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

In summary, the conversation is about finding the maximum and minimum of the function f(x,y) = x^2 + xy + y^2 on the circle x^2 + y^2 = 1. The discussion includes solving a system of equations, considering possible cases, and using a common method for handling such equations. Ultimately, the solution is obtained by recognizing the constraint and simplifying the function.
  • #1
Benny
584
0
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.
 
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  • #2
Add equations (1) and (2).

Is your constraint g(x,y)=1 or g(x,y)=2?
 
  • #3
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.
 
  • #4
Thanks for the help Galileo and HallsofIvy.

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

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

1. What is the purpose of using Lagrange multipliers in finding the maximum and minimum of a function?

Lagrange multipliers are used to find the extreme values (maximum or minimum) of a function subject to one or more constraints. This method allows us to incorporate the constraints into the optimization problem and find the optimal solution.

2. How do Lagrange multipliers work?

Lagrange multipliers use a mathematical technique called the Lagrange multiplier method, which involves finding the critical points of the function and the constraint equations. These critical points are then solved simultaneously to find the maximum or minimum value of the function.

3. What are the necessary conditions for using Lagrange multipliers?

The necessary conditions for using Lagrange multipliers are that the function must be differentiable, the constraints must be differentiable, and the constraints must be independent. Additionally, the constraints must be satisfied at the maximum or minimum point.

4. Can Lagrange multipliers be used for functions with more than two variables?

Yes, Lagrange multipliers can be used for functions with any number of variables. The method remains the same, but the calculations may become more complex with more variables.

5. Are there any limitations or drawbacks to using Lagrange multipliers?

One limitation of using Lagrange multipliers is that they can only be used for finding the maximum or minimum of a function subject to equality constraints. Also, the method may not always yield a solution, and in some cases, the solution may not be unique.

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