338 ACTIVITY 21: Optima of Convex Functions

WHY:

This activity is designed to review the ways to determine where a function is convex or concave and also to find the absolute maxima and minima of these functions. Convex and concave functions are especially nice for optimization purposes since they have a unique relative min and max respectively and hence this relative min or max is also its absolute min or max. Although functions encountered in practice may not be convex or concave, they can sometimes be closely approximated by convex or concave functions.

LEARNING OBJECTIVES:

  1. Understand how to find stationary points for a function of several variables and determine which of its stationary points are max and min points.
  2. Discover the role of the Hessian in showing that a function of several variables is convex.
  3. Understand how group roles enhance group learning

PERFORMANCE CRITERIA:

  1. Quality of the answers to the Critical Thinking Questions.
  2. Diligence with which each member of the team performed her role.

RESOURCES:

  1. Section 4.7, Strategic Mathematics
  2. 30 minutes

MODEL:

Determine if the following function is convex and find its optimal solution: 


          minimize f(x, y) = x2 + 6xy - 2y3 + 24y

     First find the gradient vector and then the Hessian matrix.

           f(x, y) = (2x + 6y, 6x - 6y2 + 24)

		| 2    6 |
          H =   | 6  -12y|

          Solving the gradient for point(s) which make it zero we get:

          2x + 6y = 0  or  x = -3y
          6(-3y) - 6y2 + 24 = 0 or  y2 + 3y - 4 = 0
          y2 + 3y - 4 = (y + 4)(y - 1) = 0  or  y = -4,  y = 1
          When y = -4, x = 12, and when y = 1, x = -3.

It is easy to check that H is positive definite at (12, -4), and thus the function has a relative minimum there. At (-3, 1) the Hessian is indefinite, so we do not know what type of point we have. In fact, the diagonal matrix we get when reducing H is 


					| 1     0  |
					| 0 -12y-18|

Thus, as long as -12y-18 > 0, or y < -3/2, the function is convex and in this half plane, the absolute minimum is at (12, -4). f(12, -4) = -112. If y = 5, and x = 0, f(x, y) = -250 + 120 = -130 < -112. Thus over the entire plane, the function has no absolute minimum, but decreases to - infinity.

CRITICAL THINKING QUESTIONS:

These will be provided in class.

SKILL EXERCISES:

  1. Let f(x1,x2,x3) = 4x12 + x22 + x32 - 2x1x2 - 2x1x3 - 2x1 + 8

    Find the gradient vector for this function, its Hessian matrix, its stationary point(s), and determine whether or not the function is convex. Classify the stationary point(s) as relative min or relative max or saddle point(s). Find the absolute min for this function over the first octant. (The first octant is {X | X >= 0})

Math 338 Activity 21 -- Revised 11/20/98

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