The given function is f(x, y) = x²y + y³ - 48y.
To find the critical points of this function, we need to take the partial derivatives with respect to x and y, and set them equal to zero.
1. Partial derivative with respect to x:
∂f/∂x = 2xy
2. Partial derivative with respect to y:
∂f/∂y = x² + 3y² - 48
Setting the partial derivatives equal to zero, we get:
1. 2xy = 0
- This equation is satisfied when either x = 0 or y = 0.
2. x² + 3y² - 48 = 0
To further analyze the critical points, we can substitute the values of x and y into the original function and evaluate the results.
Case 1: x = 0
Substituting x = 0 into the original function:
f(0, y) = 0 + y³ - 48y = y³ - 48y
Case 2: y = 0
Substituting y = 0 into the original function:
f(x, 0) = x²(0) + (0)³ - 48(0) = 0
Now, let's solve the equation x² + 3y² - 48 = 0 to find the critical points.
Rearranging the equation, we get:
x² = 48 - 3y²
We can substitute this value of x² into the first equation (2xy = 0) to find the corresponding values of y.
Case 3: x² = 48 - 3y²
Substituting x² = 48 - 3y² into 2xy = 0:
2(48 - 3y²)y = 0
96y - 6y³ = 0
6y(16 - y²) = 0
From this equation, we can solve for y to find the corresponding x-values.