Find the global extrema of this function over the given region

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Discussion Overview

The discussion revolves around finding the global extrema of a function over a specified region, focusing on critical points and boundary behavior. Participants explore the nature of critical points and the implications of boundary conditions on the existence of global maxima and minima.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants identify the critical points as (0,0) and (2,1) and suggest that the global maximum occurs at (2,1) while asserting there is no global minimum due to unbounded behavior on the boundary x = 0.
  • Others inquire whether the second partials test has been applied to determine the nature of the critical points, suggesting they may be saddle points.
  • There is a discussion about the boundaries, with some participants agreeing that the relevant boundaries are x = 0 and y = 1.
  • One participant notes that the limit of the function as y approaches negative infinity leads to no global minimum, while another suggests that the limit as x approaches infinity may indicate no global maximum.
  • A later reply introduces a caution regarding the interpretation of the domain, suggesting that the critical point (2,1) may be outside the intended region of analysis, which consists of the unit square.

Areas of Agreement / Disagreement

Participants express differing views on the existence of global extrema, with some suggesting there is no global minimum or maximum, while others highlight the need to consider the specified domain, leading to unresolved questions about the critical point's relevance.

Contextual Notes

There is uncertainty regarding the definitions and boundaries of the region under consideration, particularly whether the critical point (2,1) falls within the specified domain. The implications of boundary behavior on the extrema are also not fully resolved.

Vishak95
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Hi MHB. Can someone help me with this one please?

View attachment 1595I've worked out that the critical points are (0,0) and (2,1). But looking at the boundary x = 0, there seems to be no limit to the minimum value. Also, on the boundary y = 1, the value of f(x,1) = -1.

So, would I be correct in saying that the global maximum value occurs at (2,1) and that there is no global minimum?
 

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Re: Find the global maxima of this function

Before we look at the boundaries, have you used second partials test for relative extrema to determine that nature of the critical points (which you have correctly determined)?
 
Re: Find the global maxima of this function

The critical points both appear to be saddle points (the derivative D = -1 for both)
 
Re: Find the global maxima of this function

Vishak said:
The critical points both appear to be saddle points (the derivative D = -1 for both)

Yes, I get the same result. :D

So now you want to look at the boundaries. What are they?
 
Re: Find the global maxima of this function

Great :D

I think the only real boundaries would be x = 0 , and y = 1
 
Re: Find the global maxima of this function

Vishak said:
Great :D

I think the only real boundaries would be x = 0 , and y = 1

Yes, those are the real boundaries. You have already found that:

$$f(0,y)=-y^2$$

and because:

$$\lim_{y\to-\infty}\left(-y^2 \right)=-\infty$$

You correctly conclude that the function has no global minimum. Suppose we choose an arbitrary value for $y$, such as $y=-k$ where $0<k$. Then the function is:

$$f(x,-k)=\left(k^2+k \right)x-k^2$$

What is the limit of this as $$x\to\infty$$?
 
Re: Find the global maxima of this function

The limit appears to be going to $$\infty $$. So then I guess that there is no global maximum either?
 
Re: Find the global maxima of this function

Vishak said:
The limit appears to be going to $$\infty $$. So then I guess that there is no global maximum either?

Yes, I come to the same conclusion. :D
 
Re: Find the global maxima of this function

Awesome! Thank you Mark :D
 
  • #10
Re: Find the global maxima of this function

Vishak said:
Hi MHB. Can someone help me with this one please?

View attachment 1595I've worked out that the critical points are (0,0) and (2,1). But looking at the boundary x = 0, there seems to be no limit to the minimum value. Also, on the boundary y = 1, the value of f(x,1) = -1.

So, would I be correct in saying that the global maximum value occurs at (2,1) and that there is no global minimum?
Caution: some people use $\{(x,y)\ |\ 0\leq x,y\leq1\}$ as an abbreviation for $\{(x,y)\ |\ 0\leq x\leq1 \text{ and }0\leq y\leq1\}$. I suspect that is what is intended here. If so, you are being asked for the extreme values of $f(x,y)$ on the unit square. Then the critical point $(2,1)$ is outside the domain $D$, and the boundary consists of four line segments.
 
  • #11
Re: Find the global maxima of this function

Opalg said:
Caution: some people use $\{(x,y)\ |\ 0\leq x,y\leq1\}$ as an abbreviation for $\{(x,y)\ |\ 0\leq x\leq1 \text{ and }0\leq y\leq1\}$. I suspect that is what is intended here. If so, you are being asked for the extreme values of $f(x,y)$ on the unit square. Then the critical point $(2,1)$ is outside the domain $D$, and the boundary consists of four line segments.

That makes much more sense as a problem. :D
 

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