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

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Homework Help Overview

The discussion revolves around finding the minimum and maximum values of the function f(x,y) = xy under the constraint 4x² + 9y² = 32. Participants are exploring the application of Lagrange multipliers in this context.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the method of Lagrange multipliers but encounters difficulty in determining the minimum value. They express confusion regarding the signs of x and y in their calculations. Other participants suggest considering additional combinations of x and y values to explore the function further.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the signs of variables and discussing the setup for problems involving multiple variables and constraints. There is no explicit consensus on the minimum value, but suggestions for further exploration have been made.

Contextual Notes

There is mention of potential confusion arising from the textbook's presentation of the problem, particularly regarding the signs of the variables. The discussion also touches on extending the method to problems with three variables and one constraint.

snoggerT
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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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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.
 
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?
 
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.
 

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