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Optimization z=xy^2 - 5

  • #1

Homework Statement



Find the global max/min for z=xy^2 - 5 on the region bounded by y=x and y=1-x^2 in the xy-plane.

Homework Equations





The Attempt at a Solution



I found the critical point of z=xy^2 - 5 at (0,0), but I do not know how to relate this to the boundary.
 

Answers and Replies

  • #2
Dick
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Homework Statement



Find the global max/min for z=xy^2 - 5 on the region bounded by y=x and y=1-x^2 in the xy-plane.

Homework Equations





The Attempt at a Solution



I found the critical point of z=xy^2 - 5 at (0,0), but I do not know how to relate this to the boundary.
How did you get just (0,0) for a critical point? I think you have a whole line of critical points. Can you show us? And to handle the boundary find the intersection between the two boundary curves. Then substitute the boundary condition into your expression for z and extremize that. That will give you special points to look at along the boundary.
 
  • #3
Ray Vickson
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Homework Statement



Find the global max/min for z=xy^2 - 5 on the region bounded by y=x and y=1-x^2 in the xy-plane.

Homework Equations





The Attempt at a Solution



I found the critical point of z=xy^2 - 5 at (0,0), but I do not know how to relate this to the boundary.
Why would you look at critical points of z? They have nothing to do with the problem here.

Let me expand: you have a "feasible region" F consisting of points inside or on the boundary of some figure. First: what is that figure? (Yes, I mean draw a picture.) If an optimum happens to lie INSIDE the figure, then, indeed, it must be a critical point. However, if an optimum (x,y) lies on the boundary of the figure, then there is no reason at all to suppose that it is a critical point.

Your case is trickier than some, because you may have a solution on both parts of the boundary, or on only one part; that is, you may have y = x or y = 1-x^2 or both.

RGV
 
  • #4


How did you get just (0,0) for a critical point? I think you have a whole line of critical points. Can you show us? And to handle the boundary find the intersection between the two boundary curves. Then substitute the boundary condition into your expression for z and extremize that. That will give you special points to look at along the boundary.
dz/dx= y^2
dz/dy= 2xy

Then I set each one equal to zero and solved the system. (0,0) was the solution and so the critical point. (Critical points exist where the gradient is equal to zero.)

Will the intersection of the boundary be something like x=((sqrt5)-1)/2 or -((sqrt5)+1)/2?
 
  • #5
Dick
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dz/dx= y^2
dz/dy= 2xy

Then I set each one equal to zero and solved the system. (0,0) was the solution and so the critical point. (Critical points exist where the gradient is equal to zero.)

Will the intersection of the boundary be something like x=((sqrt5)-1)/2 or -((sqrt5)+1)/2?
Yes, I think you've got the boundary correct. But (x,0) is also a critical point for any value of x, isn't it?
 
  • #6


Yes, I think you've got the boundary correct. But (x,0) is also a critical point for any value of x, isn't it?
Yes.

After plugging x into the expression for z, would I just set y=0 and solve for z?
 
  • #7
Dick
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Yes.

After plugging x into the expression for z, would I just set y=0 and solve for z?
Sure. That should be pretty easy, right? It's z=(-5) all along the critical line, yes? Now work on the boundaries.
 

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