Absolute Max Min Problem on Region Bounded by x = \sqrt{1-y^2}, y=x, and y=0

[jegues]

Homework Statement

Find the absolute maximum and the absolute minimum of the function,

$$f(x,y) = x^2 + 2xy - y^2$$

on the region bounded by, $$x = \sqrt{1-y^2},y=x \text{ and } y=0$$

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

Everything seems to be going fine until I get to the point where I have to find critical points on the edge of C3.

I can't think of what I should be substituting into the function. I tried,

$$y^{2} = 1 - x^{2}$$

but I still end up with nasty square roots in my new function.

Any ideas?

 

[Mamooie312]

Use Lagrange Multipliers on the boundaries.

 

[jegues]

Lagrange Multipliers were not covered in my course. My professor recommended I use trig functions for the edge C3 but I'm still not seeing it.

Can someone nudge me in the right direction?

 

[Dick]

Lagrange Multipliers were not covered in my course. My professor recommended I use trig functions for the edge C3 but I'm still not seeing it.

Can someone nudge me in the right direction?

You can parametrize the unit circle using x=cos(t), y=sin(t).

 

[jegues]

You can parametrize the unit circle using x=cos(t), y=sin(t).

Okay so,

$$x = cos(t), y=sin(t) \text{ then,}$$

$$p(t) = cos(2t) + sin(2t)$$

$$\text{So,}$$

$$p'(t) = -2sin(2t) + 2cos(2t) = 0$$

From here we obtain,

$$t = \frac{tan^{-1}(\frac{1}{2})}{2}$$

I'm stuck again now.

How do I go from here? I need to get the actual value of this so I can see how big it is in order to classify it.

EDIT: Since,

$$tan(2t) = 1/2$$

I can draw a equilateral triangle and get the values for

$$cos(2t),sin(2t)$$

and plug them into p(t)

giving me $$\frac{3}{\sqrt{5}}$$ as a critical point on the edge of C3.

Is this correct?

I can also consider the endpoints of C3, i.e.

$$p(0) = 2, p(\frac{\pi}{2}) = -2$$

Can I conclude that absolute max/min's in the region R are z=2 and z= -2?

 

[Dick]

Okay so,

$$x = cos(t), y=sin(t) \text{ then,}$$

$$p(t) = cos(2t) + sin(2t)$$

$$\text{So,}$$

$$p'(t) = -2sin(2t) + 2cos(2t) = 0$$

From here we obtain,

$$t = \frac{tan^{-1}(\frac{1}{2})}{2}$$

I'm stuck again now.

How do I go from here? I need to get the actual value of this so I can see how big it is in order to classify it.

EDIT: Since,

$$tan(2t) = 1/2$$

I can draw a equilateral triangle and get the values for

$$cos(2t),sin(2t)$$

and plug them into p(t)

giving me $$\frac{3}{\sqrt{5}}$$ as a critical point on the edge of C3.

Is this correct?

I can also consider the endpoints of C3, i.e.

$$p(0) = 2, p(\frac{\pi}{2}) = -2$$

Can I conclude that absolute max/min's in the region R are z=2 and z= -2?

I get tan(2t)=1 for a critical point. And the endpoints of t for C3 are t=0 and t=pi/4, aren't they? That makes p(0)=1 and p(pi/4)=(-1), which it should be if you look at C1 and C2. You've got the right ideas, but you are scrambling some stuff up.

 

[jegues]

I get tan(2t)=1 for a critical point. And the endpoints of t for C3 are t=0 and t=pi/4, aren't they? That makes p(0)=1 and p(pi/4)=(-1), which it should be if you look at C1 and C2. You've got the right ideas, but you are scrambling some stuff up.

Thanks for helping me clear things up Dick.

Due to all my little mistakes I decided to redo the entire problem from the start again.

Here are my results/work. (See figure attached)

Do things look better this time?

EDIT: ***NOTE***: The conclusion I made at the bottom of the first page should really be at the bottom of the second page.

 

[Dick]

Thanks for helping me clear things up Dick.

Due to all my little mistakes I decided to redo the entire problem from the start again.

Here are my results/work. (See figure attached)

Do things look better this time?

EDIT: ***NOTE***: The conclusion I made at the bottom of the first page should really be at the bottom of the second page.

Better, but still sloppy errors. Try and at least do a consistancy check between your curves, h(1) should be equal to p(0) and g(1) should be equal to p(pi/4), right? (I would also check if 1 is the correct value to substitute into g(x) to get the endpoint). Can you figure out what's causing them to disagree? (One thing to watch out for is if you are trying to find the value of f(a) DON'T substitute a into f'(x), substitute it into f(x)!).

 

[jegues]

Better, but still sloppy errors. Try and at least do a consistancy check between your curves, h(1) should be equal to p(0) and g(1) should be equal to p(pi/4), right? (I would also check if 1 is the correct value to substitute into g(x) to get the endpoint). Can you figure out what's causing them to disagree? (One thing to watch out for is if you are trying to find the value of f(a) DON'T substitute a into f'(x), substitute it into f(x)!).

Okay I can see now that something is wrong since my end points on each curve are not in agreement so I'm going to try to work things out piece by piece.

First,

I would also check if 1 is the correct value to substitute into g(x) to get the endpoint

Okay so,

$$g(x) = 2x^{2}$$

Since g(x) is always positive in the region R, g reaches at maximum at the maximum value of x in R on C1.

This x value should be at, $$\frac{1}{\sqrt{2}}$$ not 1.

So, $$g(\frac{1}{\sqrt{2}}) = \framebox{1}$$

One thing to watch out for is if you are trying to find the value of f(a) DON'T substitute a into f'(x), substitute it into f(x)

Whoops! I see what I did wrong now,

$$p(0) = \framebox{1},p(\frac{\pi}{4}) = \framebox{1}$$

Things are looking better now I hope.

 

[Dick]

Okay I can see now that something is wrong since my end points on each curve are not in agreement so I'm going to try to work things out piece by piece.

First,



Okay so,

$$g(x) = 2x^{2}$$

Since g(x) is always positive in the region R, g reaches at maximum at the maximum value of x in R on C1.

This x value should be at, $$\frac{1}{\sqrt{2}}$$ not 1.

So, $$g(\frac{1}{\sqrt{2}}) = \framebox{1}$$



Whoops! I see what I did wrong now,

$$p(0) = \framebox{1},p(\frac{\pi}{4}) = \framebox{1}$$

Things are looking better now I hope.

Much better. I think all that's left is stating the conclusion about absolute max and min.

 

[jegues]

Much better. I think all that's left is stating the conclusion about absolute max and min.

The absolute max should be when z=2 and absolute min when z=0.

Is it necessary to state the x and y values as well when making this conclusion?

 

[Dick]

The absolute max should be when z=2 and absolute min when z=0.

Is it necessary to state the x and y values as well when making this conclusion?

I would definitely do that since I don't see any (x,y) point where f(x,y)=2 on C1, C2 or C3. Do you? Now I think you are forgetting you changed your mind about some earlier answers you gave. That's kind of annoying. Are you just looking at this stuff once a day and forgetting everything that came before?

 

[jegues]

I would definitely do that since I don't see any (x,y) point where f(x,y)=2 on C1, C2 or C3. Do you? Now I think you are forgetting you changed your mind about some earlier answers you gave. That's kind of annoying. Are you just looking at this stuff once a day and forgetting everything that came before?

Whoops! I guess looking at work that is incorrect isn't a good idea.

That max should be when f(x,y) = 1.

 

[Dick]

Whoops! I guess looking at work that is incorrect isn't a good idea.

That max should be when f(x,y) = 1.

Didn't you once upon a time find an (x,y) point where f(x,y)=sqrt(2)? I really recall that pretty distinctly. Uh, yes, discard old work. Or at least correct it. Sheesh. I think you deleted it from the problem thread. Wasn't it along C3 where tan(2*t)=1? You know, sloppiness is your enemy much more than actually having problems with knowing how to solve these things. Ok, it's still in the blurry thumbnail in post 7.