Mathematical methods for economic theory

Martin J. Osborne
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6.1.1 Exercises on optimization with an equality constraint: necessary conditions for an optimum for a function of two variables

  1. Consider the problem
    maxx,y x2 + y2 subject to x2 + xy + y2 = 3,
    and the similar problem in which max is replaced by min. Does the extreme value theorem show that these problems have solutions? Use the Lagrangean method to find the solutions.

    Solution

    The objective function is continuous (it is a polynomial). We can write the constraint as (x + (1/2)y)2 + (3/4)y2 = 3. Each of the terms on the left hand side is nonnegative, so certainly at any point that satisfies the constraint we have (3/4)y2 ≤ 3, or −2 ≤ y ≤ 2. At any point that satisfies the constraint we have also (x + (1/2)y)2 ≤ 3, so that −√3 ≤ x + (1/2)y ≤ √3, and hence, given −2 ≤ y ≤ 2, we have −√3 − 1 ≤ x ≤ √3 + 1. Thus the constraint set is bounded. The constraint set is also closed (it is defined by an equality). Thus the extreme value theorem shows that the problems have solutions.

    The first-order conditions and the constraint are

    2x − λ(2x + y)  =  0
    2y − λ(x + 2y)  =  0
    x2 + xy + y2  =  3
    The first two equations imply that x2 = y2, so that either x = y or x = −y. Now substitute for y in the third equation to conclude that the three equations have four solutions: (1, 1, 2/3), (−1, −1, 2/3), (31/2, −31/2, 2), (−31/2, 31/2, 2). The unique solution of g'x'(x, y) = g'y(x, y) = 0 (where g is the function in the constraint) is (x, y) = (0, 0), which does not satisfy the constraint, so the only candidates for solutions of the problems are the four solutions of the first-order conditions.

    The values of the objective function at these four points are 2, 2, 6, and 6. Thus, given that the two problems have solutions, the solutions of the minimization problem are (x, y) = (1, 1) and (x, y) = (−1, −1), and the solutions of the maximization problem are (x, y) = (31/2, −31/2) and (x, y) = (−31/2, 31/2).

  2. Find the solution of
    minx,y (x−1)2 + y2 subject to y2 − 8x = 0,
    taking as given that the problem has a solution.

    Solution

    The first-order conditions and the constraint are
    2(x − 1) + 8λ  =  0
    2y − 2λy  =  0
    y2 − 8x  =  0.
    The unique solution of these conditions is (x, y, λ) = (0, 0, 1/4). The equations g'x(x, y) = g'y(x, y) = 0 have no solution. Thus if the problem has a solution, it is (x, y) = (0, 0).
  3. What does the Lagrangean procedure say about the solutions of each of the following problems?
    maxx,y y − 2x2 + x subject to (x + y)2 = 0
    and
    maxx,y y − 2x2 + x subject to x + y = 0

    Solution

    First problem: The first-order conditions and the constraint are
    −4x + 1 − 2λ(x + y)  =  0
    1 − 2λ(x+y)  =  0
    (x + y)2  =  0
    These equations have no solution. Let g(x, y) = (x + y)2, the constraint function. The equations g'x(x, y) = g'y(x, y) = 0 have (x, −x) as solutions for any value of x, and all such solutions satisfy the constraint. Thus the Lagrangean procedure says nothing about the solution (except that it satisfies the constraint!).

    Second problem: The first-order conditions and the constraint are

    −4x + 1 − λ  =  0
    1 − λ  =  0
    x + y  =  0
    These equations have the unique solution (x, y, λ) = (0, 0, 1). Let g(x, y) = x + y, the constraint function. The equations g'x(x, y) = g'y(x, y) = 0 have no solutions. Thus the Lagrangean procedure says that if the problem has a solution (which in fact it does) then that solution is (x, y) = (0, 0).

    NOTE: The two problems are equivalent (any point that satisfies the constraint in the first problem satisfies the constraint in the second problem, and vice versa), so in fact the solution of the second problem is the unique solution of the first problem.

  4. Solve the problem
    maxx,yxy subject to x2 + y2 = 2c2,
    where c > 0 is a constant. Be sure to lay out all the steps of your argument!

    Solution

    The objective function is continuous, and the constraint set is closed and bounded (it is a circle), so by the extreme value theorem the problem has a solution. Any solution of the problem satisfies the constraint and either the first-order conditions or the condition that g'1(x, y) = g'2(x, y) = 0, where g is the constraint function. Now, g'1(x, y) = 2x and g'2(x, y) = 2y, so that the only solution of g'1'(x, y) = g'2(x, y) = 0 is (x, y) = (0, 0). This solution does not satisfy the constraint, so we conclude that any solution of the problem satisfies the first-order conditions. The first-order conditions and the constraint are
    y − 2λx  =  0
    x − 2λy  =  0
    x2 + y2  =  2c2.
    From the first-order conditions, either y = 0 or λ2 = 1/4. If y = 0 then x = 0, so that the constraint is not satisfied. Thus we have λ2 = 1/4, so that either λ = 1/2 or λ = −1/2. Thus the first-order conditions have four solutions: (c, c), (c, −c), (−c, c), and (−c, −c). The value of the objective function at (c, c) and (−c, −c) is c2, while its value at (c, −c) and (−c, c) is −c2. Thus the problem has two solutions: (c, c) and (−c, −c).
  5. Solve the problem
    maxx,yx + 2y subject to 2x2 + y2 = 18.
    Be sure to lay out carefully the steps in your argument.

    Solution

    The constraint set is compact and the objective function is continuous, so the problem has a solution. Suppose that (x*, y*) solves the problem. Then we know that g(x*, y*) = c and either there exists λ such that (x*, y*, λ) satisfies the first-order conditions or ∇g(x*, y*) = (0, 0). The first-order conditions are
    1 − 4λx  =  0
    2 − 2λy  =  0
    which imply that 4x = y. Combined with the constraint we deduce that (x, y) = (1, 4) or (x, y) = (−1, −4). We have ∇g(x, y) = (4x, 2y), which is (0, 0) only if (x, y) = (0, 0), which violates the constraint. Since f(1, 4) = 9 > f(−1, −4) = −9, the unique solution of the problem is (x, y) = (1, 4).