Question

Minimize \\[ \begin{array}{l} C=x_{1} \\ x_{1}^{2}-x_{2} \geq 0 \end{array} \\] subject to and \\[ x_{1}, x_{2} \geq 0 \\] Solve graphically. Does the optimal solution occur at a cusp? Check whether the optimal solution satisfies \((a)\) the constraint qualification and \((b)\) the Kuhn-Tucker minimum conditions.

Short answer

The minimum value occurs at the point (0,0), satisfying constraint qualification and Kuhn-Tucker conditions. No cusp present.

Step-by-step solution

Understand the Objective Function and Constraints

We want to minimize the function \( C = x_{1} \) with respect to the constraints \( x_{1}^{2} - x_{2} \geq 0 \) and \( x_{1}, x_{2} \geq 0 \). This means we are searching for the smallest value of \( x_1 \) while ensuring all constraints are satisfied.

Graph the Constraints

Plot the constraint \( x_{1}^{2} - x_{2} \geq 0 \). This can be rewritten as \( x_{2} \leq x_{1}^{2} \), indicating the region below the parabola \( x_{2} = x_{1}^{2} \). Also, plot the constraints \( x_{1} \geq 0 \) and \( x_{2} \geq 0 \), which confine the solution to the first quadrant of the graph.

Identify the Feasible Region

The feasible region is the part of the first quadrant that is below the curve \( x_{2} = x_{1}^{2} \). This is the shaded area bounded by the parabola and the axes. The feasible set is essentially all the points \((x_1, x_2)\) where \( x_1 \geq 0 \) and \( 0 \leq x_2 \leq x_1^2 \).

Find the Optimal Point Graphically

To minimize \( C = x_1 \), we aim to find the smallest \( x_1 \) value in the feasible region. The minimum occurs at the point where the objective function first intersects the feasible region, which is the origin \((0,0)\).

Check for a Cusp at the Optimization Point

At the point \((0,0)\), the feasible region's boundary created by the curve is touching the x-axis, but not changing slope or direction swiftly, meaning it is a smooth boundary. This confirms there is no cusp at the optimum point.

Verify Constraint Qualification

Constraint qualification generally demands a linearly independent gradient of active constraints at optimum. At \((0,0)\), the constraint \( x_1^2 - x_2 = 0 \) holds (active), with gradient \((2x_1, -1) = (0, -1)\). Since there are no dependent constraints here, qualification is satisfied.

Verify Kuhn-Tucker Conditions

The Kuhn-Tucker conditions include the need for a Lagrangian with zero gradient at optimum. The Lagrangian is \( L = x_1 + \lambda (x_1^2 - x_2) \), where \( \lambda \) is a multiplier. At \((0,0)\), \( \partial L/\partial x_1 = 1 + 2\lambda x_1 = 1 \), \( \partial L/\partial x_2 = -\lambda = 0 \). Setting \( \lambda = 0 \) satisfies without violating any regularity. Hence, Kuhn-Tucker conditions hold.

Key concepts

Constraint Qualification

Constraint qualification is an important concept when dealing with optimization problems, especially those involving inequalities. It helps ensure that the optimum solution is well-defined and that certain mathematical properties hold.
In simple terms, constraint qualification is about checking if the constraints are nicely behaved at the optimum solution. For our exercise, the constraint is given by \( x_1^2 - x_2 \geq 0 \), which is active at the optimal point \((0,0)\).

• The active constraint at this point is \( x_1^2 - x_2 = 0 \), which translates to the equation of the parabola \( x_2 = x_1^2 \).
• We look at the gradient of this constraint, which is \( (2x_1, -1) \), and at \((0,0)\), it becomes \((0, -1)\).
• Since there are no other active constraints at \((0,0)\), there is no dependency between constraints, which means they are linearly independent.

Meeting these criteria satisfies the constraint qualification, meaning our problem is well-posed at the solution point.

Objective Function

The objective function is what we aim to optimize, either minimize or maximize, depending on the problem's requirements. In this exercise, our goal is to minimize the function \( C = x_1 \).
Minimizing \( C = x_1 \) translates to finding the smallest possible value of \( x_1 \) without violating the constraints. The lower \( x_1 \)'s value, the better it is for our objective.

• The objective function is simple here: it is linear with a single variable and no coefficients besides 1. This makes it a straightforward task in terms of mathematical manipulation.
• While minimizing \( C = x_1 \), we must ensure that \( x_1^2 - x_2 \geq 0 \) and \( x_1, x_2 \geq 0 \) remain true. This confines us to certain regions on the graph.

This minimal intersection with the feasible region at \((0,0)\) shows the efficiency of reaching the objective.

Feasible Region

The feasible region is the area of interest where all constraints are satisfied simultaneously. This region is where possible solutions and the optimum lie in any constrained optimization problem.
For this exercise, we have two main sets of constraints: \( x_1^2 - x_2 \geq 0 \) and \( x_1, x_2 \geq 0 \). Here's how they shape the feasible region:

• The first constraint, \( x_2 \leq x_1^2 \), creates a region below or on the parabola \( x_2 = x_1^2 \).
• The second set of constraints, \( x_1, x_2 \geq 0 \), restricts solutions to the first quadrant of the graph, representing only positive values for both variables.

The feasible region is thus, the area in the first quadrant where the points satisfy both constraints simultaneously, which is below the parabola and above the y-axis. It's important as it shows all the potential places where the objective function can achieve its minimum value.

Graphical Optimization

Graphical optimization is a visual method to solve optimization problems, especially useful when dealing with two-variable systems. It allows you to see and understand the feasible region and discover optimal points through intersections and boundaries.
To solve graphically, you graph the functions and constraints, then identify the feasible region and optimal points:

• Start by plotting the parabola \( x_2 = x_1^2 \) which determines one boundary of the feasible region.
• Next, plot the lines \( x_1 = 0 \) and \( x_2 = 0 \) which limit the region to the first quadrant.

Once the feasible region is identified, find where the objective function \( C = x_1 \) achieves its minimum. The intersection of the objective function's line with the feasible region, in this case, occurs at the origin \((0,0)\). Thus, graphically, it is simple yet powerful for visualizing solutions.