The critical points of the function f(x, y) = x³ + y³ + 3xy are (0, 0) and (√(-1), 1). The critical point (0, 0) is inconclusive, and the critical point (√(-1), 1) is a local minimum.
To find the critical points of the function f(x, y) = x³ + y³ + 3xy, we need to find the values of x and y where the partial derivatives with respect to x and y are equal to zero. Let's start by finding the partial derivatives:
∂f/∂x = 3x² + 3y
∂f/∂y = 3y² + 3x
To find the critical points, we set both partial derivatives equal to zero and solve the resulting system of equations:
3x² + 3y = 0
3y² + 3x = 0
Simplifying the equations, we get:
x² + y = 0 ...(1)
y² + x = 0 ...(2)
Now, let's solve the system of equations. We can start by isolating x in equation (1):
x² = -y
x = ±√(-y)
Substituting this into equation (2), we have:
y² + ±√(-y) = 0
Simplifying further, we get:
y² = ±√(-y)
To make the equation easier to work with, let's square both sides:
y⁴ = y
Now we have a polynomial equation. To solve for y, we can set y equal to 0 or use factoring. By inspection, we can see that y = 0 is one solution.
Now let's factor the equation y⁴ - y = 0:
y(y³ - 1) = 0
This gives us two additional solutions:
y = 0
y³ - 1 = 0
For y³ - 1 = 0, we can solve for y:
y³ = 1
y = 1
So the critical points of the function are (x, y) = (0, 0) and (x, y) = (√(-1), 1).
To identify the local maximum values, minimum values, and saddle points at these critical points, we can use the second partial derivatives test. Let's find the second partial derivatives:
∂²f/∂x² = 6x
∂²f/∂y² = 6y
∂²f/∂x∂y = 3
Evaluating the second partial derivatives at the critical points, we have:
At (0, 0):
∂²f/∂x² = 6(0) = 0
∂²f/∂y² = 6(0) = 0
∂²f/∂x∂y = 3
The determinant of the Hessian matrix is zero:
D = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = 0
Since D = 0, the second derivative test is inconclusive at (0, 0). Therefore, we cannot determine if it is a local maximum, minimum, or saddle point.
At (√(-1), 1):
∂²f/∂x² = 6(√(-1)) = 6i√1 = 6i
∂²f/∂y² = 6(1) = 6
∂²f/∂x∂y = 3
The determinant of the Hessian matrix is D = (6i)(6) - (3)² = 36i - 9 = 36(i - 1/4).
Since D ≠ 0 and the second partial derivatives are both positive (6 and 6i), (√(-1), 1) is a local minimum.
In summary, the critical points of the function f(x, y) = x³ + y³ + 3xy are (0, 0) and (√(-1), 1). The critical point (0, 0) is inconclusive, and the critical point (√(-1), 1) is a local minimum.