LESSON 2.06:Modeling Linear Relations

MFM1P

 COSTS that may be associated with renting a hall for a wedding or banquet.

Discuss some cost that may be associated with renting a hall for a wedding or banquet.

 From your above list which costs are fixed and which are variable. Fixed costs are those which are set and do not change, whereas variable cost may vary depending on how many people show up to your event.

Complete the following table of values. You should be able to notice a pattern.

Explain the pattern.

Number of Guests

Total Cost ($)

0 5001 5502 600 34567

EXAMPLE 1:Ralph’s Rental has many tools that are available for rent. They charge $20 fixed price plus $5 per hour for the rental of their post hole digger. This situation can be represented by the equation

C = 5 n + 20 .

Number of Hours Rented

Rough Work Cost ($)

0 C = 5 n + 20 201 C = 5 n + 20 252 C = 5 n + 20 303 C = 5 n + 20 4 C = 5 n + 20 5 C = 5 n + 20 6 C = 5 n + 20 7 C = 5 n + 20 8 C = 5 n + 20

Number of Hours Rented

Rough Work Cost ($)

0 C = 5 (0) + 20 201 C = 5 (1) + 20 252 C = 5 (2) + 20 303 C = 5 (3) + 20 354 C = 5 (4) + 20 405 C = 5 (5) + 20 456 C = 5 (6) + 20 507 C = 5 (7) + 20 558 C = 5 (8) + 20 60

COMPLETE THE FIRST DIFFERENCES

FIRST DIFFERENCES

25 - 20= 5

30 - 25= 5

35 - 30= 5

40 - 35= 5

45 - 40= 5

50 - 45= 5

55 - 50= 5

60 - 55= 5

Number of Hours Rented

Cost ($)

0 201 252 303 354 405 456 507 558 60

In general, we can make a few notes about our equation and how it relates to the description given in the example.

Summary:Equations can be developed from written descriptions using the general formula below.

COST = Price per unit x Variable + Fixed Cost

EXAMPLE 2:Brandon and Beyonce are looking at renting a hall for her their wedding. The hall charges a maintenance fee of $300 and a cost of $25 per plate for supper. Write an equation to model this situation.

C =25 n + 300 .

Number of Plates

Rough Work Cost ($)

0 C =25 n + 300

10 C =25 n + 300

20 C =25 n + 300

30 C =25 n + 300

40 C =25 n + 300

50 C =25 n + 300

60 C =25 n + 300

70 C =25 n + 300

80 C =25 n + 300

Number of Plates

Rough Work Cost ($)

0 C =25 (0) + 300 30010 C =25 (10) + 300 55020 C =25 (20) + 300 80030 C =25 (30) + 300 105040 C =25 (40) + 300 130050 C =25 (50) + 300 155060 C =25 (60) + 300 180070 C =25 (70) + 300 205080 C =25 (80) + 300 2300

COMPLETE THE FIRST DIFFERENCES

FIRST DIFFERENCES

250

250

250

250

250

250

250

250

Number of Plates

Cost ($)

0 30010 55020 80030 105040 130050 155060 180070 205080 2300

Change in costChange in plates

=250 10

So, the cost increases by $250 for every 10 more plates.

RATE OF CHANGE=25 1

So, the cost increases by $25 for every plate added