JNTU B.Tech III OPTIMIZATION TECHNIQUES Previous Question Paper (October - 2020 - R16)

Multivariable optimization without equality constraints involves finding the values of independent variables that minimize or maximize a function f(x₁, x₂, ..., x_n) defined over an open set. The necessary condition for a local extremum is that the gradient of the function must be zero at the point, expressed as ∇f(x) = 0.

The steps to solve such problems are:

1. Calculate the partial derivatives of the function with respect to each variable: ∂f/∂x_i = 0 for all i = 1, 2, ..., n.
2. Solve the resulting system of simultaneous equations to find the stationary points.
3. Evaluate the Hessian matrix H, where H_ij = ∂²f/(∂x_i∂x_j).
4. Check the nature of the stationary points using the principal minors of the Hessian matrix: if all principal minors are positive, the point is a local minimum; if they alternate in sign starting with negative, it is a local maximum.

Example: Minimize f(x, y) = x² + y² - 2x - 4y + 10.

1. First-order conditions:

The stationary point is (1, 2).

2. Second-order conditions using the Hessian matrix:

The principal minors are |H₁| = 2 > 0 and |H₂| = 4 > 0. Since all principal minors are positive, the function has a local minimum at (1, 2). The minimum value is f(1, 2) = 1 + 4 - 2 - 8 + 10 = 5.

To maximize z = -x1² - x2² - x3² + 4x1 + 6x2 subject to g1(x) = x1 + x2 - 2 ≤ 0, g2(x) = 2x1 + 3x2 - 12 ≤ 0, and x1, x2, x3 ≥ 0, we define the Lagrangian function L(x, λ) = -x1² - x2² - x3² + 4x1 + 6x2 - λ1(x1 + x2 - 2) - λ2(2x1 + 3x2 - 12).

The Kuhn-Tucker conditions are:

1. ∂L/∂x1 = -2x1 + 4 - λ1 - 2λ2 = 0
2. ∂L/∂x2 = -2x2 + 6 - λ1 - 3λ2 = 0
3. ∂L/∂x3 = -2x3 = 0 => x3 = 0
4. λ1(x1 + x2 - 2) = 0, λ2(2x1 + 3x2 - 12) = 0
5. λ1, λ2 ≥ 0

Case 1: λ1 = 0, λ2 = 0. Then -2x1 + 4 = 0 => x1 = 2 and -2x2 + 6 = 0 => x2 = 3. Checking constraints: x1 + x2 = 5 > 2. This violates the first constraint.

Case 2: λ1 > 0, λ2 = 0. Then x1 + x2 = 2. From ∂L/∂x1: -2x1 + 4 - λ1 = 0 => λ1 = 4 - 2x1. From ∂L/∂x2: -2x2 + 6 - λ1 = 0 => λ1 = 6 - 2x2. Equating: 4 - 2x1 = 6 - 2x2 => 2x2 - 2x1 = 2 => x2 - x1 = 1. Solving x1 + x2 = 2 and x2 - x1 = 1 gives 2x2 = 3 => x2 = 1.5 and x1 = 0.5. Since λ1 = 4 - 2(0.5) = 3 > 0, this is a feasible solution.

Case 3: λ1 = 0, λ2 > 0. Then 2x1 + 3x2 = 12. From ∂L/∂x1: -2x1 + 4 - 2λ2 = 0 => λ2 = 2 - x1. From ∂L/∂x2: -2x2 + 6 - 3λ2 = 0 => 3λ2 = 6 - 2x2 => λ2 = 2 - 2/3x2. Equating: 2 - x1 = 2 - 2/3x2 => x1 = 2/3x2. Substituting into 2x1 + 3x2 = 12: 2(2/3x2) + 3x2 = 12 => 4/3x2 + 9/3x2 = 12 => 13/3x2 = 12 => x2 = 36/13 ≈ 2.77, x1 = 24/13 ≈ 1.85. Constraint x1 + x2 = 60/13 ≈ 4.6 > 2. Violates constraint 1.

The optimal solution is x1 = 0.5, x2 = 1.5, x3 = 0.