mark44]

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.

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.

 

[Amaelle]

thanks a million!

 

[Amaelle]

What do you mean by "already included"?

I meant they belong to that line, so there is no need to check them independently.

 

[PeroK]

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.

 

[Amaelle]

thanks a lot!