Finding Local Maximum and Minimum Values and Saddle Points
The function in question is:
f(x, y) = x³ + y³ – 3x² – 3y² + 9x
To find the local maxima, minima, and saddle points of this function, we must follow these steps:
1. Find the Critical Points
First, we need to compute the first partial derivatives of the function with respect to x and y:
fx = 3x² – 6x + 9
fy = 3y² – 6y
Next, we set these partial derivatives equal to zero to find the critical points:
- For fx: 3x² – 6x + 9 = 0
- For fy: 3y² – 6y = 0
2. Solving the Equations
We solve each equation to find the critical points:
3x² – 6x + 9 = 0 can be simplified to:
x² – 2x + 3 = 0
Using the quadratic formula:
x = &frac{-b ± √(b² – 4ac)}{2a} = &frac{2 ± &radic{(-4)}}{2} = 1 ± i (no real solutions)
Now, solving 3y² – 6y = 0:
3y(y – 2) = 0 gives us:
- y = 0
- y = 2
3. Determine the Nature of Critical Points
Since we have critical points with real y-values at (1+i, 0) and (1+i, 2), we need to check the second derivatives:
Calculate the second partial derivatives:
fxx = 6x – 6, fyy = 6y – 6, fxy = 0
Now, we evaluate the Hessian determinant, H, at each critical point:
H = fxx * fyy – (fxy)²
4. Classify Each Critical Point
Now we inspect the signs of H and fxx:
- If H > 0 and fxx > 0, we have a local minimum.
- If H > 0 and fxx < 0, we have a local maximum.
- If H < 0, we have a saddle point.
- If H = 0, the test is inconclusive.
5. Conclusion
After evaluating, we will find that our critical points lead us to understand local behavior in the function. Since the only real critical points we identify are in terms of y-values only and complex for x, we usually focus on local minima and maxima in those contexts involving real parts.
Finally, the behavior of the function in its domain can be represented more fully by evaluating it with nearby values or through graphical analysis to confirm the presence of local minima, maxima, and saddle points in their respective context.