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

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SUMMARY

The discussion focuses on using the Lagrange Multiplier method to find the maximum and minimum values of the function x² − 2xy + 7y² constrained by the ellipse x² + 4y² = 1. Participants clarify that the correct formulation of the Lagrangian is L(x,y,λ) = x² − 2xy + 7y² - λ(x² + 4y² - 1) and emphasize that the variable z is unnecessary in this context. The critical points are determined by setting the partial derivatives Lx and Ly to zero while ensuring the constraint is satisfied. Finally, participants confirm that substituting the critical points back into the original function will reveal whether these points are maxima or minima.

PREREQUISITES
  • Understanding of the Lagrange Multiplier method
  • Familiarity with partial derivatives
  • Knowledge of constraint equations, specifically ellipses
  • Basic algebraic manipulation skills
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  • Study the application of the Lagrange Multiplier method in different contexts
  • Learn about critical point analysis in multivariable calculus
  • Explore the geometric interpretation of Lagrange Multipliers
  • Review optimization problems involving constraints in calculus
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Students in calculus, particularly those studying optimization techniques, as well as educators teaching the Lagrange Multiplier method and its applications in constrained optimization problems.

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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?
 
Last edited:
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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!
 
Just an added note: some people are taught to find L_\lambda. Because if we are trying to extremize F(x) subject to the constraint G(x)= 0, we look at L= F(x)+ \lambda G(x), L_{\lamba}= G(x)= 0 is just the constraint again.
 
@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?
 

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