How Do Constraints Affect Finding Stationary Points and Global Extrema?

[Pulty]

The function g is defined by g(x,y) = 3 + x^3 - x^2 - y^2 on the domain D given by points in the xy-plane satisfying x^2 + y^2 <=1 and x >= 0.

So I need to find and classify the stationary points of g, and find the global extreme points of g in D.

Do I start with taking partial derivatives and how do the constraints affect the problem?

Thanks

 

[MarkFL]

First, you want to find any critical points of $g$ in the interior of $D$. So, solve the system:

$$g_x(x,y)=0$$

$$g_y(x,y)=0$$

And discard any points not in the interior. What do you find?

 

[Pulty]

Taking the partials i get;

gx(x,y) = 3x^2 - 2x = 0

solving this I get x = 0, or x = 2/3

gy(x,y) = -2y = 0

solving this I get y = 0

The only point that falls in the domain is (2/3, 0)

 

[MarkFL]

Yes, the origin is on the boundary, not within $D$. So, use the second partials test on your interior critical point to determine its nature. :)

 

[Pulty]

second partials:

gxx(x,y) = 6x - 2

gxy(x,y) = 0

gyy(x,y) = -2

gxx(x,y)gyy(x,y) - [gxy(x,y)]^2 = (2)(-2) - (0)^2 = -4

So the stationary point is a saddle point?

What does finding the global extreme points of g in D mean exactly?

Thanks!

 

[MarkFL]

I agree, the interior critical point is not an extremum. What we are trying to do is find the global extrema for $g$ where $(x,y)$ satisfy the given constraints.

So, now let's look at the left boundary, which is $x=0$ where $-1\le y\le1$. We have:

$$g(0,y)=3-y^2$$

This is a function in one variable only, and so we use the techniques we learned in Calculus I to find the extrema. What are the extrema here, and what are their nature? Recall functions in one variable with a restricted domain must also be checked on the boundaries.

 

[Pulty]

Do i take the derivative of the single variable function and solve it at 0?

y=0?

 

[MarkFL]

Do i take the derivative of the single variable function and solve it at 0?

y=0?

Yes, and don't forget the boundaries $y=\pm1$.

Next, you will want to look at the circular portion of the boundary, and to make things easier we can use parametric equations:

$$x=\cos(t)$$

$$y=\sin(t)$$

$$-\frac{\pi}{2}<t<\frac{\pi}{2}$$

And so we can write $g$ as a function of one variable, $t$. What do you get?

 

[Pulty]

So what I've done so far is:

g(x,y) = 3 + x^3 - x^2 - y^2

x^2+y^2 <= 1, x>=0

gx(x,y) = 3x^2 - 2x = x(3x - 2) = 0

x = 0, or x = 2/3

gy(x,y) = -2y = 0

y = 0

(2/3, 0) interior: saddle point using second partials

x [0, 1], y [-1, 1]

(0, 0) boundary: 3
(1, 0) boundary: 3
(0, -1) boundary: 2 - Min
(0, 1) boundary: 2 - Min

Boundary of S consists of the circle x^2 + y^2 = 1

around boundary: 3 + x^3 (?)

(0, 0): 3 + 0^3 = 3
(1, 0): 3 + 1^3 = 4 - Max

Not sure if this works or I am completely lost.

Thanks

 

[MarkFL]

On the circular portion of the boundary, we may write:

$$g(t)=3+\cos^3(t)-\cos^2(t)-\sin^2(t)=2+\cos^3(t)$$

We don't need to worry about the boundaries of $t$ because we already covered those points when we explored the linear portion of the boundary.

So, we need to set $g'(t)=0$, and solve for $t$ in the given range to find the critical values. What do you find?

 

[Pulty]

g'(t) = -3sin(x) cos^2(x) = 0

x = 0 in the given range?

 

[MarkFL]

g'(t) = -3sin(x) cos^2(x) = 0

x = 0 in the given range?

Replace $x$ with $t$ and that's correct. So, using our parametrizations, to which point does this correspond?

 

[Pulty]

is that just substituting t = 0 back into the function g(t)

g(t) = 2

 

[MarkFL]

is that just substituting t = 0 back into the function g(t)

g(t) = 2

No, recall we used:

$$x=\cos(t)$$

$$y=\sin(t)$$

in order to express $g$ as a function in one variable along the circular portion of the boundary.

edit: Belay that...yes you can use $t=0$ in $g(t)$...sorry for the confusion. Either way works, though. But $g(0)\ne2$.

 

[Pulty]

so,

x = cos(t) = cos(0) = 1
y = sin(t) = sin(0) = 0

(1.0) for the boundary?

 

[MarkFL]

so,

x = cos(t) = cos(0) = 1
y = sin(t) = sin(0) = 0

(1.0) for the boundary?

Yes, $(x,y)=(1,0)$ is a critical point on the circular boundary, so what are you global max and min?

 

[Pulty]

So the circular boundary:

g(1,0) = 3

linear boundary:

g(0,0) = 3 (?)
g(0,-1) = 2 (?)
g(0, 1) = 2 (?)

Interior

(2/3, 0) = saddle

Can there be this many points with the same values and still be extrema?

 

[MarkFL]

So the circular boundary:

g(1,0) = 3

linear boundary:

g(0,0) = 3 (?)
g(0,-1) = 2 (?)
g(0, 1) = 2 (?)

Interior

(2/3, 0) = saddle

Can there be this many points with the same values and still be extrema?

Yes, for example consider the 2D function $y=\sin(x)$...it has an infinite number of relative extrema, however it has only one 1 global minimum and one global maximum. We choose the smallest value from all the critical points as the global minimum and the largest as the global maximum. In the case of the problem you are working on, this means we may write:

$$g_{\min}=2$$

$$g_{\max}=3$$