How Do You Identify Critical Points in a Bounded Rectangle?

[cdotter]

Homework Statement

f(x,y)=xy-y^2 is bound by a rectangle with vertices at (2,-1), (2,2), (-1,2), and (-1,-1.)

Find all 2D, 1D, and 0D critical points.

Homework Equations

The Attempt at a Solution

2D Points:
[tex]f(x,y)=xy-y^2[/tex]
[tex]f_x=y=0 \Rightarrow y=0[/tex]
[tex]f_y=x-2y=0 \rightarrow x-2(0)=0 \Rightarrow x=0[/tex]
Therefore the critical point is (0,0)

1D Points:

The domain is a rectangle. The edges are
Top: (x,2) [itex]\rightarrow (x)(2)-2^2=2x-4; \frac{d}{dx}=2[/itex]
Bottom: (x,-1) [itex]\rightarrow(x)(-1)-(-1)^2=-x-1; \frac{d}{dx}=-1[/itex]
Left: (-1,y) [itex]\rightarrow(-1)(y)-y^2=-x-1; \frac{d}{dy}=-y-y^2=0 \Rightarrow y=-\frac{1}{2}[/itex]
Right: (2,y) [itex]\rightarrow(2)(y)-y^2=2y-y^2; \frac{d}{dy}=2-2y=0 \Rightarrow y=1[/itex]

That gives me two 1D points: (-1, -1/2) from the left edge and (2,1) from the right edge. My professor says the left edge is (-1/2, 2.) What am I doing wrong? That doesn't even make sense because 2 is a y coordinate which would be a top or bottom edge.

0D Points: The rectangle vertices.

 

[hunt_mat]

What about saddle points?

 

[cdotter]

What about saddle points?

Sorry, I'm not sure how to do those?

 

[HallsofIvy]

Since the problem asked for max and min, saddle points are irrelevant.

Surely your professor did NOT say "the left edge is (-1/2, 2.)". That makes no sense- the left edge is a line segment, not a point. He/she may have said that "a critical point on the left edge is (-1/2, 2)", the vertex where the top and left edges join.

The difficulty is that you checked the "2 dimensional" case, finding critical points in the interior of the square and then its boundary, the "1 dimensional case", the four edges. But you did not check the boundaries of those edges, the "0 dimensional case, the vertices of the square.

Since the derivative of the given function always exists, max and min may occur at any of three kinds of points:

1) In the interior where the gradient is 0. Here, (0, 0).
2) On the edges where the derivative is 0. Here, (-1, -1/2) and (2, 1).
3) At endpoints of the edges (whether a derivative there is 0 or not). Here, (2,-1), (2,2), (-1,2), and (-1,-1).

Evaluate the target function at each of those 7 points.

 

[cdotter]

Yes, he has (-1/2, 2) listed as a 2-d critical point on the left edge.

My confusion is that (-1/2, 2) does not lie on the left edge of the rectangle. The left edge should have the form (-1, y), right? How did he calculate (-1/2, 2)?

 

[cdotter]

Bump.

 

[cdotter]

Does anyone know how he calculated (-1/2, 2)? I still can't figure it out. Maybe it's a mistake.

 

[SammyS]

Yes, he has (-1/2, 2) listed as a 2-d critical point on the left edge.

My confusion is that (-1/2, 2) does not lie on the left edge of the rectangle. The left edge should have the form (-1, y), right? How did he calculate (-1/2, 2)?

"My confusion is that (-1/2, 2) does not lie on the left edge of the rectangle." You are correct that (-1/2, 2) does not lie on the left edge!

"The left edge should have the form (-1, y), right?" Correct again.

"How did he calculate (-1/2, 2)?" I have no idea regarding how he calculated (-1/2, 2). Clearly it's a mistake.

HallsofIvy did a thorough job of analyzing this problem.

(Added in Edit): Those of us who teach, do occasionally make mistakes. Students need to gain enough confidence in their own abilities to recognize these as mistakes. Then, being very tactful, ask the teacher how he/she arrived at such & such a result.