To find the local minimum value of the function f(x, y) = x^2 + 2y^2 subject to the constraint g(x, y) = x + y = 3, we can use the method of Lagrange multipliers.
Step 1: Define the Lagrangian function
The Lagrangian function is defined as L(x, y, λ) = f(x, y) - λg(x, y), where λ is the Lagrange multiplier.
L(x, y, λ) = x^2 + 2y^2 - λ(x + y - 3)
Step 2: Calculate the partial derivatives
Take the partial derivatives of L with respect to x, y, and λ:
∂L/∂x = 2x - λ
∂L/∂y = 4y - λ
∂L/∂λ = -(x + y - 3)
Step 3: Set the partial derivatives to zero
Set ∂L/∂x = 0, ∂L/∂y = 0, and ∂L/∂λ = 0, and solve the resulting system of equations to find the critical points.
2x - λ = 0
4y - λ = 0
x + y - 3 = 0
From the first equation, we have λ = 2x.
Substituting this value of λ into the second equation, we get 4y - 2x = 0, which simplifies to y = x/2.
Substituting these values of λ and y into the third equation, we have x + x/2 - 3 = 0, which simplifies to x = 2.
Substituting x = 2 into y = x/2, we get y = 1.
So, the critical point is (2, 1).
Step 4: Determine the nature of the critical point
To determine whether the critical point is a local minimum, maximum, or saddle point, we need to evaluate the second partial derivatives.
∂^2L/∂x^2 = 2
∂^2L/∂y^2 = 4
∂^2L/∂x∂y = 0
D = (∂^2L/∂x^2)(∂^2L/∂y^2) - (∂^2L/∂x∂y)^2 = (2)(4) - (0)^2 = 8.
Since D > 0 and (∂^2L/∂x^2) > 0, the critical point (2, 1) is a local minimum.
Therefore, the local minimum value of f(x, y) = x^2 + 2y^2 subject to g(x, y) = x + y = 3 is obtained at the point (2, 1).