REVIEW OF MATHEMATICAL CONCEPTS

Transcription

1 REVIEW OF MATHEMATICAL CONCEPTS Variables, functions and slopes: A Variable is any entity that can take different values such as: price, output, revenue, cost, etc. In economics we try to 1. Identify the variables. Measure them 3. Study their behavior in relation to other variables to understand how and why these variables change 4. Find the conditions that will lead to the optimal values of the variable. We express the change in the variable in response to a change in another variable in term of a function A function refers to the relationship between one (dependent) variable s value and the value of one or more (independent) variable(s). Functions are expressed in functional forms The functional relationship can be expressed in schedules (tables), graphs, or equations The general form of an equation looks as the following Y = ƒ(x) Where, Y is the dependent variable, and X is the independent variable. Independent variable is the variable whose value is determined independently of any other variable under consideration Dependent variable is the variable whose value depends on some other variable. Page 1 of 17

2 To explain the components of the linear function equation Where, Y = a + bx Y is the dependent variable, X is the independent variable, a is the intercept, and b is the coefficient of X, which measures the slope of the line. The slope of the line measures the rate of change of the dependent variable with respect to a change in some independent variable. Slope = Y X = Y X Y X 1 1 Slope is very important in economics, since much of economic analysis is based on the incremental or marginal analysis of variables. Example: TR = P x Q Assume P is known so TR = f (Q) If P = 5, Tabular form of the function TR Q Page of 17

3 Graphical representation of the function 5 0 TR 15 TR 10 5 Q Slope = b = TR / Q Q Equation form of the function Y = a + bx TR = a + bq TR = 0 + 5Q TR = 5Q o From the above table, graph, and equation, the intercept = a =0; and the coefficient = slope = b = TR / Q = (15-10) / (3-) = 5/1 = 5 o Slope of the function is critical to economic analysis. It tells us what would be the change in the dependent variable as a result of change in the dependent variable. Page 3 of 17

4 Marginal Analysis Marginal analysis refers to o The change in dependent variable as a result of a one unit change in an independent variable (discrete change definition) o The consideration of a small change around some given point (continuous change definition). If the relationship between dependent and independent variables is linear, the change is represented by the slope. In the above example, TR is expected to change by $5 for every one unit change in Q. Most of economic analyses attempt to study how a change in one or several variables affects the change in another variable and in what direction. For example, P Q R and C π. So the change in price ultimately results in change in profit, which may be positive or negative depending on other factors. P Q (demand function) Q R (TR function) Q C (TC function) Q π (π function) Many other economic decisions rely on managerial analysis, such as hiring additional worker, purchasing additional machines, etc. However, many other decisions are also taken in incremental way. Incremental analysis is used when total change is considered For example if P and as a result Q by 10 and TR by 0 then Marginal Revenue= TR/ Q = 0/10 = Incremental Revenue = 0 Page 4 of 17

5 Functional Forms For the purpose of illustration and simplicity we will often use a linear function, but there are many instances when a linear function is not the proper representation for changes between variables. So we use non-linear functions. For example, most common cost functions are non-linear functions, and nonlinear TR functions are also common, Most famous non-linear functions in economics are quadratic functions and cubic functions o Linear (or straight line) function: Y = a + bx o Quadratic function: Y = a + bx + cx o Cubic Function: Y = a + bx + cx + dx 3 Example: Consider the following demand schedule and demand curve Demand Schedule P Qd P Q Both the demand schedule and demand curve show a linear relationship between P and Q. Q = f (P) demand function Q = a + b P Intercept: when P = 0, a = Q = 700 Coefficient: when P by one ( P= -1), Q by 100 ( Q= 100), b = ( Q / P) = (100) / (-1) = -100 Therefore, this demand function could be written as Page 5 of 17

6 Q = P demand function. (1) But TR = P x Q and it is a function of Q TR = f (Q) So we have to express P in terms of Q in order to eliminate P From equation (1) 100 P = 700 Q P = Q inverse demand function () Substituting in TR equation, TR = P x Q = (7-0.01Q) Q = 7Q 0.01Q TR equation.... (3) Q TR TR (100) = (00) = (300) = (400) = (500) = (600) = (700) =0 0 TR 700 Q As we can see, a linear demand function results in a nonlinear TR function (quadratic equation If the independent variable is raised to a third power we will have a cubic function such as TC TC = Q -5Q +1.5Q 3 In both cases (quadratic, cubic) the curves TC will not be a straight line. Q Page 6 of 17

7 There are other forms used in economics such as a. Exponential: Y = X a Y X b. Logarithm: Y = log X Y X c. Reciprocal: Y = 1/X Y X Continuous vs. Discreet We will assume for the purpose of the analysis that all economic variables are related to each other in a continuous fashion although many of these variables are discrete such as people, output, and machines. Page 7 of 17

8 USING CALCULUS We know that a marginal is the change ( ) in a dependent variable associated with a one-unit change in an independent variable. If: Y = ƒ(x), then: Y = Y/ X = (Y Y1) / (X X1) Calculus: is a mathematical technique that enables one to find instantaneous rate of change of a continuous function. That is, instead of finding the rate of change between two points ( Y/ X), we can find it at any given point on the function. It can be applied only if the function is continuous Calculus is actually a slope-finding technique. Finding the slope for linear function is easy because the slope is the same at any point and between any two points. It is equal to the coefficient b. So, no need for calculus. Calculus (slope-finding) is important if the function is nonlinear because slope is different at every point. Using calculus, it is possible to measure the slope exactly at one point. The slope here is a measure of the change in Y relative to a very small change in X. To know how very small, we need to use the concept of derivative The derivative (the slope) measure of Y relative to a very small X. It is written as: dy/dx = lim x 0. Y/ X d means Y relative to very small X. Derivative turns out to be the slope of a line that is tangent to some given point on a curve. Calculus helps obtaining optimum values of function (maximum or minimum). Page 8 of 17

9 Rules of differentiation Differentiation is the process of determining the derivative of a function, i.e. finding dy/dx (ory ) when Y = ƒ(x). There are 6 main rules to find a derivative: 1. Constant Rule o If Y = f (X) = C, and C is a constant, then dy/dx = 0 o If Y = 10, then dy/dx = 0 o The derivative of a constant is zero. Derivatives involve rates of changes. o A constant by definition never change in value o The graph of a constant function is a horizontal line, which has zero slope.. Power function Rule o If Y = ax n where a is constant and n is the exponent (power), then dy/dx = (n)(a) X (n-1) o If Y = X, then dy/dx = 1X 1-1 = 1X 0 = 1 o If Y = 10X 3, dy/dx = 3(10)X (3-1) = 30X o If Y = 0 X -4 = -80X Sum and Difference Rule Sum: o U = g(x) and V = h(x). U and V are functions of X, then: o If Y = U + V, dy/dx = du/dx + dv/dx o The derivative of the sum is the sum of the derivatives. o U = 3X, V = 4X 3, then Y = U + V = 3X + 4X 3, and dy/dx = 6X + 1X o If Y = 15X +7X, then dy/dx = 30X +7 Page 9 of 17

10 Difference: o If Y = U V, where U = g(x) and V = h(x), then dy/dx = du/dx - dv/dx o The derivatives of the difference is the difference of the derivatives o U = X, V= 4 X, then Y = U V = X (4 - X) = X 4 + X, and dy/dx = X + o If Y = 5X 10X, then dy/dx = 5-0X 4. Product Rule o Y = UV where U and V are functions of X, then dy/dx = U (dv/dx) + V (du/dx) o The first times the derivative of the second plus the second times the derivative of the first. o U = 5X, V = 7 X, then Y = 5X (7 X), and dy/dx = 5X (-1) + (7 X) (10X) = - 5X + 70X -10 X dy/dx = 70X - 15 X 5. Quotient Rule: V(dU / dx) U(dV / o If Y = U / V, then dy/dx = V o Y = (5X -9) / 10X, then dx) dy dx = (10X )(5) (5X 9)(0X) (10X ) = 50X 100X 100X X 180X 50X = 4 100X 18 5X = 3 10X 6. Chain Function Rule o If Y = ƒ(u) and U = ƒ(x), then dy/dx = dy/du * du/dx o Y = U , U = X ; so, Y = (X ) , and then dy/du = 3 (X ), and du/dx = 4X dy/dx = 3 (X ) * 4X = 3 (4X 4 ) * 4X = 1 X 4 * 4X = 48 X 5 Page 10 of 17

11 Economic Application: Slope If demand function is Q = P find the slope Slope = b = dq/dp = -100 So, no matter what is the value of P, Q with respect to P is always the slope = b = -100 That is why there is no need for calculus in linear equations. Marginal Revenue: If demand function is Q = P, find MR function. We have to find TR function but since TR is a function of Q we should first find the inverse demand function P = f (Q) 0.004P = 10 Q P = (10/0.004) (1/0.004)Q = Q TR = P x Q = ( Q) Q = 500Q 50 Q MR = dtr/dq = Q Marginal Cost: If TC = Q 5Q + 1.5Q 3, find the MC function. MC = dtc/dq = Q + 4.5Q Partial Derivatives and Multivariate Functions Multivariate functions have more than one unknown variable, so the maximization procedure is different than for single variable equations Most economic variables depend on more than one economic variable. Y = f (X, Y, W, Z,.) For example, demand equation Q d = f (P, I, P s, N,..) Maximization Procedure 1. Take partial derivative of each unknown variable. Page 11 of 17

12 . Set each partial derivative equal to zero. 3. Solve the resulting system of simultaneous equations for all unknown variables. The partial derivative, Y/ X, measures the marginal change in Y associated with a very small change in X, holding constant all other factors. For example, if Q = -100P + 50I +P s + 4N Where, Q = Quantity demanded; P = Price of the good; I = Customers income; P s = Price of substitute goods; N = Number of Customers To know the change in Q as a result of a change in one of the independent variables, take the partial derivative of Q with the independent variable and held all other variables constant (the derivative of a constant is zero) Q/ P = -100P 1-1 = -100; Q/ N = 4 Another example, Suppose TR = f(x, Y) where X = sales of X and Y = sales of Y, and given: TR = 80X X - XY 3Y + 100Y, Find X * and Y * that maximize TR, and find the maximum TR First, find partial derivatives: TR/ X = 80 4X Y (treat Y as a constant). TR/ Y = -X 6Y (treat X as constant). Second, set partial derivatives equal to zero and solve for X * and Y * : 80 4X Y = 0 Y = 80 4X 100 X 6Y = 0 X = 100 6Y Thus, Y = 80 4 (100 6Y) Y = Y -3Y = -30 Y* = 30 / 3 = Substituting into X equation, X* = 100 6(13.913) = 16.5 Page 1 of 17

13 Therefore, the maximum TR (TR * ) is: TR * = 80(16.5) (16.5) 16.5(13.913) 3(13.913) + 100(13.913) = Second Derivative So far we have discussed only the first derivative (first-order condition) (dy/dx or y ), but it is important in many economic applications to find the second derivative (second-order condition) Y = X 4 dy/dx = 8X 3 dy /d X = 4 x First-order Condition Second-order Condition Maximum and Minimum Values of a Function The main objective of managerial economics is to find the optimal values of key variables. This means finding the best value under certain conditions. In economic analysis, finding the optimum means finding either the maximum or minimum value of a variable. Y A 0 Y B X X At both pints of A and B, dy/dx = 0 To determine whether the optimal value is at maximum or minimum,take first derivative of the equation; then set dy/dx equal to zero and solve for the optimal value of the independent variable (Q*) that maximizes or minimizes the objective function Page 13 of 17

14 To distinguishing maximum from minimum, find the second derivative of the equation (d Y/dX ). o If d Y/dX < 0 (if slope changes from positive to negative), the value is a maximum. o If d Y/dX > 0 (if slope changes from negative to positive), the value is a minimum. dy/dx gives the slope of the function. Set dy/dx = 0 to determine the extreme of the function Taking the second derivative tells you whether the extreme is a maximum or a minimum. Maximum Minimum First Derivative dy/dx = 0 dy/dx = 0 Second Derivative d Y/dX < 0 d Y/dX > 0 Examples 1. TR Maximization If Q = P, what is the Q* and price P* that maximize TR? First find the inverse demand function: P = (450/16) (1/16)Q So, TR equation is: TR = PQ = (450/16)Q (1/16)Q dtr/dq = MR = (450/16)Q 1-1 (1/16)Q -1 = (450/16) (/16)Q = (450/16) (1/8)Q Set MR equation (dtr/dq) equal to zero, which means the last unit produced will not add any additional revenue dtr/dq = (450/16) (1/8)Q = 0 Q* = (450/16) x (8/1) = 5 P* = (450/16) (1/16)(5) = To check whether Q* and P* are in max or min conduct the second derivative d Y/dX = - 1/8 < 0 maximum Maximum TR = (450/16)(5) (1/16)(5) = MR = (450/16) (1/8)(5) Page 14 of 17

15 P 8 MR D 0 TR TR Q Q. Profit Maximization Suppose Q = P, and TC = Q +Q, find Q* and P* that will maximize profit. Since π = TR TC, first find TR equation. So, Inverse demand function: P = 17 10Q TR = PQ = (17 10Q)Q = 17Q 10Q π = TR TC = (17Q 10Q ) ( Q +Q ) = 17Q 10Q Q - Q = Q -11Q dπ/dq = 107 Q = 0 Q* = 107/ = 4.86 P* = (4.86) = 13.4 d π/dq = - < 0 maximum profit OR Page 15 of 17

16 π = TR TC dπ/dq = (dtr/dq) (dtc/dq) = 0 dtr/dq = dtc/dq MR = MC 17-0Q = 65 + Q Q = 107 Q* = 107/ = 4.86 To check second-order condition o For TR: d TR/dQ = -0 < 0 maximum TR o For TC: d TC/dQ = + > 0 minimum TC 3. Quadratic Equation if you can t find the roots, you can use the equation below to solve for the roots: If Y = ax + bx + c, then X = b ± b a 4ac Suppose TR = 50Q; and TC = Q 3Q Q 3. What is the level of output that will maximize profit? π = TR TC = 50Q Q + 3Q Q 3 = Q - +3Q Q 3 dπ/dq = Q 0.30Q Re-arrange to fit the general formula of a quadratic function Y = ax + bx + c So, dπ/dq = 0.30Q + 6Q -10 = 0 The values of Q that set this quadratic function equal to zero can be found using the formula for the roots of a quadratic function X = b ± b a 4ac In our example: a = -0.30; b = +6; c = ± Q = 6 4(0.30)( 10) ( 0.30) = 6 ± = 6 ± = 6 ± Page 16 of 17

17 Q 1 * = ( ) / = 1.84 Q * = ( ) / = Both quantities satisfy the first order conditions but only one satisfies the second order condition d π/dq = Q + 6 substitute Q 1 * and Q * For Q 1 *: d π/dq = (1.84) + 6 = 4.90 > 0 For Q *: d π/dq = (18.16) + 6 = < 0 Thus, only Q * satisfies maximization conditions Classical production function (Cobb-Douglus) Q = AL α K β Transfer into log log Q = log A + α log L + β log K Page 17 of 17