For f(x,y)= 3 - x^2 +2x -y^2 - 4, find all critical points of f and determine their nature.
To find the critical points of the function f(x, y) = 3 - x^2 + 2x - y^2 - 4, we need to calculate the partial derivatives of f with respect to x and y, and then solve the resulting equations simultaneously for x and y.
Step 1: Calculate the partial derivative with respect to x (denoted as ∂f/∂x):
∂f/∂x = -2x + 2
Step 2: Calculate the partial derivative with respect to y (denoted as ∂f/∂y):
∂f/∂y = -2y
Step 3: Set the partial derivatives equal to zero and solve for x and y:
∂f/∂x = -2x + 2 = 0
Solving this equation, we get:
-2x = -2
x = 1
∂f/∂y = -2y = 0
Solving this equation, we get:
y = 0
So, the critical point is (x, y) = (1, 0).
Step 4: Determine the nature of the critical point using the second partial derivatives test:
To use the second partial derivatives test, we need to calculate the second partial derivatives of f.
The second partial derivative with respect to x (denoted as ∂²f/∂x²) is:
∂²f/∂x² = -2
The second partial derivative with respect to y (denoted as ∂²f/∂y²) is:
∂²f/∂y² = -2
The second partial derivative with respect to both x and y (denoted as ∂²f/∂x∂y) is:
∂²f/∂x∂y = 0
Now, substitute the values x = 1 and y = 0 into the second partial derivatives:
∂²f/∂x² = -2
∂²f/∂y² = -2
∂²f/∂x∂y = 0
We can apply the second partial derivatives test:
1. If ∂²f/∂x² > 0 and ∂²f/∂y² > 0, then the critical point is a local minimum.
2. If ∂²f/∂x² < 0 and ∂²f/∂y² < 0, then the critical point is a local maximum.
3. If ∂²f/∂x² and ∂²f/∂y² have different signs or ∂²f/∂x² = ∂²f/∂y² = 0, then the test is inconclusive.
In this case, ∂²f/∂x² = -2 and ∂²f/∂y² = -2, both of which are less than 0. Therefore, the critical point (1, 0) is a local maximum.