Multivariable maximum/minimum problem

In summary, the purpose of this problem is to find the minimum and maximum points of a function f(x,y) in an interval R, which consists of the interior and the boundary of R. The problem asks for the points on the boundary, but if R is an open interval then it doesn't really have a boundary to check points on. The gradient is found to be (x+1)i + (y+1)j, and if R is closed then the point (x+1)i + (y+1)j will be on the boundary. The problem also says that if R is an open interval then it is guaranteed to have at least one point on the boundary. However, this is not the case
  • #1
InsaneBraine
9
0

Homework Statement



The function is f(x,y) = xy + x + y. The region R is defined as all the points lying inside x^2 + y^2 < 8. The problem asks us to find all the points in the region R where f may have a minimum or a maximum.

Homework Equations



Partial derivatives etc

The Attempt at a Solution



I only have one point so far which is (-1, -1) which I got from setting the partial derivatives of f equal to zero. I'm not quite sure how to obtain the rest. Lagrange multipliers would only be appropriate if x^2 + y^2 = 8, but over here its less than 8 (not even less than or equal to). The other thing I have in mind is trying the points on the boundary but the problem with that is that the actual boundary is not inside the region. So how can I obtain those points?
 
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  • #2
InsaneBraine said:

Homework Statement



The function is f(x,y) = xy + x + y. The region R is defined as all the points lying inside x^2 + y^2 < 8. The problem asks us to find all the points in the region R where f may have a minimum or a maximum.

Homework Equations



Partial derivatives etc

The Attempt at a Solution



I only have one point so far which is (-1, -1) which I got from setting the partial derivatives of f equal to zero. I'm not quite sure how to obtain the rest. Lagrange multipliers would only be appropriate if x^2 + y^2 = 8, but over here its less than 8 (not even less than or equal to). The other thing I have in mind is trying the points on the boundary but the problem with that is that the actual boundary is not inside the region. So how can I obtain those points?

So far as I know, an open interval like you have doesn't have local maxima or minima along the edges, so you don't need to worry about boundary points. You will want to remember that the point you have might be a saddle point, though.
 
  • #3
If a region is closed and bounded then it is guaranteed to have such points. However, this is not the case, so you'll have to check where (as you did) the gradient is 0.
 
  • #4
So what approach should I take if the points aren't on the boundary? How do I find them?
 
  • #5
InsaneBraine said:
So what approach should I take if the points aren't on the boundary? How do I find them?

The same way that you did. Ask when does the gradient equal 0, and then check to see if the points you find are saddle points or minima/maxima.
 
  • #6
I don't think I have to do the latter part, since the problem only asks the point where it MIGHT have max/mins.

So the gradient is:

(x+1)i + (y+1)j.

Now do I just set it equal to zero? If yes, then the only point I get is (-1, -1) and the region doesn't play a role. I could also plug in values of x and y using x^2 + y^2 < 8 but that would give me inequalities.
 
  • #7
Indeed it would. I would say that (-1,-1) is your only point then.
 
  • #8
So basically the region R doesn't play any role in this problem even though we're supposed to find the max/min points of the function within that interval? Could you please explain why that is so?
 
  • #9
It's because the region R is an open region, and so it doesn't really have a boundary to check points on. If you were dealing with all the points such that x2+y2≤8, then you would have a legitimate boundary to check points on.
 
  • #10
I see. But the thing is in the next part for this problem we're supposed to use Lagrange multipliers to find the other points on the actual boundary x^2 + y^2 = 8. I found 4 other points. At the end of the problem it says that in total, you should have found 5 points inside R, but two of the points I found using Lagrange are (2,2) and (-2,-2) which aren't actually inside R.
 
  • #11
Correction: Actually all 4 points I found using Lagrange lie on the boundary and not inside R (by definition).

So what does the problem mean when it says that I should have found 5 points inside R?
 
  • #12
Likely, the problem is printed incorrectly, and meant to use the ≤ sign.
 
  • #13
So here's the original problem:

[PLAIN]http://img141.imageshack.us/img141/2196/problemaz.jpg
 
Last edited by a moderator:
  • #14
Ah! There it says that R is the set of points such that x2+y2≤8. So yes, the lagrange multipliers are used in this problem.
 
  • #15
I feel so stupid for misreading the problem. It clearly says that R consists of the interior AND the boundary. I somehow thought that R is only the interior. So the 5 points its asking for are basically (-1,-1) and the 4 points I found using Lagrange, correct?
 
  • #16
That is correct.
 
  • #17
Thanks so much for your help!
 

FAQ: Multivariable maximum/minimum problem

What is a multivariable maximum/minimum problem?

A multivariable maximum/minimum problem is a mathematical problem that involves finding the maximum or minimum value of a function with multiple independent variables. This type of problem is commonly encountered in the field of optimization.

How is a multivariable maximum/minimum problem solved?

To solve a multivariable maximum/minimum problem, the first step is to find the critical points of the function by taking the partial derivatives with respect to each variable and setting them equal to zero. Then, the second derivative test is used to determine whether the critical points are maximum, minimum, or saddle points.

What is the difference between a local and global maximum/minimum?

A local maximum/minimum is the highest or lowest point in a specific region of a function, while a global maximum/minimum is the highest or lowest point across the entire domain of the function.

Can multivariable maximum/minimum problems have multiple solutions?

Yes, multivariable maximum/minimum problems can have multiple solutions. This can occur when there are multiple critical points that satisfy the second derivative test, or when there are multiple regions with local maximum/minimum points.

How are multivariable maximum/minimum problems applied in the real world?

Multivariable maximum/minimum problems have various applications in fields such as economics, engineering, and physics. For example, they can be used to optimize production processes, maximize profits, or minimize energy consumption in a system.

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