Mathematical methods for economic theory

Martin J. Osborne
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6.2 Exercises on optimization with equality constraints: n variables, m constraints

  1. Solve the problem
    maxx,y,z (x + y) subject to x2 + 2y2 + z2 = 1 and x + y + z = 1
    (assuming, without checking, that the "nondegeneracy condition" is satisfied).

    Solution

    By the extreme value theorem the problem has a solution. (The objective function is continuous because it is a polynomial (in fact, it is a linear function), and the constraint set is compact because it is the intersection of a plane and the surface of a sphere.)

    The first-order conditions are

    1−2λ1x − λ2  =  0
    1−4λ1y − λ2  =  0
    −2λ1z − λ2  =  0
    and the constraints are x2 + 2y2 + z2 = 1 and x + y + z = 1. To find values of x, y, z, λ1, and λ2 that solve these equations, we can first use the third first-order condition to eliminate λ2, and then use the second constraint to eliminate z. Then we obtain
    1 − 2λ1[2x + y − 1]  =  0
    1 − 2λ1[x + 3y − 1]  =  0
    2x2 + 3y2 − 2y + 2xy − 2x  =  0
    The first two equations yield x = 2y, so that the third equation is 3y(5y − 2) = 0, so that either y = 0 or y = 2/5.

    Thus there are two solutions of the first-order conditions and the constraints, namely (0, 0, 1) with λ1 = −1/2 and λ2 = 1, and (4/5, 2/5,−1/5) with λ1 = 1/2 and λ2 = 1/5.

    Now, the objective function x + y is concave and each constraint is convex, so that λjgj is convex for each j for the second solution. Thus the second solution is the solution of the problem. [Alternatively, you can check that the value of the function is higher at the second solution.]

  2. Consider the problem
    maxxf(x) subject to p·x = c
    where x = (x1, ..., xn), f is concave, p = (p1, ..., pn), pi > 0 for i = 1, ..., n, and c is a number.
    1. Write down the first-order conditions for a solution of this problem.
    2. What is the relation (if any) between an interior solution of the first-order conditions and a solution of the problem?
    3. Solve the problem in the case n = 2 and f(x1, x2) = −x2
      1       − 2x2
      2       (which is concave).

    Solution

    1. f'i(x) − λpi = 0 for some λ, for i = 1, ..., n.
    2. Since pi > 0 for all i, there is no point at which the gradient of the constraint function is zero. Thus if x* solves the problem then there exists λ such that (x*, λ) solve the first-order conditions. Since f is concave and g is linear, if (x*, λ) solves the first-order conditions, then x* solves the problem. Thus x* solves the problem if and only if there exists λ such that (x*,λ) solves the first-order conditions.
    3. The first-order conditions are
      −2x1 − λp1  =  0
      −4x2 − λp2  =  0
      and the constraint is p1x1 + p2x2 = c. The unique solution of these three equations is
      (x1, x2) = (2p1c/(2p2
      1       + p2
      2      ), p2c/(2p2
      1       + p2
      2      )).
  3. Consider the problem
    maxtW(t)−c(t) subject to g(t) = k,
    where t = (t1, ..., tn), W is differentiable and concave, and c is differentiable and convex, g is differentiable and increasing (for every i, g'i(t) > 0 for all t), and k is a constant.
    1. Write down the first-order conditions for a solution of this problem.
    2. What is the relation (if any) between a solution of the first-order conditions and a solution of the problem?
    3. Solve the problem for n = 2, W(t1, t2) = −(t1 − 1)2 − t2
      2      , c(t1, t2) = t1 + t2, g(t1, t2) = t1 + t2, and an arbitrary value of k. Give an interpretation to the value of the Lagrange multiplier you find.

    Solution

    1. W'i(t) − c'i(t) − λg'i(t) = 0 for some λ for i = 1, ..., n.
    2. If t* solves the problem then g(t*) = k and there exists λ such that the first-order conditions are satisfied. (Note that there is no value of t for which ∇g(t) = (0, ..., 0).) If (t*,λ) solves the first-order conditions, g(t*) = k, and λg is convex, then t* is a solution of the problem (given the concavity of W(t) − c(t) and the convexity of g).
    3. The first-order conditions are
      −2(t1 − 1) − 1 − λ  =  0
      −2t2 − 1 − λ  =  0.
      We deduce that t1 − 1 = t2, so that from the constraint t1 + t2 = k we have t1 = (1/2)(k + 1) and t2 = (1/2)(k − 1). We have also λ = −k. Because g is linear, λg is convex, so the solution of the problem is (t1, t2) = ((1/2)(k + 1), (1/2)(k − 1)). The Lagrange multiplier is equal to the rate of change of the maximal value of the objective function, W(t*) − c(t*), with respect to k.