ECON2285: Mathematical Economics

ECON2285: Mathematical Economics Yulei Luo Economics, HKU September 17, 2018 Luo, Y. (Economics, HKU) ME September 17, 2018 1 / 46

Static Optimization and Extreme Values In this topic, we will study goal equilibrium, in which the equilibrium state is de ned as the outcome of the optimizing agents (consumers, rms, and governments). Our primary focus will be on the classical techniques -di erential calculus- for nding optimal positions. The essence of Economics is about how consumers and rms make optimal choices in a certain environment such that limited resources can be allocated e ciently across agents and over time. The most common criterion of choice among alternatives in economics is the goal of maximizing sth. or minimizing sth. They can be categorized under general heading of optimization. From a purely mathematical point of view, maximum or minimum are also called extreme values. Luo, Y. (Economics, HKU) ME September 17, 2018 2 / 46

(Conti.) In formulating an optimization problem, we need rst de ne an objective function in which the dependent variable represents the object of maximization or minimization and in which the set of independent variables represents the objects whose values the agents can choose. Those independent variables are referred to as choice variables (also called decision, control, or policy variables). E.g., a rm may seek to maximize pro t, that is, to maximize the di erence between total revenue R and total cost C. Given a market demand, R and C are both functions of the output level Q: π (Q) = R (Q) C (Q), (1) in which π is the objective function and Q is the choice variable. The optimization problem is to choose Q to maximize π: π = max π (Q) = R (Q) C (Q), (2) Q where π is the optimal level of π, i.e., π s maximal level. Luo, Y. (Economics, HKU) ME September 17, 2018 3 / 46

Relative Maximum and Minimum: First-Derivative Test We consider the optimizing problem with the general function y = f (x),and attempt to develop a procedure to nd the optimal value of x that will maximize or minimize y. See Figure 9.1. (c) in CW. The points E and F are relative (or local) extremum, in the sense that each of these points represents an extremum in some neighborhood of the point only. Our discussion will mainly focus on relative extrema since an absolute (or global) extrema must be either a relative extrema or one of the ends of the function. If we know all the relative extrema, it is easy to get the absolute extrema by selecting the largest or smallest of these and comparing it with the end points. Luo, Y. (Economics, HKU) ME September 17, 2018 4 / 46

(Conti.) For y = f (x), the rst derivative f 0 (x) plays a major role in nding extreme values. For smooth functions (e.g., f is continuously di erentiable. For non-smooth functions, the derivative doesn t exist at some point), relative extreme values can be only occur where f 0 (x) = 0, (3) which is a necessary (but not su cient) condition for a relative extremum (either maximum or minimum). Luo, Y. (Economics, HKU) ME September 17, 2018 5 / 46

Theorem (First-derivative test for relative extremum) If f 0 (x 0 ) = 0, then the value of the function at x 0, f (x 0 ), will be (a) A relative maximum if f 0 (x) changes its sign from positive to negative from the immediate left of the point x 0 to immediate right. (b) A relative minimum if f 0 (x) changes its sign from negative to positive from the immediate left of x 0 to immediate right. (c) No extreme point if f 0 (x) has the same sign on some neighborhood. We call the value of x 0 a critical value of x if f 0 (x 0 ) = 0, and refer to f (x 0 ) as a stationary value of y. The point with coordinates x 0 and f (x 0 ) can be called a stationary point (wherever the slope is 0, the point in question is never situated on an upward or downward incline, but is rather at a standstill point). Luo, Y. (Economics, HKU) ME September 17, 2018 6 / 46

(Conti.) Example: For y = f (x) = (x 1) 3, (4) x = 1 is not an extreme point even f 0 (1) = 0 since f 0 (x) has the same sign on some neighborhood of 1. Example: For y = f (x) = x 3 12x 2 + 36x + 8, (5) set f 0 (x) = 3x 2 24x + 36 = 0 (6) to get the critical values. Solving it gives x = 2 or 6.It is easy to verify that f 0 (x) > 0 for x < 2, and f 0 (x) < 0 for x > 2. Thus x = 2 is a maximum point and the corresponding maximum value of the function f (2) = 40. Similarly, we can also verify that x = 6 is a minimum point. Luo, Y. (Economics, HKU) ME September 17, 2018 7 / 46

Second and Higher Derivatives The derivative of f 0 (x) is called second derivative. Similarly, we can nd derivatives of even higher orders. These will enable us to develop alternative criteria for locating the relative extrema of a function. The SD of f is denoted by f 00 or d 2 y/dx 2. If f 00 exists for all x, f (x) is said to be twice di erentiable; if f 00 is continuous, f (x) is said to be twice continuously di erentiable. The higher order derivatives can be obtained and symbolized along the same line: f 000 (x), f (4) (x),, f (n) (x) or (7) d 3 y dx 3, d 4 y dx 4,, d n y dx n. (8) Example: y = f (x) = 4x 4 x 3 + 17x 2 + 3x 1, then f 0 (x) = 16x 3 3x 2 + 34x + 3, f 00 (x) = 48x 2 6x + 34, f 000 (x) = 96x 6, f (4) (x) = 96, f (5) (x) = 0. Luo, Y. (Economics, HKU) ME September 17, 2018 8 / 46

Interpretation of the Second Derivative The second derivative f 00 measures the rate of change of the rst derive f 0. In other words, f 00 measures the rate of change of the rate of change of the function f. With a given in nitesimal increase in x from x = x 0, f 0 (x 0 ) > 0 increase f 0 means that the value of f tends to (x 0 ) < 0 decrease ; f 00 (x 0 ) > 0 increase f 00 means that the slope of f tends to (x 0 ) < 0 decrease. For example, when f 0 (x 0 ) > 0 and f 00 (x 0 ) > 0, the slope of the curve is positive and increasing. Likewise, f 0 (x 0 ) > 0 and f 00 (x 0 ) < 0 means that the slope of the curve is positive but decreasing the value of the function is increasing at a decreasing rate. See Figure 9.5 (a) and (b) in the book. Luo, Y. (Economics, HKU) ME September 17, 2018 9 / 46

(Conti.) f 00 (x 0 ) relates to the curvature of a graph; it determines how the curve tends to bend itself. There are two di erent types of curvature: Strictly concave (Figure 9.5 (a)) Strictly convex (Figure 9.5 (b)) A function whose graph is strictly concave (strictly convex) is called a strictly concave (strictly convex) function. Precise geometric characterization of a strictly concave (convex) function is as follows. If we pick any pair of points M and N on its curve and joint them by a straight line, the line segment MN must lie entirely below (above) the curve, except at points M and N. Note that if the condition is relaxed somewhat, so that the line segment MN is allowed to either below (above) the curve or along (coinciding with) the curve, then we can describe instead a concave (convex) function, without the adverb strictly. Luo, Y. (Economics, HKU) ME September 17, 2018 10 / 46

(Conti.) Note that a strictly concave or convex curve can never contain a linear segment anywhere. While a strictly concave (convex) function is automatically a concave (convex) function, the converse is not true. Example: Consider a quadratic function y = ax 2 + bx + c (a 6= 0). (9) it is obvious that the second derivative of y w.r.t. x, a, determines whether this function will have a strictly convex (U-shaped) or a strictly concave (inverse U-shaped) graph. Luo, Y. (Economics, HKU) ME September 17, 2018 11 / 46

Application to the Expected Utility Model Assume that a potential gamble player has a strictly concave utility function u = u (x) where x is the payo, with u (0) = 0, u 0 (x) > 0 (positive marginal utility of income or payo ), and u 00 (x) < 0 (diminishing marginal utility) for all x. The decision facing this player involves the choice between two actions: First, by not playing the game, he saves $15 and enjoys the utility u (15). Second, by playing, the person has a 1/2 probability of receiving $10 and thus enjoying u(10), plus a 1/2 probability of receiving $20 and thus enjoying u(20). The expected utility from playing is EU = 1 2 u (10) + 1 2 u (20), (10) which is less than u (15) since the utility function is strictly concave (See Figure 9.6 (a)). Hence, this risk averse player will always avoid this game. Luo, Y. (Economics, HKU) ME September 17, 2018 12 / 46

Theorem (Second-derivative test for relative extremum) If f 0 (x 0 ) = 0, then the value of the function at x 0, f (x 0 ), will be (a) A relative maximum if f 00 (x 0 ) < 0. (b) A relative maximum if f 00 (x 0 ) > 0. Note that this test is more convenient to use than the rst derivative test since it doesn t require us to check the derivative sign to both the left and right of x. Example: Given that y = f (x) = 4x 2 x, f 0 (x) = 8x 1 and f 00 (x) = 8 > 0, we know f (x) reaches its minimum at x = 1/8. Example: Given that y = g (x) = x 3 3x 2 + 2, we have g 0 (x) = 3x 2 6x and g 00 (x) = 6x 6. Setting g 0 (x) = 0, we obtain two critical values x = 0 or 2, which in turn yield the two extreme values g (0) = 2 (a maximum because g 00 (0) = 6 < 0) and g (2) = 2 (a minimum because g 00 (0) = 6 > 0). Luo, Y. (Economics, HKU) ME September 17, 2018 13 / 46

Necessary versus Su cient Conditions As was the case with the rst derivative test, f 0 (x) = 0 also plays the role of a necessary condition in the second derivative test. Since this condition is based on the FO derivative, it is often referred to as the rst order condition (FOC). Once we nd the FOC satis ed at x = x 0, the negative (positive) sign of f 00 (x 0 ) is su cient to establish the stationary value in question as a relative maximum (minimum). These su cient conditions are often referred to as second order conditions (SOC). Note that a relative maximum (minimum) can occur not only when f 00 (x 0 ) is negative (positive), but also when f 00 (x 0 ) is zero. Consequently, second order necessary conditions must be couched in terms of weak inequalities: for a stationary value f (x 0 ) to be maximum (minimum), it is necessary that f 00 (x 0 ) 0( 0). Luo, Y. (Economics, HKU) ME September 17, 2018 14 / 46

Condition for Pro t Maximization Suppose that R = R (Q) is the total revenue function and C = C (Q) is the total cost function where Q is the level of output. The pro t function is de ned as π (Q) = R (Q) C (Q). (11) The pro t-maximizing output Q must satisfy π 0 (Q ) = R 0 (Q ) C 0 (Q ) = 0 or R 0 (Q ) = C 0 (Q ), (12) which means that the marginal revenue (MR) of output is equal to the marginal cost (MC) of output. To be sure that the above FOC leads to a maximum, we require: π 00 (Q ) = R 00 (Q ) C 00 (Q ) < 0, (13) which means that if the rate of change of MR is less than the rate of change of MC at Q, then output Q will maximize pro t. Luo, Y. (Economics, HKU) ME September 17, 2018 15 / 46

(Conti.) Example: Given that R (Q) = 1200Q 2Q 2 and C (Q) = Q 3 61.25Q 2 + 1528.5Q + 2000, then the pro t function is π (Q) = Q 3 + 59.25Q 2 328.5Q 2000 (14) and setting the π 0 (Q) = 0 gives the rst order condition (FOC) which means that π 0 (Q) = 3Q 2 + 118.5Q 328.5 = 0, (15) Since π 00 (Q ) = 6Q + 118.5 = the pro t-maximizing output is Q = 36.5. Q = 3 or 36.5. (16) 100.5 > 0 when Q = 3 100.5 < 0 when Q = 36.5, (17) Luo, Y. (Economics, HKU) ME September 17, 2018 16 / 46

Taylor Series and the Mean-Value Theorem Now we consider how to expand a function y = f (x) around any point x = x 0 into what is known as Taylor series. This expansion means to transform the function into a polynomial form, in which the coe cients of the various terms are expressed in terms of the derivative values f 0 (x 0 ), f 00 (x 0 ), etc. evaluated at the point of expansion x 0. Theorem (Taylor s Theorem) Given an arbitrary function y = f (x), if we know the values of all derivatives: f 0 (x 0 ), f 00 (x 0 ), etc., then this function can be expanded around the point x 0 as follows: f (x) = f (x 0 ) + f 0 (x 0 ) (x x 0 ) + f 00 (x 0 ) 2! + + f (n) (x 0 ) (x x 0 ) n + R n n! = P n + R n (x x 0 ) 2 (18) (19) Luo, Y. (Economics, HKU) ME September 17, 2018 17 / 46

Theorem (conti.) where n! = n(n 1) 1 is the n factorial, P n is the n-th degree polynomial and R n denotes a remainder which can be expressed in the Lagrange form: R n = f (n+1) (p) (x x 0 ) n+1 (20) (n + 1)! where p is some number between x and x 0. Note that when n = 0, the Taylor series reduce to the so-called mean-value theorem. Luo, Y. (Economics, HKU) ME September 17, 2018 18 / 46

Theorem (Mean-value Theorem) f (x) = f (x 0 ) + f 0 (p) (x x 0 ) or f (x) f (x 0 ) = f 0 (p) (x x 0 ), (21) which means that the di erence between the value of the function f at x 0 and at any other x value can be expressed as the product of the di erence x x 0 and f 0 (p) with p being some point between x and x 0. If x 0 = 0, the Taylor series reduce to the Maclaurin series: where p 2 (0, x). f (x) = f (0) + f 0 (0) x + f 00 (x 0 ) x 2 2! (22) + + f (n) (x 0 ) x n + f (n+1) (p) n! (n + 1)! x n+1, (23) Luo, Y. (Economics, HKU) ME September 17, 2018 19 / 46

Nth-Derivative Test A relative extremum of the function f can be equivalently de ned as follows: De nition A function f (x) attains a relative maximum (minimum) value at x 0 if f (x) f (x 0 ) is nonpositive (nonnegative) for values of x in some neighborhood of x 0. Suppose that f (x) has nite, continuous derivatives up to the desired order at x = x 0, then the function can be expanded around x = x 0 as a Taylor series: f (x) = f (x 0 ) + f 0 (x 0 ) (x x 0 ) + f 00 (x 0 ) (x x 0 ) 2 (24) 2! + + f (n) (x 0 ) (x x 0 ) n + f (n+1) (p) (x x 0 ) n+1, n! (n + 1)! from which we have the Nth-Derivative Test. Luo, Y. (Economics, HKU) ME September 17, 2018 20 / 46

Theorem (Nth-Derivative Test) If f 0 (x 0 ) = 0, and if the rst nonzero derivative value at x 0 encountered in successive derivation is that of the Nth derivative, f (N ) (x 0 ) 6= 0, then the stationary value f (x 0 ) will be (a) A relative maximum if N is an even number and f (N ) (x 0 ) < 0. (b) A relative minimum if N is an even number and f (N ) (x 0 ) > 0. (c) An in ection point if N is odd. Example: Given that y = f (x) = (7 x) 4, since f 0 (7) = 4 (7 7) 3 = 0; (25) f 00 (7) = 12 (7 7) 2 = 0; (26) f 000 (7) = 24 (7 7) = 0; (27) f (4) (7) = 24 > 0, (28) x = 7 is a minimum point such that f (7) = 0. Luo, Y. (Economics, HKU) ME September 17, 2018 21 / 46

Derivatives of Exponential and Logarithmic Functions Both functions have important applications in economics, especially in growth theory and economic dynamics. The exponential function may be represented in the form: y = f (t) = b t (b > 1) (29) where b is a xed base of the exponent. Its generalized form: y = ab ct. If the base is the irrational number denoted by e = 2.71828..., the function y = ae rt is referred to the natural exponential function, which can be alternatively written as y = a exp (rt) (30) Luo, Y. (Economics, HKU) ME September 17, 2018 22 / 46

(Conti.) For the exponential function and the natural exponential function, taking the log of to the base and the base, respectively, we have t = log b y and t = log e y = ln y. (31) Some rules: ln (uv) = ln u + ln v (log of product) (32) ln (u/v) = ln u ln v (log of quotient) (33) ln (u a ) = a ln u (log of power) (34) log b u = (log b e) (log e u) (conversion of log base) (35) log b e = 1/ log e b (inversion of log base) (36) Luo, Y. (Economics, HKU) ME September 17, 2018 23 / 46

(Conti.) Derivatives of Exponential and Logarithmic Functions d exp (t) dt d exp (f (t)) dt The case of Base b : d ln (t) = exp (t) ; = 1 dt t = f 0 d ln (f (t)) (t) exp (f (t)) ; dt (37) = f 0 (t) f (t). (38) db t dt db f (t) dt = b t ln b; d log b (t) dt = 1 t ln b = f 0 (t) b f (t) ln b; d log b (f (t)) dt = f 0 (t) f (t) (39) 1 ln b. (40) Luo, Y. (Economics, HKU) ME September 17, 2018 24 / 46

An application: Find dy dx from y = x a e kx c. Taking the natural log of both sides gives ln y = a ln x + kx c. (41) Di erentiating both sides w.r.t. x yields 1 dy y dx = a dy a + k =) x dx = x + k x a e kx c. (42) Luo, Y. (Economics, HKU) ME September 17, 2018 25 / 46

The Case of More than One Choice Variable The objective is to nd the relative extreme values of an objective function that involves two or more choice variables. We can express the derivative version of rst and second condition in terms of di erentials. Consider the function, we have dz = f 0 (x) dx. (43) Since f 0 (x) = 0 is the necessary condition for extreme values, dz = 0 is also the necessary condition for extreme values. This FOC requires that dz = 0 as x varies. What is the su cient conditions in terms of second-order di erentials? Di erentiating dz = f 0 (x) dx gives d 2 z = d (dz) = d f 0 (x) dx = f 00 (x) dx 2, (44) where d 2 z means that the second di erential of z and dx 2 means the squaring of the rst order di erential dx. Hence, d 2 z < 0(> 0) i f 00 (x) < 0(> 0), which means that the second-order su cient condition for maximum (minimum) of z is d 2 z < 0(> 0). Luo, Y. (Economics, HKU) ME September 17, 2018 26 / 46

(Conti.) For the function z = f (x, y), the FO necessary condition for an extremum again involves dz = 0 for arbitrary values of dx and dy. In this case, the total di erential is dz = f x dx + f y dy. (45) Thus the equivalent derivative version of the FO necessary condition dz = 0 is f x = f y = 0. (46) To develop a su cient condition, we must look at the SO total di erential, which is related to the SO partial derivatives. Luo, Y. (Economics, HKU) ME September 17, 2018 27 / 46

Second-Order Partial Derivatives Since the rst-order derivatives, f x and f y, of z = f (x, y), are themselves functions of x and y, we can nd SO partial derivatives: f xx = x f x or 2 z x 2 = z (47) x x f yy = y f y (48) f xy = 2 z x y = x f yx = 2 z y x = y z y z x (49). (50) where f xy and f yx are called cross (or mixed) partial derivatives. They are identical as long as they are both continuous (Young s theorem). Luo, Y. (Economics, HKU) ME September 17, 2018 28 / 46

Second-Order Total Di erentials From the FO total di erential, dz = f x dx + f y dy, (51) we can obtain the SO total di erential as follows: d 2 z = d (dz) = (dz) dx + (dz) dy x y (fx dx + f y dy) (fx dx + f y dy) = dx + dy x y = (f xx dx + f xy dy) dx + (f yx dx + f yy dy) dy = f xx dx 2 + 2f xy dxdy + f yy dy 2 (if f xy = f yx ) (52) Example: Given z = x 3 + 5xy y 2, f xx = 6x, f xy = 5, and f yy = 2 implies that d 2 z = 6xdx 2 + 10dxdy 2dy 2. (53) Luo, Y. (Economics, HKU) ME September 17, 2018 29 / 46

Second-Order Su cient Condition Using the concept of d 2 z, we can state the SO su cient condition for: A maximum of z = f (x, y) if d 2 z < 0 for any values of dx and dy, not both zero; A minimum of z = f (x, y) if d 2 z > 0 for any values of dx and dy, not both zero. The above conditions are only su cient because it is again possible for to take d 2 z a zero value at a maximum or a minimum. For this reason, SO necessary conditions must be stated with weak inequalities: For maximum of z, if d 2 z 0 for any values of dx and dy, not both zero; For minimum of z, if d 2 z 0 for any values of dx and dy, not both zero. Luo, Y. (Economics, HKU) ME September 17, 2018 30 / 46

(conti.) The SO di erential conditions can be translated into equivalent condition on SO derivatives. The following is the main result (A detailed proof requires a knowledge of quadratic form and will be given later.): Theorem d 2 z < 0 i f xx < 0, f yy < 0, and f xx f yy > (f xy ) 2 d 2 z > 0 i f xx > 0, f yy > 0, and f xx f yy > (f xy ) 2 Conditions for maximum: f x = f y = 0 (FO necessary condition) and f xx < 0, f yy < 0, and f xx f yy > (f xy ) 2 (SO su cient condition). Theorem Conditions for minimum: f x = f y = 0 (FO necessary condition) and f xx > 0, f yy > 0, and f xx f yy > (f xy ) 2 (SO su cient condition). Luo, Y. (Economics, HKU) ME September 17, 2018 31 / 46

(Conti.) Example: Find the extreme values of z = 8x 3 + 2xy 3x 2 + y 2 + 1. First, we derive all partial derivatives: f x = 24x 2 + 2y 6x, f y = 2x + 2y f xx = 48x 6, f yy = 2, f xy = 2. Second, setting f x = f y = 0 gives 24x 2 + 2y 6x = 0, 2x + 2y = 0, (54) which gives two solutions for x: x = 0 or 1/3, and x = 0 or 1/3. Since f xx (0, 0) = 6 and f yy (0, 0) = 2, it is impossible that f xx f yy > (f xy ) 2 = 4, so that (x, y ) = (0, 0) is not extreme point. For the solution (x, y ) = (1/3, 1/3), we have f xx (1/3, 1/3) = 10, f yy (1/3, 1/3) = f xy (1/3, 1/3) = 2, and f xx f yy > (f xy ) 2, so (1/3, 1/3) is a relative minimum point. Luo, Y. (Economics, HKU) ME September 17, 2018 32 / 46

Quadratic Forms A function q with n variables is said to have the quadratic form if it can be written as q (u 1,, u n ) = d 11 u 2 1 + 2d 12 u 1 u 2 + + 2d 1n u 1 u n (55) +d 22 u 2 2 + 2d 23 u 2 u 3 + + 2d 2n u 2 u n + +d nn u 2 n If we let d ij = d ji, i < j, then q (u 1,, u n ) can be written as q (u 1,, u n ) = n n i=1 j=1 d ij u i u j = u 0 Du, (56) where the quadratic-form systematic matrix 2 3 d 11 d 1n D = 4 5 and u 0 = u 1 u n d n1 d nn (57) Luo, Y. (Economics, HKU) ME September 17, 2018 33 / 46

De nition A quadratic form q (u 1,, u n ) = u 0 Du is said to be (a) positive de nite if q (u) > 0 if all u 6= 0; (b) positive semi-de nite if q (u) 0 if all u 6= 0; (c) negative de nite if q (u) < 0 if all u 6= 0; (d) negative semi-de nite if q (u) 0 if all u 6= 0. Sometimes we say that a matrix D is positive (negative) de nite if the corresponding quadratic form is positive (negative) de nite. Luo, Y. (Economics, HKU) ME September 17, 2018 34 / 46

Determinantal Test for Sign De niteness Theorem For the quadratic form q (u) = u 0 Du, the necessary and su cient conditions for positive de niteness is that the principle minors of jdj are positive: 2 jd 1 j = d 11 > 0, jd 2 j = d 11 d 12 d 21 d 22 > 0,, D n = 4 d 11 d 1n d n1 d nn The corresponding N&S condition for negative de niteness is that the PMs alternates in sign: 3 5 > 0. jd 1 j < 0, jd 2 j > 0, jd 3 j < 0,. (58) Luo, Y. (Economics, HKU) ME September 17, 2018 35 / 46

(Conti.) Two-variable quadratic-form case: Consider the second-order total di erential d 2 z = f xx dx 2 + 2f xy dxdy + f yy dy 2 = dx dy f xx f xy f xy f yy dx dy. (59) Thus, for the function z = f (x, y), (a) d 2 z is positive de nite i f xx > 0 and f xx f yy > (f xy ) 2 ; (b) d 2 z is negative de nite i f xx < 0 and f xx f yy > (f xy ) 2. Note that since f xx f yy (f xy ) 2 > 0, f xx and f yy must have the same sign. Hence, this is just the SO su cient condition presented before. The determinant jhj = f xx f xy f xy f yy is referred to as the discriminant of the quadratic-form of f and is also called the Hessian determinant. Luo, Y. (Economics, HKU) ME September 17, 2018 36 / 46

(Conti.) Example: q = 5u 2 + 3uv + 2v 2 is positive de nite because the discriminant of q is 10 3 3 4, with leading principle minors: 10 > 0 and 10 3 3 4 = 31 > 0. Example: Given that f xx = 2, f yy = 1, and f xy = 1 at a certain point on a function z = f (x, y), does d 2 z have a de nite sign at that point regardless of the values of dx and dy? In this case, d 2 z is negative de nite because the discriminant of the quadratic-form d 2 z is in the case 2 1 1 1, with leading PMs 2 < 0 and 2 1 1 1 = 1 > 0. Luo, Y. (Economics, HKU) ME September 17, 2018 37 / 46

Objective Functions with More than Two Variables Consider the following function z = f (x 1, x 2,, x n ),the total di erential is dz = f 1 dx 1 + f 2 dx 2 + + f n dx n, (60) so that the necessary condition for extremum is dz = 0 for arbitrary dx i, which in turn means that all n FO partial derivatives are 0 : f 1 = f 2 = = f n = 0. (61) It can be veri ed that the SO di erential d 2 z is 2 3 2 d 2 z = f 11 f 1n dx 1 dx n 4 5 4 f n1 f nn dx 1 dx n = (dx) T Hdx (62) Thus the Hessian determinant is jhj and the SO su cient condition for extremum is that all n principal minors be positive for a minimum in z and that they duly alternate in sign for a maximum, the rst one being negative. Luo, Y. (Economics, HKU) ME September 17, 2018 38 / 46 3 5

Theorem Conditions for maximum: (1) f 1 = f 2 = = f n = 0 (necessary condition); (2) jh 1 j < 0, jh 2 j > 0, jh 3 j < 0,, ( 1) n jh n j > 0. d 2 z is negative de nite. Theorem Conditions for minimum: (1) f 1 = f 2 = = f n = 0 (necessary condition); (2) jh 1 j > 0, jh 2 j > 0,, jh n j > 0. d 2 z is negative de nite. Luo, Y. (Economics, HKU) ME September 17, 2018 39 / 46

(Conti.) Example: Find the extreme values of z = 2x1 2 + x 1x 2 + 4x2 2 + x 1x 3 + x3 2 + 2. Step 1: Deriving the FOCs: f 1 = f 2 = f 3 = 0 =) x 1 = x 2 = x 3 = 0, which means that there is only one stationary value z = 2. Step 2: Deriving the SO partial derivatives and forming the Hessian determinant: jhj = f 11 f 12 f 13 f 21 f 22 f 23 f 31 f 32 f 33 = 4 1 1 1 8 0 1 0 2 with jh 1 j = 4 > 0, jh 2 j = 31 > 0, jh 3 j = 54 > 0. Hence, z = 2 is a minimum., Luo, Y. (Economics, HKU) ME September 17, 2018 40 / 46

SO conditions and Concavity&Convexity SO conditions are always concerned with whether a stationary point is the peak of a hill or the bottom of a valley and are closely related to the (strictly) concave or convex functions. De nitions A function that gives rise to a hill (valley) over the entire domain is said to be a concave (convex) function. If the hill (valley) pertains only to a subset S of the domain, the function is said to be concave (convex) on S. De nitions Mathematically, the function f is concave (convex) i for any pair of distinct points u and v in the domain of f, for any θ 2 (0, 1), θf (u) + (1 θ) f (v) () f (θu + (1 θ) v). (63) Further, if (or ) is replaced by < (or >) the function is said to be strictly concave (strictly convex). Luo, Y. (Economics, HKU) ME September 17, 2018 41 / 46

Theorem (Linear functions) If f (x) is a linear function, then it is a concave function as well as a convex function, but not strictly. Theorem (Negative of a function) If f (x) is a (strictly) concave function, then function, and vice versa. f is a (strictly) convex Theorem (Sum of functions) If f (x) and g (x) are both concave (convex) functions, then f (x) + g (x) is also a concave (convex) function. Further, if either one or both are strictly concave (strictly convex), the f (x) + g (x) is strictly concave (strictly convex). Luo, Y. (Economics, HKU) ME September 17, 2018 42 / 46

Fact In view of the association of concavity (convexity) with a global hill (valley) con guration, an extremum of a concave (convex) function must be a peak-a maximum (a bottom - a minimum). Moreover, the maximum (minimum) must be an absolute maximum (minimum). Further, the maximum (minimum) is unique if the function is strictly concave (strictly convex). Fact Note that here the properties of concavity or convexity are taken to be global in scope. If they are valid only in a subset S, the associated maximum or minimum are relative to that subset. Luo, Y. (Economics, HKU) ME September 17, 2018 43 / 46

Theorem A twice continuously di erentiable function z = f (x 1, x 2,, x n ) is concave (convex) i d 2 z is everywhere negative (positive) semide nite. The said function is strictly concave (convex) if (but not only if) d 2 z is everywhere negative (positive) de nite. Hence, we can easily verify whether a function is strictly concave (strictly convex) by checking whether its Hessian matrix is negative (positive) de nite. Example: For z = x1 2 + x 2 2, since jhj = f 11 f 12 f 21 f 22 = 2 0 0 2, with jh 1 j = 2 > 0, jh 2 j = 4 > 0, the function is strictly convex. Luo, Y. (Economics, HKU) ME September 17, 2018 44 / 46

Firm s Optimal Investment Decision Problem Consider a competitive rm with the following pro t function π = R C = PQ wl rk (64) where P is the price of output, Q is the quantity of output, w and r are two factor-prices, K and L are two factors: capital and labor, respectively. Since the market is competitive, the rm cannot a ect the prices and thus P, w and r are given exogenously. Q, K and L are three endogenous variables in the model. Assume that Q is a function of K and L via a Cobb-Douglas production function Q = Q (K, L) = L α K β. For simplicity, we assume that α = β < 1 2. Substituting it into the pro t function gives π (L, K ) = PL α K α wl rk (65) Luo, Y. (Economics, HKU) ME September 17, 2018 45 / 46

(Conti.) FOCs for pro t maximization are π = αpl α 1 K α w = 0 (66) L π K = αplα K α 1 r = 0, (67) which de nes the optimal levels of K and L that maximizes pro t. Second, let s check the SO condition to verify that we do have a maximum. The Hessian for this problem is jhj = π LL π LK = α (α 1) PL α 2 K α α 2 PL α 1 K α 1 α 2 PL α 1 K α 1 α (α 1) PL α K α 2. π KL π KK Hence, the su cient condition for a maximum is that jh 1 j = α (α 1) PL α 2 K α < 0, jh 2 j = (1 2α)α 2 P 2 L 2α 2 K 2α 2 > 0, which means that when α < 1/2, the SO su cient condition is satis ed. Combining the two FOCs can easily solve for K and L. Luo, Y. (Economics, HKU) ME September 17, 2018 46 / 46