anonymous
  • anonymous
Determining global maximums and minimums
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • 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
  • anonymous
whoops i forgot to specify a number. hmmm lets do # 5

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
a graph?
anonymous
  • 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
  • anonymous
By "it" I mean the equation
anonymous
  • anonymous
0
anonymous
  • anonymous
and the greatest will be 2 but it occurs at a few points
anonymous
  • 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
  • anonymous
so i guess its no longer a global maximum
anonymous
  • anonymous
Analyzing the edges is pretty easy since all you have to do is the same thing you did in single variable calculus :-)
anonymous
  • anonymous
Ex. 1 http://tutorial.math.lamar.edu/Classes/CalcIII/AbsoluteExtrema.aspx
anonymous
  • anonymous
ya but like at 1,-1 and -1,-1 and -1,1 and 1,1 like it all equals 2
anonymous
  • anonymous
Did you analyze the edges? You have to do traces. I can compute the traces for you.
anonymous
  • anonymous
NEVERMIND!
anonymous
  • anonymous
I'm getting you some graphs, sorry I'm a bit out of it :-(
anonymous
  • anonymous
hahha i feel stupid now
anonymous
  • 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
  • anonymous
1 Attachment
anonymous
  • anonymous
This is why you need to analyze the edges of each graph ;-)
anonymous
  • anonymous
hahah but at the edges they alll equal z=2
anonymous
  • anonymous
The corners, you mean?
anonymous
  • anonymous
ya whtvr
anonymous
  • 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
  • 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
  • anonymous
but the issue i am having is that there can only be one blobal point not many
anonymous
  • 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
  • anonymous
You'll have 4 global max's, 1 global min.
anonymous
  • anonymous
ohhhh that is weird i thought u can only have one point
anonymous
  • 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
  • anonymous
I'll go work on the last one (saddle) and see what I come up with.
anonymous
  • anonymous
uhhh no cuz i dont have to do #7
anonymous
  • 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
  • anonymous
Thanks brine
anonymous
  • 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
  • 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
  • anonymous
thanks. That was very kind :D
anonymous
  • anonymous
You helped me on linear algebra so this is the least I could do. And it's fun.
anonymous
  • anonymous
hahah that was easy this is like alot more tedious. Thanks
anonymous
  • 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
  • anonymous
thanks ill take a look
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.