Método de Multiplicadores de Lagrange
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Questions and Answers

¿Cuál es la forma de la ecuación que define el conjunto de puntos en el que se optimiza la función f(x, y)?

  • $x^2 + y^2 = 3$
  • $x^2 - 2y^2 = 3$
  • $2x^2 + y^2 = 3$
  • $x^2 + 2y^2 = 3$ (correct)
  • ¿Qué variable se está optimizando en la función f(x, y) = x?

  • f(x, y)
  • x^2
  • y
  • x (correct)
  • Dentro del conjunto de puntos definido, ¿cuál es un posible valor máximo de x?

  • $1$
  • $ rac{5}{2}$
  • $ rac{3}{ rac{ u}{2}}$
  • $ rac{3}{2}$ (correct)
  • ¿Qué tipo de figura geométrica describe el conjunto de soluciones de la ecuación $x^2 + 2y^2 = 3$?

    <p>Elipse</p> Signup and view all the answers

    Si se sustituye $x = 0$ en la ecuación $x^2 + 2y^2 = 3$, ¿cuál es el valor máximo de y?

    <p>$ rac{ u}{2}$</p> Signup and view all the answers

    Study Notes

    • The objective function is f(x, y) = x.
    • The constraint is x² + 2y² = 3.
    • The problem involves maximizing/minimizing f(x, y) subject to the constraint.

    Method of Lagrange Multipliers

    • This method helps find the extreme values of a function subject to constraints.
    • A Lagrange multiplier (λ) is introduced.
    • The Lagrangian function is defined as L(x, y, λ) = f(x, y) - λ(g(x, y) - c), where g(x, y) is the constraint function (x² + 2y² = 3) and c is the constraint value (3 in this case).
    • In this case, L(x, y, λ) = x - λ(x² + 2y² - 3)
    • The partial derivatives of the Lagrangian with respect to x, y, and λ are set to zero:
      • ∂L/∂x = 1 - 2λx = 0
      • ∂L/∂y = -4λy = 0
      • ∂L/∂λ = x² + 2y² - 3 = 0

    Solving the System of Equations

    • From ∂L/∂y = -4λy = 0, it follows that either y = 0 or λ = 0.

    • Case 1: y = 0

      • Substituting y = 0 into the constraint equation, x² + 2(0)² = 3 implies x² = 3, so x = ±√3
      • If x = √3, f(x, y) = √3
      • If x = -√3, f(x, y) = -√3
    • Case 2: λ = 0

      • If λ = 0, from ∂L/∂x = 1 - 2λx = 0 , 1 = 0 which is impossible.
      • Thus, λ cannot be zero.

    Conclusion

    • The maximum value of f(x, y) = x occurs at x = √3, y = 0, with f(√3, 0) = √3.
    • The minimum value of f(x, y) = x occurs at x = -√3, y = 0, with f(-√3, 0) = -√3.
    • The points (√3, 0) and (-√3, 0) are the critical points on the constraint curve.

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    Description

    Explora el método de los multiplicadores de Lagrange para maximizar o minimizar funciones sujeta a restricciones. Este cuestionario te guiará a través de la formulación de la función de Lagrange y la resolución del sistema de ecuaciones derivadas. Practica cómo aplicar esta técnica a diferentes problemas de optimización.

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