Example 1
Find the local extrema of f (x, y) = x3 + x2y - y2 - 4y.
Solution
Step 1: Find the critical points. The derivative of f is
| Df (x, y) = |
| x(3x + 2y) = 0 | (1) |
| x2 - 2y - 4 = 0. | (2) |
Equation (1) is satified if either x = 0 or if 3x + 2y = 0, i.e., if x = 0 or if y = - 3x/2. We consider these two solutions as two separate cases. For each case, we will find soultions for equation (2).
Case 1: Let x = 0. Then we know equation (1) is satisfied. We plug x = 0 into equation (2), which becomes 0 - 2y - 4 = 0, i.e., y = - 2. If x = 0 and y = - 2, then both equations (1) and (2) are satisfied. Therefore the point (0, - 2) is a critical point.
Case 2: Let y = - 3x/2. Then we know that equation (2) is satisfied. We plug y = - 3x/2 into equation (2) and simplify:
| x2 - 2(- 3x/2) - 4 | = 0 | |
| x2 + 3x - 4 | = 0 | |
| (x - 1)(x + 4) | = 0 | |
| x = 1 or x | = - 4. |
The second solution for case 2 is when x = - 4, which means y = - 3x/2 = 6. Therefore, the point (- 4, 6) is a critical point.
To summarize the results from both case 1 and case 2, we conclude that f (x, y) has three critical points: (0, - 2), (1, - 3/2), and (- 4, 6).
You should double check that Df (x, y) = [0 0] at each of these points.
Step 2: Classify the critical points.
The Hessian matrix is
Hf (x, y) = ![]() |
For the critical point (0, - 2),
Hf (0, -2) = ![]() |
For the critical point (1, - 3/2),
Hf (1, -3/2) = ![]() |
h11 = 3 > 0 and det(Hf )= - 6 - 4 = - 10 < 0. This means Hf (1, - 3/2) is indefinite and f has a saddle at (1, - 3/2).
For the critical point (- 4, 6),
Hf (- 4, 6) = ![]() |
h11 = - 12 < 0 and det(Hf )= 24 - 64 = - 40 < 0. This means Hf (- 4, 6) is indefinite and f has a saddle at (- 4, 6).
Example 2
Identify the local extrama of f (x, y) = (x2 + y2)e-y.
Solution
Step 1: Find the critical points.
The derivative of f is
| Df (x, y) = |
The critical points are therefore (0, 0) and (0, 2).
Step 2: Classify the critical points.
The Hessian matrix is
Hf (x, y) = ![]() |
At the critical point (0, 0)
| Hf (0, 0) = |
At the critical point (0, 2)
Hf (0, 2) = ![]() |