Mathematical methods for economic theory

Martin J. Osborne
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7.3 Exercises on optimization with inequality constraints: the sufficiency of the Kuhn-Tucker conditions

  1. For each possible value of the constant a, solve the problem
    maxx,y x + ay subject to x2 + y2 ≤ 1 and x + y ≥ 0.

    Solution

    The Kuhn-Tucker conditions are necessary and sufficient because f is concave, each gj is convex, and there exists a point x such that gj(x) < cj for each j (for example, (1/4, 1/4)).

    The Kuhn-Tucker conditions are

    1 − 2λ1x + λ2  =  0
    a − 2λ1y + λ2  =  0
    λ1 ≥ 0, x2 + y2  ≤  1, and λ1(x2 + y2 − 1) = 0
    λ2 ≥ 0, x + y  ≥  0, and λ2(x + y) = 0
    From the first condition, we have λ1 > 0 (otherwise λ2 = −1). Thus x2 + y2 = 1. Now consider the two cases for the value of λ2.

    If λ2 = 0, we have x = 1/(2λ1) and y = a/(2λ1) from the first two conditions, so that λ1 = (1/2)√(1 + a2) (using x2 + y2 = 1). Thus (x, y) = (1/√(1 + a2), a/√(1 + a2)). For this pair to satisfy the constraint x + y ≥ 0 we need a ≥ −1.

    If λ2 > 0, we have x + y = 0, so that from x2 + y2 = 1 we have (x, y) = (1/√2, −1/√2) or (x, y) = (−1/√2, 1/√2). From the first two Kuhn-Tucker conditions we have λ1 = (1 − a)/(2(x − y)). Thus for (x, y) = (1/√2, −1/√2) we have λ1 = (1 − a)√2/4 and hence λ2 = −(1 + a)/2, which are both nonnegative if and only if a ≤ −1. For (x, y) = (−1/√2, 1/√2) we have λ1 = −(1 − a)√2/4 and hence λ2 = −(1 + a)/2, which are not both nonnegative for any value of a.

    We conclude that if a ≥ −1 the solution of the problem is (1/√(1 + a2), a/√(1 + a2)), with λ1 = (1/2)√(1 + a2) and λ2 = 0, and if a < −1 the solution is (1/√2, −1/√2), with λ1 = (1 − a)√2/4 and λ2 = −(1 + a)/2.

  2. Consider the following problem.
    maxx−x2
    1     − x1x2 − x2
    2     subject to x1 − 2x2 ≤ −1 and 2x1 + x2 ≤ 2
    1. Are the Kuhn-Tucker conditions necessary for a solution of this problem?
    2. Are the Kuhn-Tucker conditions sufficient for a solution of this problem?
    3. If possible, use the Kuhn-Tucker conditions to find the solution(s) of the problem.

    Solution

    1. Yes: since each gj is concave.
    2. Yes, since f is concave and each gj is convex.
    3. The Kuhn-Tucker conditions are
      −2x1 − x2 − λ1 − 2λ2  =  0
      −x1 − 2x2 + 2λ1 − λ2  =  0
      λ1 ≥ 0, x1 − 2x2  ≤  −1, and λ1(x1 − 2x2 + 1) = 0
      λ2 ≥ 0, 2x1 + x2  ≤  2, and λ2(2x1 + x2 − 2) = 0

      Consider the possible cases in turn:

      λ1 = λ2 = 0: From the first two equations we have x1 = x2 = 0, which violates the first constraint.

      λ1 > 0 and λ2 > 0, so that x1 − 2x2 = −1 and 2x1 + x2 = 2: We have x1 = 3/5 and x2 = 4/5, which do not satisfy the first first-order condition.

      λ1 > 0 and λ2 = 0: Since λ1 > 0 we have x1 = 2x2 − 1; from the first-order conditions we have x1 = −4/14, x2 = 5/14, and λ1 = 3/14. Since λ1 ≥ 0, a solution of the problem is (x1, x2) = (−4/14, 5/14).

      λ1 = 0 and λ2 > 0: Since λ2 > 0 we have x2 = 2 − 2x1; from the first-order conditions we have x1 = 1, x2 = 0, and λ2 = −1. Since λ2 < 0, this case does not yield a solution.

      Thus unique solution to the problem is (x1, x2) = (−4/14, 5/14).