Solving Maximum & Minimum of f(x,y) with Lagrange Multipliers

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Homework Help Overview

The discussion revolves around finding the maximum and minimum values of the function f(x,y) = x^2 + xy + y^2 subject to the constraint of the circle defined by x^2 + y^2 = 1. The problem involves the application of Lagrange multipliers.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to set up the Lagrange multiplier equations but expresses uncertainty about solving the resulting system. They consider special cases for x and y but encounter contradictions with the constraint.
  • Some participants suggest adding the equations derived from the Lagrange multipliers or dividing one equation by the other to simplify the problem.
  • There is a question regarding the correct formulation of the constraint function g(x,y).

Discussion Status

Participants are actively engaging with the problem, providing different approaches to simplify the equations. The original poster acknowledges a correction regarding the constraint function. There is no explicit consensus yet, but several lines of reasoning are being explored.

Contextual Notes

The original poster notes confusion regarding the values of x and y to consider, and there is an ongoing discussion about the correct constraint function.

Benny
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Hi, I'm having trouble with the following question.

Q. Find the maximum and minimum of the function f(x,y) = x^2 + xy + y^2 on the circle x^2 + y^2 = 1.

I started off by writing:

Let g(x,y) = x^2 + y^2 then [tex]\nabla f = \lambda \nabla g,g\left( {x,y} \right) = 1[/tex]

[tex] \Rightarrow 2x + y = 2\lambda x...(1)[/tex]

[tex] 2y + x = 2\lambda y...(2)[/tex]

[tex] x^2 + y^2 = 2...(3)[/tex]

I'm not sure how to solve this system of equations. I've got the impression that generally an explicit value for lamda is not needed. I'd normally start off by considering the possible cases.

Here, if x = 0 in (1) then from (2) I get y = 0 so that (x,y) = (0,0). However this contradicts (3) and it isn't on the circle so I ignore this point. But now I'm stuck. Which values of x or y should I try now? Any help would be good thanks.
 
Last edited:
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Add equations (1) and (2).

Is your constraint g(x,y)=1 or g(x,y)=2?
 
A fairly common way of handling equations like that, where you don't need to find [itex]\lambda[/itex] is to divide one equation by the other.
You get [itex]\frac{2x+y}{2y+x}= \frac{x}{y}[/itex] or y(2x+y)= x(2y+ x). That obviously gives x2= y2 so that either y= x or y= -x and that, together with x2+ y2= 1 gives you the answer.

By the way, since we have the constraint x2+ y2= 1, f(x,y)= x2+ xy+ y2 immediately reduces to
f(x,y)= 1+ xy on that circle. That makes the calculations a little easier.
 
Thanks for the help Galileo and HallsofIvy.

The constraint is g(x,y) = 1, I'll fix it up now.
 

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