1. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 1 Applications of Linear Equations

2. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 2 After completing this chapter, you will be able to: Learning Objectives Solve two linear equations with two variables Solve problems that require setting up linear equations with two variables LO 2. LO 1. Also

3. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 3 Perform linear C ost-Volume-Profit and break-even analysis employing: Learning Objectives - The c ontribution margin approach - The algebraic approach of solving the cost and revenue functions A. B. LO 3.

4. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 4 2x – 6 = – 6 Solving Two Equations with Two Unknowns LO 1. 2x – 3y = – 6 x + y = 2 Equations (A) Solve for y 2x – 3y = – 6 x + y = 4 Multiply by 2 2x + 2y = 4 Subtract - 5y = - 10 y = 2 y = 2 Divide by -5 (B) Solve for x (A) Solve for y (B) Solve for x2x – 3y = – 6 Substitute y = 2 2x – 3(2) = – 6 2x = + 6 – 6 x = 0 Check…

5. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 5 2x – 3y = – 6 x + y = 2 Equations You should always check your answer by substituting the values into each of the equations! x = 0 y = 2 = x + y = 2x – 3y Solving Two Equations with Two Unknowns LS = RS Left SideRight Side LS = RS Left Side Equation 1 Equation 2 Left SideRight Side = = – 6 Substituting = 2 Right Side = – 6 = 2(0) – 3(2) = 0 + 2 = 2

6. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 6 LO 2.

7. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 7 York Daycare purchases the same amount of milk and orange juice each week. After price increases from $1.10 to $1.15 per litre for milk, and from $0.98 to $1.14 per can of frozen orange juice, the weekly bill rose from $84.40 to $91.70. How many litres of milk and cans of orange juice are purchased each week?

8. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 8 Let x = # litres of milk Let y = # cans of orange juice Let x = # litres of milk Let y = # cans of orange juice Purchases Equations After price increases from $1.10 to $1.15 per litre of milk, 1.10x +0.98y= 84.40 A. B. C. and from $0.98 to $1.14 per can of frozen orange juice, the weekly bill rose from $84.40 to $91.70. 1.15x +1.14y= 91.70 (1) Development of… (2) Solving…

9. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 9 Let x = # litres of milk Let y = # cans of orange juice 1.10 x + 0.98y = 84.40 Eliminate x by Dividing by 1.10 Equation (1) (1.10 x + 0.98y) / 1.10 = 84.40/1.10 x + 0.8909y = 76.73 Equation (2) 1.15 x + 1.14y = 91.70 Eliminate x by Dividing by 1.15 (1.15 x + 1.14y)/1.15 = 91.70/1.15 x + 0.9913y = 79.74 …continue

10. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 10 Equation (1) Equation (2) x + 0.8909y = 76.73 x + 0.9913y = 79.74 Subtract.1004y = 3.01 y = 29.98 i.e. 30 cans Substitute into 1.10 x + 0.98y = 84.40 Equation (1) 1.10 x + 0.98(29.98) = 84.40 1.10 x + 29.38 = 84.40 1.10 x = 84.40 - 29.38 1.10 x = 55.02 x = 50.02 i.e. 50 litres

11. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 11 Cans of Orange Juice Litres of Milk Quantity 50 30 Price$ $1.15$57.50 1.14 34.20 $91.70= New Weekly Cost to Purchase

12. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 12 Analysis Analysis Cost LO 3.

13. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 13 T erminology Fixed Costs Business Costs Business Expenses Variable Costs …do NOT change if sales increase or decrease e.g. rent, property taxes, some forms of depreciation … do change in direct proportion to sales volume e.g. material costs, direct labour costs

14. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 14 T erminology B reak E ven P oint … is the point at which neither a Profit or Loss is made

15. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 15 T erminology Contribution Margin …is the dollar amount that is found by deducting ALL Variable Costs from Net Sales and ‘contributes’ to meeting Fixed Costs and making a ‘Net Profit’. …is the dollar amount that is found by deducting ALL Variable Costs from Net Sales and ‘contributes’ to meeting Fixed Costs and making a ‘Net Profit’. Contribution Rate …is the dollar amount expressed as a percent (%) of Net Sales …is the dollar amount expressed as a percent (%) of Net Sales A Contribution Margin statement

16. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 16 $ % Net Sales(Price * # Units Sold) x 100 Less: Variable Costs x x Net Income x x Less: Fixed Costs x x Contribution Margin x x T erminology A Contribution Margin Statement

17. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 17 Market research for a new product indicates that the product can be sold at $50 per unit. Cost analysis provides the following information: Fixed Costs per period = $8640 Variable Costs = $30 per unit. Production Capacity per period = 900 units S cenario 1 uestion: How much does the sale of an additional unit of a firm’s product contribute towards increasing its net income?

18. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 18 Formulae CM = S - VC CR = CM/S * 100% - To Find - Contribution Margin Contribution Rate *Break Even Point:...in Units (x) x = FC / CM...in Sales $ $x = (FC / CM) * S * At Break Even, Net Profit or Loss = 0 Applying Formulae...in % of Capacity BEP in Units /PC*100

19. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 19 As in the previous scenario, the new product can be sold at $50 per unit. Costs are as follows: Fixed Costs are $8640 for the period, Variable Costs are $30 per unit, and the Production Capacity is 900 units per period. A pplying the Formulae CM = S - VC CR = CM/S * 100% Units x = FC / CM Break Even Point: In $ x = (FC / CM) * S = $50 - $30 = $20 = $20/$50 * 100 = 40% = $8640/$20 = 432 Units = ($8640/$20)* $50 = $21,600 = 432/ 900*100 = 48% of Capacity BEP in units PC*100

20. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 20 The Lighting Division of Seneca Electric Co. plans to introduce a new street light based on the following accounting information: S cenario 2 FC = $3136 VC = $157. S= $185 Capacity = 320 units uestion: Calculate the breakeven point (BEP) …in units …in dollars …as a percent of capacity

21. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 21 S cenario 2 FC = $3136 VC = $157. S= $185 Capacity = 320 units Break Even Point …in units …as a percent of capacity …in dollars = FC / CM = (FC / CM) * S = BEP in units /PC*100 = $3136 / S – VC = CM $185 – 157 = $28 S – VC = CM $185 – 157 = $28 28 =112 Units = ($3136 / 28) * $185 = $20720 = 112/320 * 100 = 35% of Capacity

22. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 22 FC = $3136 VC = $157. S= $185 Capacity = 320 units S cenario 2 -1 $2688 Determine the BEP as a % of capacity if FC are reduced to $2688. =BEP in units /PC*100 Formula Step 1 … Find CM Step 2 … Find BEP in units S = $185 VC = - 157 CM $ 28 = FC/CM = = $2688/ $28 = 96 Units =BEP in units /PC*100 Step 3 … Find % of Capacity = 96/320*100 = 30% of Capacity

23. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 23 FC = $3136 VC = $157 S= $185 Capacity = 320 units S cenario 2 -2 $4588 Determine the BEP as a % of capacity if FC are increased to $4588, and VC reduced to 80% of S. = BEP in units /PC*100 Formula Step 1 … Find CM Step 2 … Find BEP in units S = $185 VC = - 148 CM $ 37 = FC/CM = $4588/ $37 = 124 Units =BEP in units /PC*100 Step 3 … Find % of Capacity = 124/320*100 = 39% of Capacity VC =S*80% = $148 $148

24. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 24 FC = $3136 VC = $157 S= $185 Capacity = 320 units S cenario 2 -3 Determine the BEP as a % of capacity if S is reduced to $171. = BEP in units /PC*100 Formula Step 1 … Find CM Step 2 … Find BEP in units S = $ 171 VC = -157 CM $ 14 = FC/CM = $3136/ $14 = 224 Units =BEP in units /PC*100 Step 3 … Find % of Capacity = 224/320*100 = 70% of Capacity $171

25. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 25 FC = $3136 VC = $157 S= $185 Capacity = 320 units S cenario 2 -4 Determine the NI if 134 units are sold! Step 1 … Find CM Step 2 … Find BEP in units S = $185 VC = - 157 CM $ 28 = FC/CM = $3136/$28 = 112 Units Units Sold 134 BEP 112 Over BEP 22 CM of $28 per unitCompany had a NI of 22* $28 = $616. NI = #Units above BEP*CM Formula

26. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 26 FC = $3136 VC = $157 S= $185 Capacity = 320 units S cenario 2 -5 What unit sales will generate NI of $2000? #Units above BEP = NI/CM Formula Step 1 … Find CM Step 2 … Find BEP in units S = $185 VC = - 157 CM $ 28 = FC/CM = $3136/$28 = 112 Units NI/CM NI/CM CM of $28 per unit = $2000/$28 per Unit = 72 Units above Break Even 72 Units + 112 BEP Units = Total Sales Units = 184 72 Units + 112 BEP Units = Total Sales Units = 184

27. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 27 FC = $3136 VC = $157 S= $185 Capacity = 320 units S cenario 2 -6 # Units below BEP = (NI)/CM Formula Step 1 … Find CM Step 2 … Find BEP in units S = $185 VC = - 157 CM $ 28 = FC/CM = $3136/$28 = 112 Units (NI)/CM (NI)/CM CM of $28 per unit = 12 Units below Break Even What are the unit sales if there is a Net Loss of $336? = ($336)/$28 per Unit 112 BEP - 12 Units Below = Total Sales Units = 100 112 BEP - 12 Units Below = Total Sales Units = 100

28. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 28 S cenario 2 -7 Step 1 … Find CM Step 2 … Find BEP in units S = $185 VC = - 157 CM $ 28 = FC/CM = $3136/$28 = 112 Units CM of $28 per unit The company operates at 85% capacity. Find the Profit or Loss. FC = $3136 VC = $157 S= $185 Capacity = 320 units 320*.85 = 272 320*.85 = 272 Units Production 272 BEP 112 Over BEP 160 Units Production 272 BEP 112 Over BEP 160 # units above BEP *CM = NI Formula 160 Units * $28 = Profit $4480 160 Units * $28 = Profit $4480 272

29. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 29 Case The Marconi Co. year end operating results were as follows: Total Sales of $375000 Operated at 75% of capacity Total Variable Costs were $150000 Total Fixed Costs were $180000 What was Marconi’s BEP expressed in dollars of sales ?

30. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 30 Case The Marconi Co. year end operating results were as follows: Total Sales of $375000 Operated at 75% of capacity Total Variable Costs were $150000 Total Fixed Costs were $180000 What was Marconi’s BEP expressed in dollars of sales ? What information is needed to calculate the $BEP? What information is needed to calculate the $BEP? 2. VC per Unit 1. Number of Units sold 3. CM 4. Total Costs 5. BEP in $

31. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 31 The Marconi Co. year end operating results were as follows: Total Sales of $375000 Operated at 75% of capacity Total Variable Costs were $150000 Total Fixed Costs were $180000 What was Marchoni’s BEP expressed in dollars of sales ? Case 2. VC per Unit 1. Number of Units sold 3. CM Let S = $1 and X be the Number of $1 Units sold Sales of $375 000 = 375000 Total Units sold $150000 375000 Total VC Total Unit Sales = = $0.40pu S $1.00 VC.40 CM $.60

32. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 32 The Marconi Co. year end operating results were as follows: Total Sales of $375000 Operated at 75% of capacity Total Variable Costs were $150000 Total Fixed Costs were $180000 What was Marchoni’s BEP expressed in dollars of sales ? Case 4. Total Costs 5. BEP in $ TC = FC + VC= $180 000 + 0.40X $BEP = (FC/CM)*S = ($180000 / 0.60)*$1.00 = (300000)*$1.00 # Of Units = $300000 $BEP

33. Linear Equations Apps. Linear Equations Apps. McGraw-Hill Ryerson© 5 - 33 This completes Chapter 5