Maximizing z=xy^2-5 on a Bounded Region in the xy-Plane

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In summary, optimization is the process of finding the best solution to a problem with given constraints and objectives. This involves finding the maximum or minimum value of a function. The function z=xy^2 - 5 represents a system with two independent variables and one dependent variable, often used in optimization problems. To find the optimal solution for z=xy^2 - 5, various techniques such as gradient descent, Newton's method, or simplex method can be used. Constraints in optimization problems are limitations that must be considered, such as physical or financial restrictions. Optimization is widely used in real-world applications to design efficient systems, maximize profits, minimize costs, and improve performance. Examples include optimizing traffic flow, designing supply chains, and finding the best
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countzander
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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.
 
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  • #2


countzander said:

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


countzander said:

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


Dick said:
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


countzander said:
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


Dick said:
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


countzander said:
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.
 

FAQ: Maximizing z=xy^2-5 on a Bounded Region in the xy-Plane

What is optimization?

Optimization is the process of finding the best solution to a problem, given a set of constraints and objectives. In mathematical terms, it involves finding the maximum or minimum value of a function.

What does z=xy^2 - 5 represent?

The function z=xy^2 - 5 is a mathematical representation of a system or process that has two independent variables, x and y, and one dependent variable, z. It is often used in optimization problems to represent a quantity that needs to be maximized or minimized.

How do you find the optimal solution for z=xy^2 - 5?

To find the optimal solution for z=xy^2 - 5, you can use various optimization techniques such as the gradient descent method, the Newton's method, or the simplex method. These methods involve finding the critical points of the function and evaluating them to determine the optimal solution.

What are the constraints in an optimization problem?

Constraints in an optimization problem are restrictions or limitations that must be considered when finding the optimal solution. These can include physical, financial, or other limitations that the solution must adhere to. In the function z=xy^2 - 5, the constraints could be limits on the values of x and y, such as x>0 and y<10.

How is optimization used in real-world applications?

Optimization is used in a wide range of real-world applications, from engineering and physics to economics and business. It is used to design efficient systems, maximize profits, minimize costs, and improve performance. Some examples include optimizing traffic flow, designing efficient supply chains, and finding the best investment portfolio.

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