338 ACTIVITY 23: Constrained Optimization: Equality Constraints

WHY:

This activity is designed to help you discover how to solve a nonlinear programming problem with equality constraints. We apply Theorem 5.2.1 to these types of problems, as illustrated in the model below. Although inequality constraints are much more common than equality constraints, it is important to fully understand the equality constraint problem before tackling the harder inequality constraint problem.

LEARNING OBJECTIVES:

1. Discover necessary and sufficient conditions for finding relative minima under equality constraints.
2. Discover reasoning behind Theorem 5.2.1.
3. Realize the importance of getting back up to speed quickly after a vacation.

PERFORMANCE CRITERIA:

1. Quality of the answers to the Critical Thinking Questions.
   • Completeness
   • Clarity
   • Depth
2. Speed and accuracy exhibited by the team during the activity.

INFORMATION:

Lagrangian Function

The Lagrangian Function F(X, k) = f(X) - k g(X). A necessary condition for f to have a stationary point at X₀ subject to the constraint g(X) = 0 is that grad F(X₀) = 0

THEOREM 5.2.1

Let f and g be twice continuously differentiable defined on a neighborhood of a point X₀ for which g(X₀) = 0 and suppose there exists a number k such that grad f(X₀) - k grad g(X₀) = 0 and the matrix L(X₀) = H(X₀) - G(X₀) is positive definite where H is the Hessian for f and G is the Hessian for g. Then X₀ is a relative minimum for f subject to g(X) = 0.

RESOURCES:

1. Section 5.2: Strategic Mathematics
2. 30 minutes

MODEL:

minimize f(x, y) = x² + y²
subject to x + y² - 5 = 0

Let g(x, y) = x + y² - 5

The Lagrangian function is F(x, y, k) = x² + y² - k (x + y² - 5)

grad F(x, y, k) = (2x - k, 2y - 2ky, -g(x, y)). Setting each component to 0, we get the following equations:

k = 2x, 2y - 4xy = 0, x + y² = 5

Thus 2y - 4(5 - y²)y = 0 = -18y + 4y³ = y(-18 + 4y²)

The possible optimal points are y = 0, x = 5, k = 10 or y = +- 3/(2)^.5, x = 1/2, k = 1.

The Hessian of the Lagrangian is Hessian f - Hessian g =

			| 2  0 |    | 0  0 |   | 2   0   |
			| 0  2 | - k| 0  2 | = | 0  2-2k |

If k = 1, then the Hessian of the Lagrangian is positive semidefinite (Why?). This is strong evidence that the second set of points are relative minima. When k = 10, the Hessian of the Lagrangian is indefinite. Thus we don't know, but from the graph we can tell that (5, 0) is a maximum point.

CRITICAL THINKING QUESTIONS:

These will be provided in class.

SKILL EXERCISES:

1. Find the relative minima for the following function subject to the given constraint. First compute the Lagrangian function F(X) and then find points for which F(X) = 0 and then plug these points into the Hessian for the Lagrangian function to determine whether or not they are relative mins.

   f(x, y) = 2x² + y² - xy - 8x - 2y
   subject to g(x, y) = 3x + y - 10 = 0
   
   

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