Determining global maximums and minimums

- anonymous

Determining global maximums and minimums

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- anonymous

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- anonymous

Are you supposed to do computations for this one? The directions say that you should just be able to tell from looking at a graph, so I think that's what you should do, right?

- anonymous

whoops i forgot to specify a number. hmmm lets do # 5

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- anonymous

a graph?

- anonymous

Well, you don't need a graph. #5 says z = x^2 + y^2. Can you tell just by looking at it what the smallest value of z is?

- anonymous

By "it" I mean the equation

- anonymous

0

- anonymous

and the greatest will be 2 but it occurs at a few points

- anonymous

1. Analyze the overall max's and mins with your formula fxxfyy - (fxy)^2.
2. Analyze the edges. Do traces at when x=1, x=-1, y=1, y=-1 and determine the max's and mins of those traces.
3. Analyze the corners. Evaluate the derivative at when f(1,1), f(1,-1), f(-1,1), and f(-1,-1).

- anonymous

so i guess its no longer a global maximum

- anonymous

Analyzing the edges is pretty easy since all you have to do is the same thing you did in single variable calculus :-)

- anonymous

Ex. 1 http://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx

- anonymous

ya but like at 1,-1 and -1,-1 and -1,1 and 1,1 like it all equals 2

- anonymous

Did you analyze the edges? You have to do traces. I can compute the traces for you.

- anonymous

NEVERMIND!

- anonymous

I'm getting you some graphs, sorry I'm a bit out of it :-(

- anonymous

hahha i feel stupid now

- anonymous

I just messed up on the traces.... forgot the that the -1 is squared lol :-).
We need to get visual here, so I'll have these graphs up here in a sec.

- anonymous

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- anonymous

This is why you need to analyze the edges of each graph ;-)

- anonymous

hahah but at the edges they alll equal z=2

- anonymous

The corners, you mean?

- anonymous

ya whtvr

- anonymous

The edges are going to be a trace. And then you need to treat the those like what you had to do in single var. calculus.

- anonymous

For x=1 and x=-1, your trace will be z=1+y^2.
For y=1 and y=-1, your trace will be z=x^2+1
Now you just need to take the derivative to see where your slope will be zero. Those edges are going to have a slope of zero when x=0 and y=0. Therefore, your minimums will be at f(1,0), f(-1,0), f(0,1), and f(-1,0).
Your absolute maxima will be at those corners. f(1,1), f(1,-1), f(-1,1), and f(-1,-1). Your abs. minimum will be at (0,0).

- anonymous

but the issue i am having is
that there can only be one blobal point not many

- anonymous

The question is poorly worded. You are not required to find a UNIQUE global maximum. You can have many, if there are many.

- anonymous

You'll have 4 global max's, 1 global min.

- anonymous

ohhhh that is weird i thought u can only have one point

- anonymous

Since this is within a specified boundary, the rules change a bit. You have to do 3 different things instead of just one.

- anonymous

I'll go work on the last one (saddle) and see what I come up with.

- anonymous

uhhh no cuz i dont have to do #7

- anonymous

It's a really nice problem to look at. B/c you have to use fxxfyy-fxy^2 to see if it's a max, min, or saddle. I'll do it just for fun.

- anonymous

Thanks brine

- anonymous

First step: To analyze the function as a whole.
f(x,y) = x^2-y^2
fx(x,y) = 2x , fy(x,y) = -2y
Set both derivatives equal to zero.
2x = 0 , x=0
2y = 0 , y=0 (we normally don't have things this easy, we usually have to solve a system of equations)
Therefore, our critical point is at f(0,0)
But we're not done! We must analyze if it's a saddle, min, or max.
D=fxxfyy - (fxy)^2
fxx = 2, fxx(0,0) = 2
fyy = -2, fyy(0,0) = -2
fxy = 0, fxy(0,0) = 0
D=(2)(-2) - 0 = -4
D<0 and is a saddle point.
Step 2: Analyze the edges. We need to find traces of each side of our boundary.
For x=1 and x=-1 traces, it will be the same eqn.
f(y) = 1-y^2
f'(y) = -2y
0 = -2y
y=0
Critical points: f(1,0), f(-1,0)
f(1,0) = 1, f(-1,0) = 1
For y=1 and y=-1 traces, it will be the same.
f(x) = x^2 -1
f'(x) = 2x
0=2x
x=0
Critical points: f(0,1), f(0,-1)
f(0,1) = -1, f(0,-1) = -1
Step 3. Analyze the corners.
f(1,1) = 0, f(1,-1) = 0, f(-1,1) = 0, f(-1,-1) = 0
So in summary, we have:
Saddle point: f(0,0)
Global max's: f(1,0) and f(-1,0) = 1
Global Mins: f(0,1) = -1, f(0,-1) = -1

- anonymous

LOL this is actually my favorite problem out of the 3! So all you ever need to remember is 1. analyze whole function, 2. analyze edges (4 traces if square boundary), 3. corners. And you've got it!

- anonymous

thanks. That was very kind :D

- anonymous

You helped me on linear algebra so this is the least I could do. And it's fun.

- anonymous

hahah that was easy this is like alot more tedious. Thanks

- anonymous

With the first step, it wasn't so bad. Normally, you have to take the derivative fx and fy and then set each equal to zero. And then you have to solve for a system of equations to get y. And then you have to plug y into the first equation. Doesn't make much sense with me explaining it, so here's Paul!
Ex. 1: http://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx

- anonymous

thanks ill take a look

- anonymous

Sorry, I meant... plug y into the first eqn to get x.
Lol I hope some of this is useful, I just love to blabber when I actually know what I'm doing.

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