Max and Min Values of f(x,y) = x^2 + 2y^2 on Domain [0,1]

[snoggerT]

determine the global extreme values of the function on the given domain:

f(x,y)=x^2+2y^2 , 0<=x,y<=1

The Attempt at a Solution

- I know you need to evaluate the critical points first and then use the boundary to find the maximum and minimum values, but I can't seem to get the right answer. I think I'm just misunderstanding the way you go about a problem like this. Please help.

 

[Defennder]

Finding critical points is one way to solve it, but it's a little tedious and unnecessary. Just start by examining the function [tex]f(x,y) = x^2 + 2y^2 [/tex].

Note that the function is greater or equal to 0. That should tell you something about the extreme values. And note that while x is bounded below, it's not bounded above, what does that say about the function?

 

[snoggerT]

Finding critical points is one way to solve it, but it's a little tedious and unnecessary. Just start by examining the function [tex]f(x,y) = x^2 + 2y^2 [/tex].

Note that the function is greater or equal to 0. That should tell you something about the extreme values. And note that while x is bounded below, it's not bounded above, what does that say about the function?

- would the function always being great than or equal to zero mean that the global minimum would be 0?

with x not being bounded above, that would mean that it contains points arbitrarily far from the origin, but I'm not sure what that would mean for finding the maximum.

 

[Defennder]

Yes, for the first part.

For the second, does the question say anything about the function necessarily having a maximum?

 

[snoggerT]

Yes, for the first part.

For the second, does the question say anything about the function necessarily having a maximum?

- no, but the answer in the back of the book has a maximum and a minimum (0). I just don't see how they got the answer they got for the maximum.

 

[Dick]

If the function has a max, it must be on the boundary of 0<=x,y<=1, yes?

 

[snoggerT]

If the function has a max, it must be on the boundary of 0<=x,y<=1, yes?

- right. so how would you set up the problem to find that max? the problem before it, I wasn't sure if I did it right, but I got the right answer. I did this problem the same way and got the wrong answer, so I just think I got lucky doing the problem before incorrectly. So I guess I'm not really sure how to set a problem like this up.

 

[HallsofIvy]

- no, but the answer in the back of the book has a maximum and a minimum (0). I just don't see how they got the answer they got for the maximum.

What does your book say that maximum is?

 

[Dick]

- right. so how would you set up the problem to find that max? the problem before it, I wasn't sure if I did it right, but I got the right answer. I did this problem the same way and got the wrong answer, so I just think I got lucky doing the problem before incorrectly. So I guess I'm not really sure how to set a problem like this up.

The boundary consists of four line segments. One of them is x=0, 0<=y<=1. What's the maximum along that segment? What about the other three?

 

[snoggerT]

What does your book say that maximum is?

- the book has the max at 3, but I got 2 for my max.

 

[Dick]

- the book has the max at 3, but I got 2 for my max.

Try ALL of the boundary segments. How about x=1, 0<=y<=1?

 

[snoggerT]

Try ALL of the boundary segments. How about x=1, 0<=y<=1?

- this is what I did to try and find my maximum value:

y=0, 0<=x<=1 >>> f(x,0)=x^2 >>> f(1,0) = 1

y=1, 0<=x<=1 >>> f(x,1)=x^2+2 >>> f(0,0)=2

x=0, 0<=y<=1 >>> f(0,y)=2y^2 >>> f(0,1)=2

x=1, 0<=y<=1 >>> f(1,y)=1+2y^2 >>> f(1,0)=1

I have no clue if that's the correct way of setting these problems up because the book isn't very clear on it, but that's how I got the answer for the first problem and it was right.

 

[Dick]

- this is what I did to try and find my maximum value:

y=0, 0<=x<=1 >>> f(x,0)=x^2 >>> f(1,0) = 1

y=1, 0<=x<=1 >>> f(x,1)=x^2+2 >>> f(0,0)=2

x=0, 0<=y<=1 >>> f(0,y)=2y^2 >>> f(0,1)=2

x=1, 0<=y<=1 >>> f(1,y)=1+2y^2 >>> f(1,0)=1

I have no clue if that's the correct way of setting these problems up because the book isn't very clear on it, but that's how I got the answer for the first problem and it was right.

Somehow, you are completely missing the possibility that x=1 and y=1. f(1,1)=3.

 

[HallsofIvy]

Basically, then, your error was not looking at the "boundaries of the boundaries"- that is, the endpoints of the line segments.

 

[snoggerT]

Somehow, you are completely missing the possibility that x=1 and y=1. f(1,1)=3.

- alright, I see now. I was going a lot off the one example in the book and they didn't use that possibility (just the 4 I used). Can somebody explain the proper way to set these problems up?

 

[rootX]

So, we need to look at total of 12 cases (some will be duplicates)?

4 cases he described and 2 more for each of those cases
1)y=1, 0<=x<=1 >>> f(x,1)=x^2+2 >>> f(0,0)=2
2) y =1, x = 0
3) y = 1,x = 1

In case of circular dimension/any dimension that can be described by a function, we do two times?
>find min/max of area
>find min/max along the circular line (using deltas/something)

 

[Defennder]

Oh wow, I thought he meant x>=0 and y<=1 rather than both x,y in [0,1].