Step 1: Extrema Identification:
The second partial derivative test classifies extrema (maxima, minima) and saddle points of two-variable functions. This involves finding critical points (where first partial derivatives are zero) and using second partial derivatives for classification.
Step 2: Key Steps:
1. Find Critical Points: Solve \(f_x = 0\) and \(f_y = 0\).
2. Second Derivative Test: Calculate the discriminant \(D(x,y) = f_{xx}f_{yy} - (f_{xy})^2\).
- If \(D>0\) and \(f_{xx}>0\), local minimum.
- If \(D>0\) and \(f_{xx}<0\), local maximum.
- If \(D<0\), saddle point.
- If \(D = 0\), test inconclusive.
Step 3: Example:
Consider \( f(x, y) = x^3 + y^3 - 3x - 12y + 12 \).
Part 1: Critical Points
Find partial derivatives:
\[ f_x = 3x^2 - 3 \]\
\[ f_y = 3y^2 - 12 \]\
Set to zero:
\[ 3x^2 - 3 = 0 \implies x = \pm 1 \]\
\[ 3y^2 - 12 = 0 \implies y = \pm 2 \]\
Critical points: (1, 2), (1, -2), (-1, 2), and (-1, -2).
Part 2: Second Derivatives
Find second partial derivatives:
\[ f_{xx} = 6x \]\
\[ f_{yy} = 6y \]\
\[ f_{xy} = 0 \]\
Calculate discriminant \(D\):
\[ D(x,y) = (6x)(6y) - (0)^2 = 36xy \]\
Part 3: Classify Critical Points
Test each point:
At (1, 2):
\( D(1,2) = 72>0 \).
\( f_{xx}(1,2) = 6>0 \).
Local minimum. Statement A is correct.
At (-1, -2):
\( D(-1,-2) = 72>0 \).
\( f_{xx}(-1,-2) = -6<0 \).
Local maximum. Statement B is correct.
At (1, -2):
\( D(1,-2) = -72<0 \).
Saddle point.
At (-1, 2):
\( D(-1,2) = -72<0 \).
Saddle point.
Part 4: Evaluate Statements C and D
Statement D: "the saddle points are (-1,2) and (1,-2)". Correct. Statement D is correct.
Statement C: "neither a maxima nor a minima at (1,-2) and (-1,2)". Correct (saddle points). Statement C is correct.
Step 4: Conclusion:
All statements A, B, C, and D are correct. The answer is (C).