Determinants of the Income Maximum Point: The Use of Discriminants in Economic Analysis

Extremum in Quadratic Functions: The Case of Total Company Revenue Name: Alvinska Hebat Grade: 910-Math Application of Quadratic Functions to Economic Concepts Quadratic functions are widely used in economic concepts, usually aimed at finding the highest values or lowest values of a particular economic variable where theoretically the behavior of the variable follows the form of a quadratic function equation. 1. Company Acceptance Function If the company's revenue is assumed to be obtained from the sale of its products, then the company's revenue (Total Revenue, TR) is the multiplication of the price per unit (P) with the amount of output sold (Q), or TR = P.Q. In economics, there are two possible powers of producers in determining the price of the goods they sell, namely: There are producers who have the power to set their own prices or act as price makers, this is possible because there are few or no competitors. There are producers who do not have the power to determine the price of their own goods or act as price takers, this happens because there is too much market competition. As a price maker he will determine the price depending on the supply of the amount of goods to be sold, or mathematically the price function that a producer acting as a price maker has is: P = f(Q) = a - mQ P = price/unit and Q = the amount of output So the company's revenue function is: TR = P.Q = (a - mQ).Q = aQ - mQ2 or TR = - mQ2 + aQ In the form of such a quadratic function, TR will have a maximum extreme point which can be described as follows: Example: A cement producer has the following function for the price of cement he sells: P = f(Q) = 480 0.8Q; where P = price of cement per piece; Q = quantity of cement products. (a) Determine the company's revenue function in the form TR = f(Q); where TR = company revenue (sales proceeds) of cement, (b) Calculate how much cement i s sold when the company experiences maximum cement sales? (c) What is the price per kilogram of cement that a cement producer sets when its sales yield is maximum? (d) Draw the revenue function of the company! Answer: (a) Revenue = TR = f (Q) = P . Q = (480 - 0.8Q)Q = 480Q - 0.8Q2 = - 0.8Q2 + 480Q (b) Maximum sales occur at the extreme or turning point of the TR curve, so to determine the turning point first determine the Determinant. D = b 2 - 4ac = 4802 - (4 x -0.8 x 0) = 230400 Turning point = (-b/2a ; -D/4a) = (-480/(2x-0.8) ; -230400/(4x-0.8)) = (-480/-1,6 ; -230400/ -3,2 ) = (300 ; 72000) So the maximum amount of cement is 300 zak with a total revenue of 72,000. c) The price of cement at the time of maximum sales is: P = 480 - 0.8Q = 480 - (0.8 x 300) = 240 d) Tipot (intersection point) with Y axis (or TR axis) then X (or Q) = 0, then TR = 0 Tipot (intersection point) with the Q axis: Q 1,2 = (-b ± √D) : 2a Q 1 = (-480 + √230400): (2x0,8) = (-480 + 480) : -1,6 = 0 Q 2 = (-480 - √230400): (2x-0,8) = (-480 - 480) : -1,6 = 600 Company Cost Function To understand this context, the cost function is assumed to be a description of the dependency between the amount of cost and the amount of output only. Variables other than the amount of output are assumed not to affect the cost of production. Total Cost If total cost (TC) is total fixed cost (TFC) plus total variable cost (TVC), and total cost is in the form of a quadratic function as follows: TC = f(Q) = aQ2 - bQ + c where a>0; b Example: The total cost function (TC) of a company is as follows: TC = 0.1Q2 -100Q + 12000 (Q = amount of output) (a) determine the variable cost (VC) and fixed cost (FC) equations. (b) determine the number of outputs that results in the lowest total cost and the lowest variable cost! (c) Draw the total cost and variable cost functions on a Cartesian plane! Answer: (a) TC = 0.1Q2 - 100Q + 12,000 TVC = 0.1Q2 - 100Q and TFC = c = 12000 (b) Draw the total cost and variable cost functions on a Cartesian plane! D = b 2 - 4ac = -100 2 (4 x 0.1x 12000) =10,000 - 4800 = 5200 Turning point = (-b/2a ; -D/4a) = (100/(2 x 0.1) ; -5200/(4 x 0.1)) = (100/0,2 ;- 5200/0,4) = (500; -13000) The intersection point with the Y axis (or TC axis) is when X (or Q) = 0, then TC = 12000 The intersection point with the X axis (or Q axis): X 1,2 = (-b ± √D): 2a X 1 = (100 + √5200): (2 x 0,1) = (100 + 72,11) : 0,2 = 172,11/0,2 = 860,55 X 2 = (100 - √5200): (2 x 0,1) = (100 - 72,11) : 0,2 = 27,89/0,2 = 138,45 To draw the variable cost function, do it in the same way as the total cost function, but with the function = TVC = 0.1Q2 - 100Q To draw the TVC function, the function TVC = 0.1Q2 - 100Q is used, then: a = 0.1 b = -100 c = 0 Discriminant = D = b2 - 4ac = -100 2 - (4 x 0.1x0)= 10,000 Turning Point = (-b/2a ; -D/4a) = (100/(2 x 0.1) ; -10000/(4 x 0.1)) = (100/0,2 ; -10000/0,4) = (500; -25000) The intersection point with the Y-axis (or TC axis) is when X (or Q) = 0, hence TVC = 0: Q 1,2 = (-b ± √D) : 2a Q 1 = (100 + √10000): (2 x 0,1) = (100 + 100) : 0,2 = 200/0,2 = 1000 Q 2 = (100 - √10000): (2 x 0,1) = (100 - 100) : 0,2 = 0/0.2 = 0 Do the same for the TC function equation as follows: (a) TC = - Q2 + 20Q + 5000 (b) TC = - 0.5Q2 + 120Q + 12,000