Global max/min (multivariable)

[sandy.bridge]

Homework Statement

Find the global maximum and global minimum of
f(x, y)=x^3+y^2
in the half disk x^2+y^2\leq{1}, y\geq{0}

This is a rather new topic for me, so I am looking for clarification for problems such as these.

The Attempt at a Solution

First, I determined the critical points of the function.
\nabla{f(x, y)=3x^2i+2yj} and thus has a critical point at
(x, y)=(0, 0)

Next, we will consider along the boundary of the half disk. Since the disk is a half circle, I must include the endpoints x=1, x=-1.

We have,
x=cost, y=sint, (0, \pi{)}
We have,
f(cost, sint)=cos^3t+sin^2t=g(t)
Moreover,
-3cos^2tsint+2sintcost=0\rightarrow{cost=2/3}
g(cos^{-1}2/3)=0.852
Next, checking the boundaries at t=0 and t=π
g(0)=1, g(π)=-1

Therefore, f has a global max at the boundary on the half disk of 1 at (1, 0) and a global min of -1 at (-1, 0).

However, I executed two of the tests for determining if critical points are max/mins for the point (0. 0) however all the tests were inconclusive. What else can be done here?

 

[SammyS]

You checked along the boundary, x² + y² = 1, but to be thorough, you should also check along the boundary, y = 0 .

Regarding (0,0): Are the second derivatives of f defined at (0,0) ?

 

[sandy.bridge]

Okay, not entirely sure how to show work for checking along y=0. Moreover, I believe the second derivatives are defined at (0, 0), and if they are not, I don't see why considering it's a cubic and square function.

 

[I like Serena]

Hi sandy.bridge!

Okay, not entirely sure how to show work for checking along y=0. Moreover, I believe the second derivatives are defined at (0, 0), and if they are not, I don't see why considering it's a cubic and square function.

In (0,0) you'll find that the Hessian 2nd derivative test is inconclusive.
What else can be done?
Well, since there are no other extrema nearby, simply check a few points around (0,0).
Since it's already likely that it is a saddle point, you can verify this by finding a positive and negative value near (0,0).

As for checking along y=0.
What about x=t, y=0?
In particular you will find that (0,0) is a saddle point.

 

[sandy.bridge]

Hello!
Okay, moving the left of the origin along the x-axis, we have
f(-.1, 0)=(-.1)^3
and ro the right
f(.1, 0)=0.1^3
therefore, this illuminates the fact this is a saddle point.
As for your suggestion of x=t, how can this be done? Is "t" not denoting the angle measurement with respect to the positive x-axis?

 

[I like Serena]

Good!

"t" can be anything you want. It's just an arbitrary parameter with an arbitrary range.
You could also use simply (x,0) where x is the parameter running from -1 to +1.

 

[sandy.bridge]

Alright, then the global max occurs when x=1, or t=0(angle). Also, the min occurs when x=-1 (t=pi).
max(x, y)=(1, 0)
min(x ,y)=(-1, 0)

 

[I like Serena]

And... we're done!