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 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.