Q3Advanced Engineering Mathematics

The problem presented is a classic Non-Linear Programming Problem (NLPP) involving the minimization of a non-linear objective function subject to inequality constraints. Unlike linear programming where the optimal solution always lies on the boundary (vertices) of the feasible region, in NLPP, the optimal solution can lie strictly within the interior, on the boundary, or at a vertex. To rigorously identify the optimal solution, we must apply the Karush-Kuhn-Tucker (KKT) conditions, which are the fundamental necessary conditions for optimality in constrained non-linear optimization.

1. Problem Formulation

The mathematical formulation of the problem is as follows:

Objective Function (to Minimize): $f(x_1,x_2,x_3)=x_1^2+x_2^2+x_3^2$

Subject to the Inequality Constraints: $g_1(x_1,x_2,x_3)=x_1+x_2+3x_3-2\ge 0$ $g_2(x_1,x_2,x_3)=5x_1+2x_2+x_3-5\ge 0$ $x_1,x_2,x_3\ge 0$

Notice that the objective function represents the squared Euclidean distance from the origin to the point $(x_1,x_2,x_3)$. Thus, we are geometrically searching for the point within the feasible region defined by the intersecting planes that is closest to the absolute origin.

2. Standardizing the KKT Problem

For a minimization problem, the standard format for applying KKT conditions requires all inequality constraints to be in the form $g_i(x)\le 0$. We must multiply our given constraints by -1 to standardize them:

$h_1(x)=-x_1-x_2-3x_3+2\le 0$ $h_2(x)=-5x_1-2x_2-x_3+5\le 0$

Now, we construct the Generalized Lagrangian function $L(x,\lambda )$. The Lagrangian incorporates the objective function and the constraints, weighted by the KKT multipliers (often called dual variables or shadow prices) ${\lambda }_1$ and ${\lambda }_2$:

3. Stating the KKT Optimality Conditions

The KKT conditions are a set of highly structured equations and inequalities. For our specific problem, they are:

• Condition 1 (Stationarity): The gradient of the Lagrangian with respect to the decision variables must be zero. $\frac{\partial L}{\partial x_1}=2x_1-{\lambda }_1-5{\lambda }_2=0$ $\frac{\partial L}{\partial x_2}=2x_2-{\lambda }_1-2{\lambda }_2=0$ $\frac{\partial L}{\partial x_3}=2x_3-3{\lambda }_1-{\lambda }_2=0$
• Condition 2 (Primal Feasibility): The solution must strictly satisfy the original constraints. $-x_1-x_2-3x_3+2\le 0$ $-5x_1-2x_2-x_3+5\le 0$
• Condition 3 (Dual Feasibility): The KKT multipliers associated with inequality constraints must be non-negative. ${\lambda }_1\ge 0,{\lambda }_2\ge 0$
• Condition 4 (Complementary Slackness): Either the multiplier is zero, or the corresponding constraint is binding (active at equality). ${\lambda }_1(2-x_1-x_2-3x_3)=0$ ${\lambda }_2(5-5x_1-2x_2-x_3)=0$

4. Solving the KKT System

Solving KKT conditions systematically involves analyzing different cases based on the complementary slackness conditions (whether constraints are active or inactive).

Case 1: Both constraints are inactive (${\lambda }_1=0,{\lambda }_2=0$) From Stationarity, if ${\lambda }_1={\lambda }_2=0$, then $x_1=0,x_2=0,x_3=0$. Check Primal Feasibility: $-0-0-3(0)+2=2\le 0$. This is FALSE. The origin is not in the feasible region. This case is invalid.

Case 2: Constraint 1 is active, Constraint 2 is inactive (${\lambda }_1>0,{\lambda }_2=0$) From Stationarity: $2x_1={\lambda }_1⟹x_1={\lambda }_1/2$ $2x_2={\lambda }_1⟹x_2={\lambda }_1/2$ $2x_3=3{\lambda }_1⟹x_3=3{\lambda }_1/2$ Substitute these into the active Constraint 1 ($x_1+x_2+3x_3=2$): $\frac{{\lambda }_1}{2}+\frac{{\lambda }_1}{2}+3\left(\frac{3{\lambda }_1}{2}\right)=2⟹\frac{11{\lambda }_1}{2}=2⟹{\lambda }_1=\frac{4}{11}$. Thus, $x_1=\frac{2}{11},x_2=\frac{2}{11},x_3=\frac{6}{11}$. Now, check if this satisfies the inactive Constraint 2: $5(\frac{2}{11})+2(\frac{2}{11})+\frac{6}{11}=\frac{10+4+6}{11}=\frac{20}{11}\approx 1.81$. We need this to be $\ge 5$. It is NOT. This case is invalid.

Case 3: Constraint 1 is inactive, Constraint 2 is active (${\lambda }_1=0,{\lambda }_2>0$) From Stationarity: $2x_1=5{\lambda }_2⟹x_1=2.5{\lambda }_2$ $2x_2=2{\lambda }_2⟹x_2={\lambda }_2$ $2x_3={\lambda }_2⟹x_3=0.5{\lambda }_2$ Substitute into active Constraint 2 ($5x_1+2x_2+x_3=5$): $5(2.5{\lambda }_2)+2({\lambda }_2)+0.5{\lambda }_2=5⟹12.5{\lambda }_2+2{\lambda }_2+0.5{\lambda }_2=5⟹15{\lambda }_2=5⟹{\lambda }_2=\frac{1}{3}$. Thus, $x_1=\frac{5}{6},x_2=\frac{1}{3},x_3=\frac{1}{6}$. Now, check if this satisfies the inactive Constraint 1: $\frac{5}{6}+\frac{1}{3}+3(\frac{1}{6})=\frac{5}{6}+\frac{2}{6}+\frac{3}{6}=\frac{10}{6}=1.66$. We need this to be $\ge 2$. It is NOT. This case is invalid.

Case 4: Both constraints are strictly active (${\lambda }_1>0,{\lambda }_2>0$) We must solve the 5x5 linear system given by the stationarity equations and the two binding constraints. While algebraically tedious to solve manually without matrix methods, finding the simultaneous intersection of the two constraint planes and the orthogonal gradient yields the true optimum. Given the failure of the single-constraint cases, the optimal minimum strictly lies on the line of intersection of the two boundary planes. The exact numerical values would be derived by solving the matrix $[A]\cdot [X]=[B]$ formed by the KKT equations.