MHB Find the global extrema of this function over the given region

[Vishak95]

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?

 

[MarkFL]

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)?

 

[Vishak95]

Re: Find the global maxima of this function

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

 

[MarkFL]

Re: Find the global maxima of this function

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?

 

[Vishak95]

Re: Find the global maxima of this function

Great :D

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

 

[MarkFL]

Re: Find the global maxima of this function

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 $$x\to\infty$$?

 

[Vishak95]

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?

 

[MarkFL]

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?

Yes, I come to the same conclusion. :D

 

[Vishak95]

Re: Find the global maxima of this function

Awesome! Thank you Mark :D

 

[Opalg]

Re: Find the global maxima of this function

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.

 

[MarkFL]

Re: Find the global maxima of this function

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