Question

Find the maximum value of x1 + x2 + 3x3 + 5x4 subject to the condition that x1^2 + x2^2 + x3^2 + x4^2

Answer 1

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!!!