To find the minimum value of the function f(x, y) = 2x^2 + 3y^2 - 4x - 6y + 9 subject to the constraint x^2 + y^2 = 1, we can use the method of Lagrange multipliers.
First, we need to set up the Lagrangian function L:
L(x, y, λ) = f(x, y) - λ(g(x, y))
where g(x, y) is the constraint function, which is x^2 + y^2 - 1 = 0.
Answers & Comments
Step-by-step explanation:
To find the minimum value of the function f(x, y) = 2x^2 + 3y^2 - 4x - 6y + 9 subject to the constraint x^2 + y^2 = 1, we can use the method of Lagrange multipliers.
First, we need to set up the Lagrangian function L:
L(x, y, λ) = f(x, y) - λ(g(x, y))
where g(x, y) is the constraint function, which is x^2 + y^2 - 1 = 0.
So, we have:
L(x, y, λ) = 2x^2 + 3y^2 - 4x - 6y + 9 - λ(x^2 + y^2 - 1)
Next, we need to find the partial derivatives of L with respect to x, y, and λ, and set them equal to zero:
∂L/∂x = 4x - 4λx = 0
∂L/∂y = 6y - 6λy = 0
∂L/∂λ = x^2 + y^2 - 1 = 0
Solving these equations, we get:
x = ±√(3/11)
y = ±√(8/11)
λ = 4/11
Now, we need to plug these values back into the Lagrangian function to find the minimum value of f(x, y):
L(√(3/11), √(8/11), 4/11) = -1/11
L(-√(3/11), -√(8/11), 4/11) = -1/11
So, the minimum value of the function f(x, y) = 2x^2 + 3y^2 - 4x - 6y + 9 subject to the constraint x^2 + y^2 = 1 is -1/11.