Mathematical methods for economic theory

Martin J. Osborne
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5.3 Exercises on conditions under which a stationary point is a global optimum

  1. Find all the local maxima and minima (if any) of the following functions, and determine whether each local extremum is a global extremum.
    1. f(x, y) = −x3 + 2xy + y2 + x.
    2. f(x, y, z) = 1 − x2 − y2 − z2.

    Solution

    1. First-order conditions are
      f'1(x, y) = −3x2 + 2y + 1 =  0
      f'2(x, y) = 2x + 2y =  0
      From the second we have y = −x, so that the first condition is 3x2 + 2x−1 = 0, which has solutions x = 1/3, x = −1. Thus there are two solutions of the first-order conditions: (1/3, −1/3) and (−1, 1).

      The Hessian of f is

      left parenthesis −6x 2 right parenthesis
      2 2
      which is indefinite for x = 1/3, with f"11(1/3, −1/3) < 0 and f"22(1/3, −1/3) > 0, and positive definite for x = −1. Thus (1/3, −1/3) is a saddle point and (−1, 1) is a local minimizer.

      (−1, 1) is not a global minimizer: for example, f(2,0) = −6 < f(−1, 1) = −1.

    2. First-order conditions: f'1(x, y, z) = −2x = 0, f'2(x, y, z) = −2y = 0, f'3(x, y, z) = −2z = 0. There is a unique solution, (x, y, z) = (0, 0, 0). The Hessian matrix is
      left parenthesis −2 0 0 right parenthesis
      0 −2 0
      0 0 −2
      The leading principal minors are −2 < 0, 4 > 0, and (−2)(4) = −8 < 0, so that (x1, x2, x3) = (0, 0, 0) is a local maximizer; the value at this maximum is 1.

      The Hessian is negative definite for all (x, y, z), so that f is concave. Thus (0, 0, 0) is a global maximizer.

  2. A competitive firm receives a price p > 0 for each unit of its output, pays a price w > 0 for each unit of its single input. Its output from using x units of the variable input is f(x). What do the results in Section 4 tell us about the value(s) of x that maximize the firm's profit in each of the following cases? Find the firm's maximum profit in each case.
    1. f(x) = x1/4. [Refer to a previous exercise.]
    2. f(x) = x. [Treat the cases p = w and p ≠ w separately.]
    3. f(x) = x2.

    Solution

    1. The firm's profit, as a function of x, is px1/4 − wx. By the previous exercise mentioned in the problem, this function is concave. Its stationary points are the solutions of p(1/4)x−3/4 − w = 0, so it has a single stationary point, x = (p/4w)4/3. This value of x is positive, so that it is in the interior of the interval on which f is defined, and hence by the first result in this section is the only positive global maximizer of the firm's profit. The firm's profit when x = (p/4w)4/3 is p(p/4w)1/3 − w(p/4w)4/3 = (1/41/3 − 1/44/3)p4/3/w1/3 = (3/44/3)(p4/3/w1/3) > 0. The firm's profit when x = 0 is 0, so the only maximizer of the firm's profit is x = (p/4w)4/3.
    2. If f(x) = x then the firm's profit is px − wx = (p − w)x, which is concave (the second derivative is zero).
      • If p = w then every value of x is a stationary point, so that by the first result in this section every value of x > 0 is a global maximizer of the firm's profit. The firm's profit at every such value of x is 0, so that x = 0 is also a global maximizer.
      • If p ≠ w the function has no stationary points. Thus the first result in the section implies that no value of x > 0 is a global maximizer. The firm's profit when x = 0 is 0, so this value of x maximizes the firm's profit if p < w (in which case the firm's profit for x > 0 is negative). If p > w, no value of x maximizes the firm's profit, which increases without bound as x increases.
    3. If f(x) = x2, the firm's profit is px2 − wx, which is not concave. By the first result in Section 4.4, if x maximizes profit then f'(x) = 0, or x = w/2p. The firm's profit at this value of x is −w2/4p < 0, while its profit for x = 0 is 0. Thus the only candidate for a maximizer of the firm's profit is x = 0. But looking at the function px2 − wx, we see that the firm's profit increases without bound, and hence has no maximum.
  3. Find all the local maximizers and local minimizers (if any) of the function f(x, y) = e3x − 3x + 4y2 − 1 (without any constraints). Determine whether each local maximizer you find is a global maximizer and whether each local minimizer is a global minimizer.

    Solution

    The first-order conditions are
    f'x(x, y) = 3e3x − 3  = 0
    f'y(x, y) = 8y  = 0.
    These equations have a single solution, (x, y) = (0, 0). The Hessian of the function is
    left parenthesis 9e3x 0 right parenthesis
    0 8
    which is positive definite for all (x, y). Thus the function is convex. Hence (x, y) = (0, 0) is the unique global (and in particular local) minimizer. The function has no maximizer.
  4. Let f(x1, x2) = x2
    1     − x1x2 + x2
    2     + 3x1 − 2x2 + 1.
    1. Is f convex, concave, or neither?
    2. Find any local maxima or minima of f. Are they global maxima/minima?

    Solution

    1. The Hessian matrix of f is
      left parenthesis 2 −1 right parenthesis
      −1 2
      This is positive definite, so f is convex.
    2. The first-order conditions are
      f'1(x1, x2)  =  2x1 − x2 + 3 = 0
      f'2(x1, x2)  =  −x1 + 2x2 − 2 = 0
      These equations have a unique solution, (x1, x2) = (−4/3, 1/3). Since f is convex, this is the global minimizer of f (and there are no other maximizers or minimizers).
  5. Using the results that relate the solutions of the first-order conditions with the solutions of maximization/minimization problems, what can you say about the local and global maxima and minima of the following functions?
    1. f(x, y, z) = x2 + 3y2 + 6z2 − 3xy + 4yz.
    2. f(x, y) = −x3 − y3 + 3xy.

    Solution

    1. The first-order conditions are
      2x − 3y = 0
      6y − 3x + 4z = 0
      12z + 4y = 0
      The unique solution is (x,y,z) = (0,0,0). The Hessian is
      left parenthesis 2 −3 0 right parenthesis
      −3 6 4
      0 4 12
      the leading principal minors of which are 2, 3, and 4. Thus the function is convex (in fact, strictly convex). Thus (0,0,0) is the global minimizer.
    2. The first-order conditions are
      −3x2 + 3y = 0
      −3y2 + 3x = 0
      These equations have two solutions, (0,0) and (1,1). The Hessian is
      left parenthesis −6x 3 right parenthesis
      3 −6y
      		.
      Thus the function is neither concave nor convex. At (0,0) the Hessian is neither positive nor negative semidefinite, so (0,0) is neither a maximizer nor a minimizer. At (1,1) the Hessian is negative definite, so that (1,1) is a local maximizer.
  6. Suppose that a firm that uses 2 inputs has the production function f(x1, x2) = 12x1/3
    1    x1/2
    2     and faces the input prices (p1, p2) and the output price q.
    1. Show that f is concave, so that the firm's profit is concave.
    2. Find a global maximum of the firm's profit (and give the input combination that achieves this maximum).

    Solution

    1. The Hessian matrix at (x1, x2) is
      left parenthesis −(8/3)x−5/3
      1      x1/2
      2      		 2x−2/3
      1      x−1/2
      2      		 right parenthesis
      2x−2/3
      1      x−1/2
      2      		 −3x1/3
      1      x−3/2
      2
      The leading principal minors are −(8/3)x−5/3
      1      x1/2
      2       < 0 and 8x−4/3
      1      x−1
      2       − 4x−4/3
      1      x−1
      2       = 4x−4/3
      1      x−1
      2       > 0. Hence the Hessian is negative definite, so that f is concave.
    2. The first-order conditions for a maximum of profit are
      qf'1(x1, x2) − p1  =  4qx−2/3
      1      x1/2
      2       − p1 = 0
      qf'2(x1, x2) − p2  =  6qx1/3
      1      x−1/2
      2       − p2 = 0
      These equations have a unique solution
      x*1  =  (24q2/p1p2)3
      x*2  =  (122q3/p1p2
      2      )2
      Since the objective function is concave, this input combination is the one that globally maximizes the firm's profit. The value of the maximal profit is obtained by substituting these optimal input values into the profit function.
  7. Suppose that a firm that uses 2 inputs has the production function f(x1, x2) = 4x1/4
    1    x1/4
    2     (where (x1, x2) is the pair of amounts of the inputs) and faces the input prices (p1, p2) and the output price q.
    1. Show that the firm's profit is a concave function of (x1, x2). (To show that f is concave, use first principles, not a general result about the concavity of Cobb-Douglas production functions.)
    2. Find the pair of inputs that maximizes the firm's profit.

    Solution

    1. First consider the production function f. The Hessian matrix at (x1, x2) is
      left parenthesis −(3/4)x−7/4
      1      x1/4
      2      		 (1/4)x−3/4
      1      x−3/4
      2      		 right parenthesis
      (1/4)x−3/4
      1      x−3/4
      2      		 −(3/4)x1/4
      1      x−7/4
      2
      The leading principal minors are −(3/4)x−7/4
      1      x1/4
      2       < 0 and (9/16)x−3/2
      1      x−3/2
      2       − (1/16)x−3/2
      1      x−3/2
      2       =       2x−3/2
      1      x−3/2
      2       > 0. Hence the Hessian is negative definite, so that f is concave. Thus given that the firm's revenue p1x1 + p2x2 is convex (and concave), the firm's profit, qf(x1, x2) − p1x1 − p2x2 is concave.
    2. The first-order conditions for a maximum of profit are
      qf'1(x1, x2) − p1  = qx−3/4
      3      x1/4
      2       − p1 = 0
      qf'2(x1, x2) − p2  = qx1/4
      1      x−3/4
      2       − p2 = 0.
      These equations have a unique solution
      x*1  = q2/(p3
      1      p2)1/2
      x*2  = q2/(p1p3
      2      )1/2
      (To obtain this solution you can, for example, isolate x2 in the second equation and plug it into the first equation.)

      The objective function is concave, so this input combination globally maximizes the firm's profit.

  8. Solve the problem
    maxx∫1
    0    −(x−y)2g(y)dy,
    where g is a function for which 0 ≤ g(y) ≤ 1 for all y and ∫1
    0    g(y)dy = 1.

    Solution

    The second derivative of the objective function is −2, so that the objective function is concave. (Use Leibniz's formula.) The unique solution of the first-order condition is x = ∫1
    0    yg(y)dy, so this is the solution of the problem. (If g is a probability density function then ∫1
    0    yg(y)dy is the mean of y.)
  9. A firm produces output using n inputs according to the differentiable production function f. It sells output at the price p and pays wi per unit for input i, i = 1, ..., n. Consider its profit maximization problem
    maxzpf(z) − w·z subject to z ≥ 0.
    Throughout the problem fix the value of w.
    1. Does the extreme value theorem imply that this problem has a solution?
    2. Denote by π(p) the firm's maximal profit, given p. Show that π'(p) = f(z*(p)). [Hint: Use the first-order conditions for the maximization problem.]
    3. Let q(p) = f(z*(p)), the output that the firm produces. By a previous problem, π is convex. What testable restrictions does this property of π imply for q, given the result in (b)? (What does the convexity of π tell you about its second derivative? How is the second derivative related to the derivative of q*?)

    Solution

    1. No, since the constraint set is not bounded.
    2. We have
      π'(p) = f(z*(p)) + ∑n
      j=1       [pf'j(z*(p))·∂z*j/∂p − wj·∂z*j/∂p].
      Now, the first-order conditions for the firm's maximization problem are pf'j(z*(p)) − wj = 0 for all j, so each term in square brackets is zero.
    3. The convexity of π means that π"(p) ≥ 0 for all p. From (b) we have π"(p) = q'(p). Thus the theory predicts that the output of a profit-maximizing firm is an increasing function of the price of output. That is, the supply function is upward-sloping.
  10. A firm produces the output of a single good in two plants, using a single input. The amount of the input it uses in plant i is zi. The output in each plant depends on the inputs in both plants (there is an interaction between the plants): the output in plant 1 is f(z1, z2) and that in plant 2 is g(z1, z2), where f and g are differentiable. The price of output is p > 0 and the price of the input is w > 0. Thus the firm's profit, which it wishes to maximize, is p[f(z1, z2) + g(z1, z2)] − w(z1 + z2).
    1. What are the first-order conditions for a solution of the firm's problem?
    2. If (z*1, z*2) maximizes the firm's profit and z*i > 0 for i = 1, 2, then does (z*1, z*2) necessarily satisfy the first-order conditions?
    3. If (z*1, z*2) satisfies the first-order conditions and z*i > 0 for i = 1, 2, then does (z*1, z*2) necessarily maximize the firm's profit? If not, are there any conditions on f and g under which (z*1, z*2) does necessarily maximize the firm's profit?

    Solution

    1. First-order conditions:
      p[f'1(z*1, z*2) + g'1(z*1, z*2)] − w = 0
      p[f'2(z*1, z*2) + g'2(z*1, z*2)] − w = 0
    2. Yes: at an interior maximum the first-order conditions must be satisfied.
    3. No: a solution of the first-order conditions is not necessarily a maximizer. If f and g are concave, however, then the objective function is concave, so the first-order conditions are sufficient.
  11. You have the following information about f, a function of n variables:
    • f is differentiable
    • f'i(x) = 0 for i = 1, ..., n if and only if x = x1, x = x2, x = x3, x = x4, or x = x5
    • the Hessian H of f is negative definite at x1, negative semidefinite at x2, positive definite at x3, negative definite at x4, and positive definite at x5
    • f(x1) = 3, f(x2) = 6, f(x3) = 2, f(x4) = 6, f(x5) = 5.
    What do you know about the local and global maximizer(s) and minimizers of f (in the absence of any constraint on x)?

    Solution

    The only possible local or global maximizers or minimizers of f are x1, x2, x3, x4, and x5. Thus if f has a maximum then x2 and x4 are its global maximizers, and if it has a minimum then x3 is its global minimizer. Whether or not it has a maximum or a minimum, x1 and x4 are local maximizers, and x3 and x5 are local minimizers.
  12. A firm faces uncertain demand D and has inventory I. It chooses its stock level Q ≥ 0 to minimize
    g(Q) = c(Q − I) + h∫Q
    0    (Q−D)f(D)dD + p∫a
    Q    (D−Q)f(D)dD,
    where c, I, h, p, and a are positive constants with p > c, and f is a nonnegative function that satisfies ∫a0f(D)dD = 1 (so that it can be interpreted as a probability distribution function).
    1. Denote by Q* the stock level that minimizes g(Q). Write down the first-order condition for Q* (in terms of c, I, h, p, and integrals involving the function f).
    2. What is the relation (if any) between an interior solution of the first-order condition and the solution of the problem? (Be precise.)
    3. Find the solution Q* in the case that f(D) = 1/a for all D.

    Solution

    1. The first-order condition is
      g'(Q*) = c + h∫Q*
      0      f(D)dD − p∫a
      Q*      f(D)dD = 0.
    2. g'(Q) = c + h∫Q
      0      f(D)dD − p∫a
      Q      f(D)dD and g"(Q) = (h + p)f(Q) ≥ 0 for all Q, so g is convex. Thus Q* > 0 solves the problem if and only if it satisfies the first-order condition.
    3. When f(D) = 1/a for all D, the first-order condition is
      c + (h/a)∫Q*
      0      dD − (p/a)∫a
      Q*      dD = 0,
      or
      c + (h/a)Q* − (p/a)(a − Q*) = 0,
      or
      c + (h+p)Q*/a − p = 0,
      or
      Q*= a(p − c)/(h + p).
  13. A politician chooses the number of hours h ≥ 0 of advertising that she buys; the cost of h hours is c(h), where c is a convex function. She wishes to maximize g(π(h))−c(h), where π(h) is her probability of winning when she buys h hours of advertising. Assume that π is a concave function, and g is an increasing and concave function.
    1. Write down the first-order condition for an interior solution (i.e. a solution h* with h* > 0) of this problem.
    2. What is the relation (if any) between an interior solution of the first-order condition and the solution of the problem? (Be precise.)

    Solution

    1. First-order condition: g'(π(h))π'(h) − c'(h) = 0.
    2. Since g is increasing and concave and π are concave, g(π(h)) is concave; since c is convex, −c is concave. Thus g(π(h)) − c(h) is the sum of two concave functions, and hence is concave. We conclude that an interior solution of the first-order condition is a solution of the problem.
  14. A firm sells goods in two markets. In each market i the price is fixed at pi, with pi > 0. The amount of the good the firm sells in each market depends on its advertising expenditure in both markets. If it spends ai on advertising in market i, for i = 1, 2, its sales in market 1 are f(a1,a2) and its sales in market 2 are g(a1, a2), where f and g are twice differentiable functions. The firm's cost of production is zero. Thus its profit is
    p1f(a1,a2) + p2g(a1, a2) − a1 − a2.
    It chooses (a1, a2) to maximize this profit.
    1. What are the first-order conditions for the firm's optimization problem?
    2. Suppose that (a*1, a*2) maximizes the firm's profit and a*i > 0 for i = 1, 2. Does (a*1, a*2) necessarily satisfy the first-order conditions?
    3. Suppose that (a*1, a*2) satisfies the first-order conditions and a*i > 0 for i = 1, 2. Under what condition is (a*1, a*2) a local maximizer of the firm's profit?
    4. Suppose that (a*1, a*2) satisfies the first-order conditions and a*i > 0 for i = 1, 2. Under what conditions on the functions f and g is (a*1, a*2) necessarily a global maximizer of the firm's profit?

    Solution

    1. The first-order conditions are
      p1f'i(a*1, a*2) + p2g'i(a*1, a*2) − 1 = 0 for all i = 1,2.
    2. Yes: any interior solution of a maximization problem in which the objective function is differentiable satisfies the first-order condition.
    3. If (a*1, a*2) satisfies the first-order conditions and the Hessian of the objective function at (a*1, a*2) is negative definite then (a*1, a*2) is a local maximizer. The Hessian is
      left parenthesis p1f"11(a*1, a*2) + p2g"11(a*1, a*2) p1f"12(a*1, a*2) + p2g"12(a*1, a*2) right parenthesis
      p1f"21(a*1, a*2) + p2g"21(a*1, a*2) p1f"22(a*1, a*2) + p2g"22(a*1, a*2)
      		.
      [Note that the condition that the Hessian be negative definite at (a*1, a*2) is sufficient, not necessary. Note also that this condition is not the same as the objective function being “concave at (a*1, a*2)". If you look at the definition of concavity, you see that every function is concave on a domain consisting of a single point!]
    4. If f and g are concave then the objective function is concave (since the sum of concave functions is concave), and hence any solution of the first-order conditions is a global maximizer.