There are two extrema for
z=2x-x^2+2y^2-y^4. One is located
at (1,1,2). Where is the second one located?
To find the second extremum of the function z=2x-x^2+2y^2-y^4, we can start by finding the critical points of the function. Critical points occur where the gradient of the function is equal to zero.
The gradient of a function is a vector that represents the rate of change of the function with respect to each variable. In this case, the gradient of z with respect to x and y is given by:
∇z = (∂z/∂x, ∂z/∂y)
Let's calculate the partial derivatives:
∂z/∂x = 2 - 2x
∂z/∂y = 4y - 4y^3
Now, set the partial derivatives equal to zero and solve for x and y:
2 - 2x = 0 ⇒ x = 1
4y - 4y^3 = 0 ⇒ 4y(1 - y^2) = 0
From the second equation, we can see that either y = 0 or 1 - y^2 = 0.
If y = 0, then x = 1. So one critical point is (1, 0).
If 1-y^2 = 0, then y^2 = 1 and y = ±1. Substituting y = ±1 into the first equation, we get x = 1. So the other two critical points are (1, 1) and (1, -1).
To determine if these critical points are extrema, we need to analyze the second partial derivatives. For a function to have an extremum at a critical point, the second partial derivatives should satisfy one of the following conditions:
1. The Hessian matrix is positive definite (all eigenvalues are positive), indicating a local minimum.
2. The Hessian matrix is negative definite (all eigenvalues are negative), indicating a local maximum.
3. The Hessian matrix has a mix of positive and negative eigenvalues, indicating a saddle point.
The Hessian matrix of z with respect to x and y is given by:
H = [∂²z/∂x² ∂²z/∂x∂y]
[∂²z/∂y∂x ∂²z/∂y²]
The second partial derivatives are:
∂²z/∂x² = -2
∂²z/∂y² = 4 - 12y^2
∂²z/∂x∂y = 0 (since the order of differentiation does not matter)
Substituting the critical points, we get:
At (1, 0):
H = [-2 0]
[0 4]
Both eigenvalues of H are positive, so this critical point represents a local minimum.
At (1, 1):
H = [-2 0]
[0 -8]
The eigenvalues of H are -2 and -8, which are both negative. Therefore, this critical point represents a local maximum.
At (1, -1):
H = [-2 0]
[0 -8]
Again, the eigenvalues of H are -2 and -8, indicating a local maximum.
Therefore, the two extrema of the function z=2x-x^2+2y^2-y^4 are located at (1, 0) (local minimum) and (1, ±1) (local maximums).