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 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, you correctly found the partial derivatives of f(x, y) and g(x, y):
fx = -2x
fy = 2y
gx = 0.5x
gy = 2y
Next, you set up the equation for the Lagrange multipliers method:
grad f(x, y) = lambda * grad g(x, y)
So, we have:
-2x * i + 2y * j = lambda * (0.5x * i + 2y * j)
From this equation, we can see that there are two possibilities:
1. lambda = -4x = 2y
2. 2x = 4y = 0
First, let's consider the case when lambda = -4x = 2y:
From -4x = 2y, we can rearrange to get:
y = -2x
Substituting this in the equation of the constraint g:
0.25x^2 + (-2x)^2 = 1
0.25x^2 + 4x^2 = 1
4.25x^2 = 1
x^2 = 1 / 4.25
x = ±sqrt(1 / 4.25)
Since y = -2x, we can find the corresponding y values:
y = -2 * ±sqrt(1 / 4.25)
Thus, we have two critical points: (±sqrt(1 / 4.25), -2 * ±sqrt(1 / 4.25))
Now, let's consider the case when 2x = 4y = 0:
From 2x = 4y = 0, we have two possibilities:
1. When x = 0:
Substituting x = 0 into the constraint g:
0.25(0)^2 + y^2 = 1
y^2 = 1
y = ±1
2. When y = 0:
Substituting y = 0 into the constraint g:
0.25x^2 + (0)^2 = 1
0.25x^2 = 1
x^2 = 1 / 0.25
x = ±2
Thus, we have two additional critical points: (0, ±1) and (±2, 0).
Now, let's evaluate the function f(x, y) at these critical points:
For the critical points (±sqrt(1 / 4.25), -2 * ±sqrt(1 / 4.25)):
f(±sqrt(1 / 4.25), -2 * ±sqrt(1 / 4.25)) = (-2 * sqrt(1 / 4.25))^2 - (±sqrt(1 / 4.25))^2
This simplifies to:
f(±sqrt(1 / 4.25), -2 * ±sqrt(1 / 4.25)) = 4 / 4.25 - 1 / 4.25 = 3 / 4.25 = 12 / 17
For the critical points (0, ±1):
f(0, ±1) = (±1)^2 - 0^2 = 1
For the critical points (±2, 0):
f(±2, 0) = 0 - (±2)^2 = -4
Now, let's analyze the nature of these critical points:
The critical points (0, ±1) are points where the function f(x, y) is equal to 1. Since f(±2, 0) = -4, it implies that (±2, 0) indeed represents the minimum points of the function.
Regarding your question about fxx(a, b) > 0 for a point to be a minimum point, we need to remember that the second partial derivative test involves the determinant of the Hessian matrix. In this case, the second partial derivatives of f(x, y) are constant, and the determinant of the Hessian matrix is equal to fxx * fyy - fxy * fyx = (-2) * 2 - 0 = -4 < 0. This indicates that (±2, 0) represents a saddle point, not a minimum point.
The book is correct in stating that (±2, 0) represents saddle points, not minimum points, with the function value f(±2, 0) = -4.