Respuesta :
Answer:
At maximum point,
x₁ = 1
x₂ = 1
x₃ = 3
x₄ = 5
F(x₁, x₂, x₃, x₄) = 36.
Step-by-step explanation:
F(x₁, x₂, x₃, x₄) = x₁ + x₂ + 3x₃ + 5x₄
Subject to the constraint
C(x₁, x₂, x₃, x₄) = x₁² + x₂² + x₃² + x₄² - 36
Using Lagrange multiplier method
L(x₁, x₂, x₃, x₄, λ) = F(x₁, x₂, x₃, x₄) + λ(C(x₁, x₂, x₃, x₄)
L(x₁, x₂, x₃, x₄, λ) = x₁ + x₂ + 3x₃ + 5x₄ + λ(x₁² + x₂² + x₃² + x₄² - 36)
At the critical points like the maximum point,
(∂L/∂x₁) = (∂L/∂x₂) = (∂L/∂x₃) = (∂L/∂x₄) = (∂L/∂λ) = 0
(∂L/∂x₁) = 1 + 2λx₁ = 0 (eqn 1)
(∂L/∂x₂) = 1 + 2λx₂ = 0 (eqn 2)
(∂L/∂x₃) = 3 + 2λx₃ = 0 (eqn 3)
(∂L/∂x₄) = 5 + 2λx₄ = 0 (eqn 4)
(∂L/∂λ) = x₁² + x₂² + x₃² + x₄² - 36 = 0 (eqn 5)
Equating (eqn 1) and (eqn 2)
1 + 2λx₁ = 1 + 2λx₂
x₁ = x₂
1 + 2λx₁ = 3 + 2λx₃
2λx₃ + 2 = 2λx₁
λx₃ + 1 = λx₁
From eqn 1,
1 + 2λx₁ = 0
2λx₁ = -1
λ = (-1/2x₁)
λx₃ + 1 = λx₁
Substituting for λ
x₃(-1/2x₁) + 1 = x₁(-1/2x₁)
x₃(-1/2x₁) + 1 = (-1/2)
x₃(-1/2x₁) = (-3/2)
(x₃/x₁) = 3
x₃ = 3x₁
1 + 2λx₁ = 5 + 2λx₄
2λx₄ + 4 = 2λx₁
λx₄ + 2 = λx₁
λ = (-1/2x₁)
x₄(-1/2x₁) + 2 = (-1/2)
x₄(-1/2x₁) = (-5/2)
(x₄/x₁) = 5
x₄ = 5x₁
x₂ = x₁
x₃ = 3x₁
x₄ = 5x₁
Substituting this into eqn 5
x₁² + x₂² + x₃² + x₄² - 36 = 0
x₁² + x₁² + (3x₁)² + (5x₁)² = 36
x₁² + x₁² + 9x₁² + 25x₁² = 36
36x₁² = 36
x₁² = 1
x₁ = 1 or -1
Since, we're looking for the maximum value of the function,
x₁ = 1 at maximum value.
x₂ = x₁ = 1
x₃ = 3x₁ = 3
x₄ = 5x₁ = 5
F(x₁, x₂, x₃, x₄) = x₁ + x₂ + 3x₃ + 5x₄
At maximum point,
F(x₁, x₂, x₃, x₄) = 1 + 1 + (3×3) + (5×5)
F(x₁, x₂, x₃, x₄) = 36.
Hope this Helps!!!
