Question

For the function \( f(x, y) = x^3 + y^3 - 3x - 12y + 12 \), which of the following are correct:
A. minima at (1,2)
B. maxima at (-1,-2)
C. neither a maxima nor a minima at (1,-2) and (-1,2)
D. the saddle points are (-1,2) and (1,-2)

• A. A, B and D only
• B. A, B and C only
• C. A, B, C and D
• D. B, C and D only

Hint:

Be systematic when classifying critical points. Create a table with columns for the point (x,y), \(f_{xx}\), \(f_{yy}\), \(f_{xy}\), the value of D, and the final classification. This organizes your work and reduces the chance of errors, especially under exam pressure.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find and classify the extrema (maxima, minima) and saddle points of a function of two variables, we use the second partial derivative test. This involves finding critical points where the first partial derivatives are zero, and then using the second partial derivatives to classify them.

Step 2: Key Formula or Approach:
1. Find Critical Points: Solve the system of equations \(f_x = 0\) and \(f_y = 0\).
2. Second Derivative Test: Calculate the discriminant (or Hessian determinant) \(D(x,y) = f_{xx}f_{yy} - (f_{xy})^2\).
- If \(D>0\) and \(f_{xx}>0\), it's a local minimum.
- If \(D>0\) and \(f_{xx}<0\), it's a local maximum.
- If \(D<0\), it's a saddle point.
- If \(D = 0\), the test is inconclusive.

Step 3: Detailed Explanation:
The given function is \( f(x, y) = x^3 + y^3 - 3x - 12y + 12 \).
Part 1: Find Critical Points
First, find the partial derivatives with respect to \(x\) and \(y\): \[ f_x = \frac{\partial f}{\partial x} = 3x^2 - 3 \] \[ f_y = \frac{\partial f}{\partial y} = 3y^2 - 12 \] Set them to zero to find the critical points: \[ 3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1 \] \[ 3y^2 - 12 = 0 \implies y^2 = 4 \implies y = \pm 2 \] The critical points are all possible combinations of these \(x\) and \(y\) values: (1, 2), (1, -2), (-1, 2), and (-1, -2).
Part 2: Second Derivative Test
Next, find the second partial derivatives: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \] Calculate the discriminant \(D\): \[ D(x,y) = f_{xx}f_{yy} - (f_{xy})^2 = (6x)(6y) - (0)^2 = 36xy \] Part 3: Classify Critical Points
Now, test each critical point:
At (1, 2):
\( D(1,2) = 36(1)(2) = 72>0 \).
\( f_{xx}(1,2) = 6(1) = 6>0 \).
Since \(D>0\) and \(f_{xx}>0\), this is a local minimum. Statement A is correct.

At (-1, -2):
\( D(-1,-2) = 36(-1)(-2) = 72>0 \).
\( f_{xx}(-1,-2) = 6(-1) = -6<0 \).
Since \(D>0\) and \(f_{xx}<0\), this is a local maximum. Statement B is correct.

At (1, -2):
\( D(1,-2) = 36(1)(-2) = -72<0 \).
Since \(D<0\), this is a saddle point.

At (-1, 2):
\( D(-1,2) = 36(-1)(2) = -72<0 \).
Since \(D<0\), this is a saddle point.
Part 4: Evaluate Statements C and D

Statement D: "the saddle points are (-1,2) and (1,-2)". This matches our findings. Statement D is correct.

Statement C: "neither a maxima nor a minima at (1,-2) and (-1,2)". A saddle point is by definition a point that is not a local extremum (not a max or min). Since (1,-2) and (-1,2) are saddle points, this statement is also true. Statement C is correct.

Step 4: Final Answer:
All four statements A, B, C, and D are correct descriptions based on our calculations. Therefore, the correct option is (C).