Max/Min of x2−2xy+7y2 on Ellipse x2+4y2=1 w/ Lagrange Multiplier

In summary, the conversation discusses the use of the Lagrange Multiplier method to find the maximum and minimum values of x2 − 2xy + 7y2 on the ellipse x2 + 4y2 = 1. The conversation also addresses some minor issues with the method, such as the inclusion of z and the handling of the constant λ. The final step is to plug the critical points back into the original function to determine if they are maximum or minimum points.
  • #1
plexus0208
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Homework Statement


Use the Lagrange Multiplier method to find the maximum and minimum values of x2 − 2xy + 7y2 on the ellipse x2 + 4y2 = 1.

Homework Equations


Lagrange multiplier method

The Attempt at a Solution


L(x,y,z,λ) = x2 − 2xy + 7y2 - λ(x2 + 4y2 - 1)
Find Lx, Ly, Lλ
Then, solve for x and y?
 
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  • #2
Where did you get the L(x,y,z,λ)? You don't need the z in there because we're dealing with functions of x and y, not functions of x, y, and z.

Also, λ is a constant: you don't need to solve for Lλ. You don't need to solve for Lz either because there's no such variable as z. Just set Lx=0 and Ly=0 and solve. Don't forget to satisfy the initial constraint of x^2+4y^2=1!
 
  • #3
Just an added note: some people are taught to find [itex]L_\lambda[/itex]. Because if we are trying to extremize F(x) subject to the constraint G(x)= 0, we look at [itex]L= F(x)+ \lambda G(x)[/itex], [itex]L_{\lamba}= G(x)= 0[/itex] is just the constraint again.
 
  • #4
@ideasrule: Sorry, ignore the "z."
@HallsofIvy: Thanks. And yes, that's how I learned it.

Also, once I find the critical points, do I just plug them back into the original function I want to extremize to see if it's a max or min?
 

FAQ: Max/Min of x2−2xy+7y2 on Ellipse x2+4y2=1 w/ Lagrange Multiplier

1. What is the purpose of using Lagrange Multiplier in this problem?

The purpose of using Lagrange Multiplier is to find the maximum or minimum value of a function subject to a constraint. In this case, we want to find the maximum or minimum value of the function x2−2xy+7y2 on the ellipse x2+4y2=1, while also satisfying the constraint of the ellipse equation.

2. How do you set up the Lagrange Multiplier equation for this problem?

To set up the Lagrange Multiplier equation, we first write out the function we want to optimize (x2−2xy+7y2) and the constraint equation (x2+4y2=1). Then, we introduce a new variable λ (the Lagrange Multiplier) and set up the following equation: ∇f(x,y) = λ∇g(x,y), where ∇f(x,y) is the gradient of the function and ∇g(x,y) is the gradient of the constraint equation.

3. How do you solve for the critical points using Lagrange Multiplier?

To solve for the critical points, we first find the partial derivatives of the function and the constraint equation with respect to x and y. Then, we set those derivatives equal to λ times the partial derivatives of the function and constraint equation with respect to λ. We can then solve this system of equations to find the critical points.

4. How do you determine if the critical points are maximum, minimum, or neither?

To determine if the critical points are maximum, minimum, or neither, we use the second derivative test. We take the second derivative of the function and evaluate it at each critical point. If the second derivative is positive, then the critical point is a minimum. If the second derivative is negative, then the critical point is a maximum. If the second derivative is zero, then we need to use another method to determine the nature of the critical point.

5. What is the significance of the solution obtained from using Lagrange Multiplier?

The solution obtained from using Lagrange Multiplier is the maximum or minimum value of the function x2−2xy+7y2 on the ellipse x2+4y2=1, while also satisfying the constraint of the ellipse equation. This solution is significant because it allows us to optimize the function while taking into account the constraint, which may not be possible using other methods.

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