Hi, this question is about Langrange multipliers.
Given f(x,y) = y^2 - x^2, subject to the constraint g(x,y) = 0.25x^2 + y^2 = 1, find the max and mins.
So I found the partial derivatives for both f(x,y) and g(x,y):
fx = -2x
fy = 2y
gx = 0.5x
gy = 2y
And set them up for the Lagrange multipliers method. In essence, grad f(x,y) = lambda times grad g(x,y).
My algebra lead me to the roots of x=0 from the x partial derivatives and y=0 from the y partial derivatives. I plugged these into g(x,y) and got the points (0, ±1) (±2, 0).
So the book agrees with my critical points being (0, ±1) (±2, 0). The book also agrees that (0, ±1) should both be a max point equal to z=1.
However, the book says (2, 0) and (-2, 0) should both be a min point at z=4. Now if I'm not mistaken, to be a min point you must have fxx(a, b) > 0, but that can't be possible if fxx = d(fx)/dx = (-2x)' = -2. Also, it says f(±2, 0) = -4. Wouldn't that also require the x root being ±16?
Any insight appreciated.
To find the extrema of the function f(x, y) = y^2 - x^2 subject to the constraint g(x, y) = 0.25x^2 + y^2 = 1 using the Lagrange multipliers method, follow these steps:
1. Define the Lagrangian function L(x, y, λ) = f(x, y) - λg(x, y), where λ is the Lagrange multiplier.
2. Calculate the partial derivatives of L with respect to x, y, and λ:
∂L/∂x = ∂f/∂x - λ(∂g/∂x)
= -2x - λ(0.5x)
= -2.5xλ
∂L/∂y = ∂f/∂y - λ(∂g/∂y)
= 2y - λ(2y)
= 2y(1 - λ)
∂L/∂λ = -g(x, y)
= -0.25x^2 - y^2
3. Set the partial derivatives equal to zero and solve the resulting equations:
∂L/∂x = 0: -2.5xλ = 0 => x = 0 or λ = 0
∂L/∂y = 0: 2y(1 - λ) = 0 => y = 0 or λ = 1
∂L/∂λ = 0: -0.25x^2 - y^2 = 0
4. Now consider the possibilities:
Case 1: x = 0
From ∂L/∂λ = 0, we have -y^2 = 0, which gives y = 0. Substituting these values of x and y into g(x, y) = 0.25x^2 + y^2 = 1, we get (0, ±1) as a critical point.
Case 2: λ = 0
From ∂L/∂x = 0, we have -2.5xλ = 0, which gives x = 0. Substituting these values of x and λ into g(x, y) = 0.25x^2 + y^2 = 1, we get (0, ±1) as a critical point.
Case 3: y = 0
From ∂L/∂λ = 0, we have -0.25x^2 = 0, which gives x = 0. Substituting these values of x and y into g(x, y) = 0.25x^2 + y^2 = 1, we get (±2, 0) as critical points.
Case 4: λ = 1
From ∂L/∂y = 0, we have 2y(1 - λ) = 0, which gives y = 0. Substituting these values of y and λ into g(x, y) = 0.25x^2 + y^2 = 1, we get (±2, 0) as a critical point.
So, the critical points are (0, ±1) and (±2, 0).
To determine the nature of these critical points, we need to examine the second-order partial derivatives:
fxx = d^2(f)/dx^2 = -2
fyy = d^2(f)/dy^2 = 2
fxy = d^2(f)/dxdy = 0
For the point (0, ±1), fxx = -2 < 0, implying it is a local maximum. f(0, ±1) = (±1)^2 - 0^2 = 1.
For the points (±2, 0), fxx = -2 < 0, indicating that they are local maxima. f(±2, 0) = 0^2 - (±2)^2 = -4.
Please note that the minima point mentioned in the book might be a typo/error because fxx < 0, which contradicts it being a minimum point.