To find the maximum and minimum points of the function f(x, y) = y^2 - x^2 subject to the constraint g(x, y) = 0.25x^2 + y^2 = 1, you correctly set up the Lagrange multipliers method, which involves finding points where the gradient of f(x, y) is parallel to the gradient of g(x, y), and also satisfies the constraint equation.
The partial derivatives you computed are correct:
fx = -2x
fy = 2y
gx = 0.5x
gy = 2y
Now, to set up the Lagrange multiplier equation, you need to equate the two gradients scaled by the Lagrange multiplier λ:
-2x i + 2y j = λ (0.5x i + 2y j)
To solve this equation, we can separate it into two component equations:
-2x = 0.5λx (1)
2y = 2λy (2)
From equation (2), we can conclude that either y = 0 or λ = 1. If y = 0, then 0.25x^2 = 1 implies x = ±2. If λ = 1, then from equation (1), we have -2x = 0.5x, which implies x = 0.
Now, let's substitute these values back into the constraint equation g(x, y) = 0.25x^2 + y^2 = 1:
For (±2, 0):
0.25(±2)^2 + 0^2 = 1
=> 1 + 0 = 1
The points (2, 0) and (-2, 0) satisfy the constraint equation.
For (0, ±1):
0.25(0)^2 + (±1)^2 = 1
=> 0 + 1 = 1
The points (0, 1) and (0, -1) also satisfy the constraint equation.
To determine whether these critical points are maximum or minimum points, we can use the second partial derivative test:
fxx = d^2(fx)/d(x)^2 = d(-2x)/dx = -2
fyy = d^2(fy)/d(y)^2 = d(2y)/dy = 2
For (0, ±1), we have fxx = -2 and fyy = 2. Since fxx < 0 and fyy > 0, this indicates that these points are saddle points, not maxima or minima.
For (±2, 0), we have fxx = -2 and fyy = 2. Again, fxx < 0 and fyy > 0, which indicates that these points are also saddle points.
Therefore, the book may have made an error when stating that (2, 0) and (-2, 0) are minimum points at f(±2,0) = -4 and z = 4. Based on the calculations, it seems that these points are saddle points.
If you suspect an error in the book, it would be worth rechecking your work and considering seeking additional resources or consulting with a tutor or instructor for further clarification.