Single & Multivariable Maxima-Minima & Saddle Points
“First/second derivative tests, Hessian matrix determinant test D = f_xx * f_yy - (f_xy)^2.”
1.Optimization Theory in One and Multiple Dimensions
§ 3.1.Optimization is fundamental across algorithm design, machine learning cost functions, and computer systems engineering.
Stationary / Critical Points: Points where the gradient vector vanishes: nabla f(x_1, ..., x_n) = 0.
Hessian Matrix: The matrix of second-order partial derivatives:
H = [ [ f_xx, f_xy ], [ f_yx, f_yy ] ]
By Clairaut's Theorem (Schwarz's Theorem), mixed partials are equal (f_xy = f_yx) whenever second derivatives are continuous.
- Det(H) = D = f_xx * f_yy - (f_xy)^2.
- If H is positive definite (eigenvalues > 0, D > 0, f_xx > 0) => strict local minimum.
- If H is negative definite (eigenvalues < 0, D > 0, f_xx < 0) => strict local maximum.
- If H is indefinite (eigenvalues of opposite signs, D < 0) => saddle point.
For f(x, y) = x^2 - y^2 at (0, 0): f_xx = 2, f_yy = -2, f_xy = 0 => D = -4 < 0. Along the x-axis, f(x, 0) = x^2 (minimum at 0). Along the y-axis, f(0, y) = -y^2 (maximum at 0). It is a classic saddle point.
2. Worked Problem: Multivariable Critical Point Classification
Ensure you find all pairs of (x, y) that simultaneously satisfy both partial equations f_x = 0 and f_y = 0.
Problem [Original Practice Problem · Modeled on GATE Pattern]:
Find and classify all critical points of the function f(x, y) = x^3 + y^3 - 3xy.
Step 1: Compute first partial derivatives and set to zero:
f_x = 3x^2 - 3y = 0 => y = x^2
f_y = 3y^2 - 3x = 0 => x = y^2
Step 2: Solve the simultaneous system:
Substitute y = x^2 into x = y^2:
x = (x^2)^2 = x^4 => x^4 - x = 0 => x(x^3 - 1) = 0.
Real solutions for x: x = 0 or x = 1.
- If x = 0 => y = 0^2 = 0. Point P1 = (0, 0).
- If x = 1 => y = 1^2 = 1. Point P2 = (1, 1).
Step 3: Compute second partial derivatives:
f_xx = 6x, f_yy = 6y, f_xy = -3.
Discriminant D(x, y) = f_xx * f_yy - (f_xy)^2 = (6x)(6y) - (-3)^2 = 36xy - 9.
Step 4: Classify each critical point:
- At P1(0, 0): D = 36(0)(0) - 9 = -9 < 0.
Since D < 0, (0, 0) is a SADDLE POINT.
- At P2(1, 1): D = 36(1)(1) - 9 = 27 > 0.
f_xx(1, 1) = 6(1) = 6 > 0.
Since D > 0 and f_xx > 0, (1, 1) is a LOCAL MINIMUM.
Local minimum value: f(1, 1) = 1^3 + 1^3 - 3(1)(1) = 2 - 3 = -1.
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