Find Min/Max of f(x,y)=xy with Constraint 4x^2+9y^2=32

In summary, the conversation discusses finding the minimum and maximum values of a function with a constraint using Lagrange problems. The speaker is having trouble getting the correct minimum and is questioning their approach. Another speaker suggests trying different combinations of positive and negative values for x and y. The conversation also briefly touches on solving for unknowns in a Lagrange problem with three variables and one constraint.
  • #1
snoggerT
186
0
find min/max:

f(x,y)=xy with constraint being 4x^2+9y^2=32

[gradient]f=[lambda]gradient g

The Attempt at a Solution



I thought I understood the Lagrange problems, but I can't seem to get the minimum right on the last few problems. I get x=+/-2 and then plug back into find y, then I use my critical points to find my min/max in f(x,y). I got 8/3 for my max on my problem (which is right), but can't get the minimum right. I set it up as such:

f(-2,-4/3)=xy and get +8/3 again, but the answer in the back of the book is -8/3 for the minimum. What am I doing wrong?
 
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  • #2
You aren't doing anything wrong. But how about f(2,-4/3) or f(-2,4/3)? Nothing in the problem forces x and y to have the same sign.
 
  • #3
Dick said:
You aren't doing anything wrong. But how about f(2,-4/3) or f(-2,4/3)? Nothing in the problem forces x and y to have the same sign.

- Alright, then that makes sense. The book isn't very good at pointing things out like that.

while on the topic of lagrange...When you get into having 3 variables and 1 constraint, would you set up the problem as [lambda]=x=y=z? If so, how would you solve for the unknowns?
 
  • #4
Well, you'll have three equations involving x,y,z and lambda coming from the gradient. Then you have the constraint equation in x,y,z. That's four equations in four unknowns. There's no special way to set it up, just derive the equations from the partial derivatives.
 

1. What is the purpose of finding the min/max of a function with a constraint?

Finding the minimum or maximum value of a function with a constraint helps us to optimize the function and find the most efficient solution to a problem. In this case, we are looking for the minimum or maximum value of f(x,y)=xy while satisfying the constraint 4x^2+9y^2=32.

2. How do you approach finding the min/max of a function with a constraint?

The first step is to find the critical points of the function by taking the partial derivatives with respect to x and y, setting them equal to 0, and solving for x and y. Then, we use the second derivative test to determine if these points are minimum or maximum values. Finally, we check if these critical points satisfy the constraint to find the minimum or maximum value.

3. What is the significance of the constraint in this problem?

The constraint 4x^2+9y^2=32 represents a boundary or limitation that the function must satisfy. It narrows down the possible values of x and y and helps us find the optimal solution for f(x,y)=xy.

4. Can there be multiple solutions for the min/max of this function with the given constraint?

Yes, there can be multiple solutions for the min/max of f(x,y)=xy with the constraint 4x^2+9y^2=32. This is because the constraint forms an ellipse, and the function can have multiple critical points within this boundary. It is important to check all critical points to find the global minimum or maximum value.

5. How can we visualize the min/max of this function with the constraint?

The min/max of f(x,y)=xy with the constraint 4x^2+9y^2=32 can be visualized by graphing the function and the constraint on a coordinate plane. The critical points will be the points where the function and the constraint intersect. We can also use contour plots to visualize the behavior of the function within the constraint.

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