Linear Functions, Simple Interest, Cost – Revenue – Profit Functions Section 1.3 – Day 1 Linear Functions, Simple Interest, Cost – Revenue – Profit Functions

Linear Function A linear function can be expressed in the form m and b are constants It can be used for Simple Depreciation Linear Supply and Demand Functions Linear Cost, Revenue, and Profit Functions

Simple Depreciation (find 2 points and write equation) Ex. A computer with original value $2000 is linearly depreciated to a value of $200 after 4 years. Find an equation for the value, V, of the computer at the end of year t. (This book tends to use for y (v – for value) and for x (t for time) – you may use x and y.) What are the two points we are given?

Simple Interest I = prt A = P(1 + rt) I = interest earned p = principal (starting amount) r = rate (as a decimal) t = time (in years) A = accumulated amount (original amount + int)

A bank pays simple int. at the rate of 8% per year A bank pays simple int. at the rate of 8% per year. If you deposit $1000 and leave it for 3 years how much will you have after 3 years? How much int will you earn? P = 1000 r = .08 t = 3 A = 1000[1 + (.08)(3)] = 1240 The interest earned is I = prt I = (1000)(.08)(3) = 240 or You could just do 1240 – 1000 = 240 (the total minus the starting amount)

Cost, Revenue, Profit Functions C(x) = cx + F R(x) = sx P(x) = R(x) – C(x) (revenue – cost) c = cost to make each item F = fixed costs s = selling price of each item

Cost, Revenue, and Profit Functions Ex. A shirt producer has a fixed monthly cost of $3600. If each shirt costs $3 and sells for $12 find: a. The cost function Cost: C(x) = 3x + 3600 where x is the number of shirts produced. b. The revenue function Revenue: R(x) = 12x where x is the number of shirts sold. c. The profit function and the profit from selling 900 shirts Profit: P(x) = Revenue – Cost = 12x – (3x + 3600) = 9x – 3600 P(900) = 9(900) – 3600 = 4500 or $4500

HOMEWORK #1 p 8 1-29 eoo, p 19 1-37 eoo #2 p20 39 – 65 odd, omit 55, on 57&59 ignore instructions referring to problem 55 #3 p 32 1 – 19 odd