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Topic 1. Topic 1 : Elementary functions Reading: Jacques Section 1.1 - Graphs of linear equations Section 2.1 – Quadratic functions Section 2.2 – Revenue, Cost and Profit. Linear Functions.
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Topic 1 Topic 1 : Elementary functions Reading: Jacques Section 1.1 - Graphs of linear equations Section 2.1 – Quadratic functions Section 2.2 – Revenue, Cost and Profit
Linear Functions • The function f is a rule that assigns an incoming number x, a uniquely defined outgoing number y. y = f(x) • The Variable x takes on different values…... • The function f maps out how different values of x affect the outgoing number y. • A Constant remains fixed when we study a relationship between the incoming and outgoing variables
Simplest Linear Relationship: y = a+bx independent dependent variable variable intercept This represents a straight line on a graph i.e. a linear function has a constant slope • b = slope of the line = change in the dependent variable y, given a change in the independent variable x. • Slope of a line = y / x = (y2-y1) / (x2-x1)
Example: y = a + bx y : is the final grade, x : is number of hours studied, a%: guaranteed Consider the function: y = 5+ 0x What does this tell us? Assume different values of x ……… Example: Student grades
y=5+0x Output = constant slope Input y a b X 5 5 0 0 5 5 0 1 5 5 0 2 5 5 0 3 5 5 0 4 5 5 0 5 Example Continued: What grade if you study 0 hours? 5 hours?
Example Continued…. • y=5+15x If x = 4, what grade will you get? Y = 5 + (4 * 15) = 65
Demand functions: The relationship between price and quantity If p =5, how much will be demanded? D = 10 - (2 * 5) = 0
Inverse Functions: • Definition • If y = f(x) • then x = g(y) • f and g are inverse functions • Example • Let y = 5 + 15x • If y is 80, how many hours per week did they study?
Example continued….. • If y is 80, how many hours per week did they study? • Express x as a function of y: 15x = y – 5.... • So the Inverse Function is: x = (y-5)/15 • Solving for value of y = 80 x = (80-5 / 15) x = 5 hours per week
An inverse demand function • If D = a – bP then the inverse demand curve is given by P = (a/b) – (1/b)D • E.g. to find the inverse demand curve of the function D= 10 -2P …… First, re-write P as a function of D 2P = 10 – D Then, simplify So P= 5 – 0.5D is the inverse function
Student grades again: y = a + bx + cz y : is the final grade, x : is number of hours studied, z: number of questions completed a%: guaranteed Example: If y = 5+ 15x + 3z, and a student studies 4 hours per week and completes 5 questions per week, what is the final grade? Answer: y = 5 + 15x + 3z y = 5 + (15*4) + (3*5) y = 5+60+15 = 80 More Variables:
The demand for a pint of Guinness in the Student bar on a Friday evening is a linear function of price. When the price per pint is €2, the demand ‘is €6 pints. When the price is €3, the demand is only 4 pints. Find the function D = a + bP 6 = a + 2b => a = 6-2b 4 = a + 3b => a = 4-3b 6-2b = 4-3b Solving we find that b = -2 If b = -2, then a = 6-(-4) = 10 The function is D = 10 – 2P What does this tell us?? Note, the inverse Function is P = 5- 0.5D Another example: Guinness Demand.
let €4000 be set as the target income. All income above the target is taxed at 40%. For every €1 below the target, the worker gets a negative income tax (subsidy) of 40%. Write out the linear function between take-home pay and earnings. Answer: THP = E – 0.4 (E – 4000) if E>4000 THP = E + 0.4 (4000-E) if E<4000 In both cases, THP = 1600 +0.6E So i) If E = 4000 => THP = 1600+2400=4000 ii) If E = 5000 => THP = 1600+3000=4600 iii) If E = 3000 => THP = 1600+1800=3400 A Tax Example….
Tax example continued…. THP = 1600 +0.6E If the hourly wage rate is equal to €3 per hour, rewrite take home pay in terms of number of hours worked? • Total Earnings E = (no. hours worked X hourly wage) • THP = 1600 + 0.6(3H) = 1600 + 1.8H Now add a (tax free) family allowance of €100 per child to the function THP = 1600 +0.6E • THP = 1600 + 0.6E + 100Z (where z is number of children) Now assume that all earners are given a €100 supplement that is not taxable, • THP = 1600 + 0.6E + 100Z + 100 = 1700 + 0.6E + 100Z
Topic 1 continued: Non- linear EquationsJacques Text Book: Sections 2.1 and 2.2
Quadratic Functions • Represent Non-Linear Relationships y = ax2+bx+cwhere a0, c=Intercept • a, b and c are constants • So the graph is U-Shaped if a>0, • And ‘Hill-Shaped’ if a<0 • And a Linear Function if a=0
Solving Quadratic Equations: 1) Graphical Approach: To find Value(s), if any, of x when y=0, plot the function and see where it cuts the x-axis • If the curve cuts the x-axis in 2 places: there are always TWO values of x that yield the same value of y when y=0 • If it cuts x-axis only once: when y=0 there is a unique value of x • If it never cuts the x-axis: when y=0 there is no solution for x
e.g. y = -x2+4x+5 Since a<0 => ‘Hill Shaped Graph’
The graph y=0, then x= +5 OR x = -1
Special Case: a=1, b=0 and c=0So y = ax2+bx+c => y = x2 Min. Point: (0,0) Intercept = 0
Practice examples • Plot the graphs for the following functions and note (i) the intercept value (ii) the value(s), if any, where the quadratic function cuts the x-axis • y = x2-4x+4 • y = 3x2-5x+6
Solving Quadratic Equations: • 2) Algebraic Approach: find the value(s), if any, of x when y=0 by applying a simple formula…
Example • e.g. y = -x2+4x+5 • hence, a = -1; b=4; c=5 • Hence, x = +5 or x = -1 when y=0 • Function cuts x-axis at +5 and –1
Example 2 • y = x2-4x+4 • hence, a = 1; b= - 4; c=4 • If y = 0 Function only cuts x-axis at one point, where x=2 x = 2 when y = 0
Example 3 • y = 3x2-5x+6 • hence, a = 3; b= - 5; c=6 • If y = 0 when y = 0 there is no solution The quadratic function does not intersect the x-axis
Understanding Quadratic Functions • intercept where x=0 is c • a>0 then graph is U-shaped • a<0 then graph is inverse-U • a = 0 then graph is linear • b2 – 4ac > 0 : cuts x-axis twice • b2 – 4ac = 0 : cuts x-axis once • b2 – 4ac < 0 : no solution
Essential equations for Economic Examples: • Total Costs = TC = FC + VC • Total Revenue = TR = P * Q • = Profit = TR – TC • Break even: = 0, or TR = TC • Marginal Revenue = MR = change in total revenue from a unit increase in output Q • Marginal Cost = MC = change in total cost from a unit increase in output Q • Profit Maximisation: MR = MC
An Applied Problem • A firm has MC = 3Q2- 32Q+96 • And MR = 236 – 16Q • What is the profit Maximising Output? Solution • Maximise profit where MR = MC 3Q2 – 32Q + 96 = 236 – 16Q 3Q2 –32Q+16Q +96 – 236 = 0 3Q2 –16Q –140 = 0 • Solve the quadratic using the formula where a = 3; b = -16 and c = -140 • Solution: Q = +10 or Q = -4.67 • Profit maximising output is +10 (negative Q inadmissable)
Another Example…. • If fixed costs are 10 and variable costs per unit are 2, then given the inverse demand function P = 14 – 2Q: • Obtain an expression for the profit function in terms of Q • Determine the values of Q for which the firm breaks even. • Sketch the graph of the profit function against Q
Solution: • Profit = TR – TC = P.Q – (FC + VC) = (14 - 2Q)Q – (2Q + 10) = -2Q2 + 12Q – 10 • Breakeven: where Profit = 0 Apply formula to solve quadratic where = 0 so solve -2Q2 + 12Q – 10 = 0 with • Solution: at Q = 1 or Q = 5 the firm breaks even
3. Graphing Profit Function • STEP 1: coefficient on the squared term determines the shape of the curve • STEP 2: constant term determines where the graph crosses the vertical axis • STEP 3: Solution where = 0 is where the graph crosses the horizontal axis
Questions Covered on Topic 1: Elementary Functions • Linear Functions and Tax…… • Finding linear Demand functions • Plotting various types of functions • Solving Quadratic Equations • Solving Simultaneous Linear (more in next lecture) • Solving quadratic functions