Mathematical methods for economic theory

Martin J. Osborne
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6.1.4 Exercises on optimization with an equality constraint: conditions under which a stationary point is a global optimum

  1. Consider the problem
    maxxy xy subject to x + y = 2.
    Show that for λ = 1 the Lagrangean has a stationary point at (x, y) = (1, 1). Show also that (x, y) = (1, 1) does not maximize the Lagrangean function L(x, y) = xy − 1·(x + y − 2).

    Solution

    For λ = 1 the derivatives of the Lagrangean are zero if and only if
    y − 1  =  0
    x − 1  =  0,
    which yield (x, y) = (1, 1).

    The point (x, y) = (1, 1) does not maximize the Lagrangean with λ = 1 because L(1, 1) = 1, whereas L(2, 2) = 2, for example.

  2. A firm's production function is f(x1, x2) and the prices of its inputs are w1 > 0 and w2 > 0; f is concave and increasing (f'i(x1, x2) > 0 for each i, for all (x1, x2)). The firm wishes to minimize the cost of producing q units of output.
    1. Write down the firm's optimization problem.
    2. Write down the first-order conditions for the problem.
    3. If (x1, x2) satisfies the first-order conditions and the constraint, is it necessarily a solution of the problem? Justify your answer carefully.

    Solution

    1. minx1,x2(w1x1 + w2x2) subject to f(x1, x2) = q.
    2. w1 − λf'1(x)  =  0
      w2 − λf'2(x)  =  0
    3. From the first-order conditions we have λ = wi/f'i(x1, x2) > 0. The objective function is convex, and the constraint function is concave, so the Lagrangean is convex. Thus if (x*1, x*2) satisfies the first-order conditions and the constraint then it is a solution to the problem.