Basic Mathematics for Economists - Rosser - Chapter 5 ppsx

5 Linear equations Learning objectives After completing this chapter students should be able to: • Solve sets of simultaneous linear equations with two or more variables using the substitution and row operations methods. • Relate mathematical solutions to simultaneous linear equations to economic analysis. • Recognize when a linear equations system cannot be solved. • Derive the reduced-form equations for the equilibrium values of dependent variables in basic linear economic models and interpret their meaning. • Derive the profit-maximizing solutions to price discrimination and multiplant monopoly problems involving linear functions • Set up linear programming constrained maximization and minimization problems and solve them using the graphical method. 5.1 Simultaneous linear equation systems ThewaytosolvesinglelinearequationswithoneunknownwasexplainedinChapter3.We now turn to sets of linear equations with more than one unknown. A simultaneous linear equation system exists when: 1. there is more than one functional relationship between a set of specified variables, and 2. all the functional relationships are in linear form. The solution to a set of simultaneous equations involves finding values for all the unknown variables. Where only two variables and equations are involved, a simultaneous equation system can be related to familiar graphical solutions, such as supply and demand analysis. For example, assume that in a competitive market the demand schedule is p = 420 −0.2q (1) and the supply schedule is p = 60 +0.4q (2) If this market is in equilibrium then the equilibrium price and quantity will be where the demand and supply schedules intersect. As this will correspond to a point which is on both © 1993, 2003 Mike Rosser thedemandscheduleandthesupplyschedulethentheequilibriumvaluesofpandqwillbe suchthatbothequations(1)and(2)hold.Inotherwords,whenthemarketisinequilibrium(1) and(2)aboveformasetofsimultaneouslinearequations. Notethatinmostoftheexamplesinthischapterthe‘inverse’demandandsupplyfunctions areused,i.e.p=f(q)ratherthanq=f(p).Thisisbecausepriceisnormallymeasuredonthe verticalaxisandwewishtorelatethemathematicalsolutionstographicalanalysis.However, simultaneouslinearequationssystemsofteninvolvemorethantwounknownvariablesin whichcasenographicalillustrationoftheproblemwillbepossible.Itisalsopossiblethat asetofsimultaneousequationsmaycontainnon-linearfunctions,buttheseareleftuntilthe nextchapter. 5.2Solvingsimultaneouslinearequations Thebasicideainvolvedinallthedifferentmethodsofalgebraicallysolvingsimultaneous linearequationsystemsistomanipulatetheequationsuntilthereisasinglelinearequation withoneunknown.ThiscanthenbesolvedusingthemethodsexplainedinChapter3.The value of the variable that has been found can then be substituted back into the other equations to solve for the other unknown values. It is important to realize that not all sets of simultaneous linear equations have solutions. The general rule is that the number of unknowns must be equal to the number of equations for there to be a unique solution. However, even if this condition is met, one may still come across systems that cannot be solved, e.g. functions which are geometrically parallel and thereforeneverintersect(seeExample5.2below). We shall first consider four different methods of solving a 2 ×2 set of simultaneous linear equations, i.e. one in which there are two unknowns and two equations, and then look at how some of these methods can be employed to solve simultaneous linear equation systems with more than two unknowns. 5.3 Graphical solution The graphical solution method can be used when there are only two unknown variables. It will not always give 100% accuracy, but it can be useful for checking that algebraic solutions are not widely inaccurate owing to analytical or computational errors. Example 5.1 Solve for p and q in the set of simultaneous equations given previously in Section 5.1: p = 420 −0.2q (1) p = 60 +0.4q (2) Solution ThesetwofunctionalrelationshipsareplottedinFigure5.1.Bothholdattheintersection point X. At this point the solution values p = 300 and q = 600 can be read off the graph. © 1993, 2003 Mike Rosser 0 p p = 420 –0.2q p =60+0.4q q X 420 300 60 600 Figure 5.1 0 5 2 y = 2+2x y = 5+2x x y Figure 5.2 A graph can also illustrate why some simultaneous linear equation systems cannot be solved. Example 5.2 Attempt to use graphical analysis to solve for y and x if y = 2 +2x and y = 5 + 2x Solution These two functions are plotted in Figure 5.2. They are obviously parallel lines which never intersect. This problem therefore does not have a solution. © 1993, 2003 Mike Rosser Test Yourself, Exercise 5.1 Solve the following (if a solution exists) using graph paper. 1. In a competitive market, the demand and supply schedules are respectively p = 9 − 0.075q and p = 2 +0.1q Find the equilibrium values of p and q. 2. Find x and y when x = 80 −0.8y and y = 10 +0.1x 3. Find x and y when y =−2 + 0.5x and x = 2y − 9 5.4 Equating to same variable The method of equating to the same variable involves rearranging both equations so that the same unknown variable appears by itself on one side of the equality sign. This variable can then be eliminated by setting the other two sides of the equality sign in the two equations equal to each other. The resulting equation in one unknown can then be solved. Example 5.3 SolvethesetofsimultaneousequationsinExample5.1abovebytheequatingmethod. Solution In this example no preliminary rearranging of the equations is necessary because a single term in p appears on the left-hand side of both. As p = 420 −0.2q (1) and p = 60 +0.4q (2) then it must be true that 420 − 0.2q = 60 + 0.4q Therefore 360 = 0.6q 600 = q © 1993, 2003 Mike Rosser The value of p can be found by substituting this value of 600 for q back into either of the two original equations. Thus from (1) p = 420 − 0.2q = 420 −0.2(600) = 420 − 120 = 300 or from (2) p = 60 + 04q = 60 + 0.4(600) = 60 + 240 = 300 Example 5.4 Assume that a firm can sell as many units of its product as it can manufacture in a month at £18 each. It has to pay out £240 fixed costs plus a marginal cost of £14 for each unit produced. How much does it need to produce to break even? Solution From the information in the question we can work out that this firm faces the total revenue function TR = 18q and the total cost function TC = 240 + 14q, where q is output. These functions are plotted in Figure 5.3, which is an example of what is known as a break-even chart. This is a rough guide to the profit that can be expected for any given production level. The break-even point is clearly at B, where the TR and TC schedules intersect. Since TC = 240 +14q and TR = 18q and the break-even point is where TR = TC, then 18q = 240 + 14q 4q = 240 q = 60 Therefore the output required to break even is 60 units. B 0 q £ 60 TR TC 1080 240 Figure 5.3 © 1993, 2003 Mike Rosser Note that in reality at some point the TR schedule will start to flatten out when the firm has to reduce price to sell more, and TC will get steeper when diminishing marginal productivity causes marginal cost to rise. If this did not happen, then the firm could make infinite profits by indefinitely expanding output. Break-even charts can therefore only be used for the range of output where the specified linear functional relationships hold. What happens if you try to use this algebraic method when no solution exists, as in Example5.2above? Example 5.5 Attempt to use the equating to same variable method to solve for y and x if y = 2 +2x and y = 5 + 2x Solution Eliminating y from the system and equating the other two sides of the equations, we get 2 + 2x = 5 + 2x Subtracting 2x from both sides gives 2 = 5. This is clearly impossible, and hence no solution can be found. Test Yourself, Exercise 5.2 1. A competitive market has the demand schedule p = 610 − 3q and the supply schedule p = 20 + 2q. Calculate equilibrium price and quantity. 2. A competitive market has the demand schedule p = 610 − 3q and the supply schedule p = 50 + 4q where p is measured in pounds. (a) Find the equilibrium values of p and q. (b) What will happen to these values if the government imposes a tax of £14 per unit on q? 3. Make up your own linear functions for a supply schedule and a demand schedule and then: (a) plot them on graph paper and read off the values of price and quantity where they intersect, and (b) algebraically solve your set of linear simultaneous equations and compare your answer with the values you got for (a). 4. A firm manufactures product x and can sell any amount at a price of £25 a unit. The firm has to pay fixed costs of £200 plus a marginal cost of £20 for each unit produced. (a) How much of x must be produced to make a profit? (b) If price is cut to £24 what happens to the break-even output? 5. If y = 16 + 22x and y =−2.5 + 30.8x, solve for x and y. © 1993, 2003 Mike Rosser 5.5 Substitution The substitution method involves rearranging one equation so that one of the unknown vari- ables appears by itself on one side. The other side of the equation can then be substituted into the second equation to eliminate the other unknown. Example 5.6 Solve the linear simultaneous equation system 20x + 6y = 500 (1) 10x − 2y = 200 (2) Solution Equation (2) can be rearranged to give 10x − 200 = 2y 5x − 100 = y (3) If we substitute the left-hand side of equation (3) for y in equation (1) we get 20x + 6y = 500 20x + 6(5x − 100) = 500 20x + 30x − 600 = 500 50x = 1,100 x = 22 To find the value of y we now substitute this value of x into (1) or (2). Thus, in (1) 20x + 6y = 500 20(22) +6y = 500 440 + 6y = 500 6y = 60 y = 10 Example 5.7 Find the equilibrium level of national income in the basic Keynesian macroeconomic model Y = C + I (1) C = 40 +0.5Y (2) I = 200 (3) Solution Substituting the consumption function (2) and given I value (3) into (1) we get Y = 40 + 0.5Y +200 © 1993, 2003 Mike Rosser Therefore 0.5Y = 240 Y = 480 Test Yourself, Exercise 5.3 1. A consumer has a budget of £240 and spends it all on the two goods A and B whose prices are initially £5 and £10 per unit respectively. The price of A then rises to £6 and the price of B falls to £8. What combination of A and B that uses up all the budget is it possible to purchase at both sets of prices? 2. Find the equilibrium value of Y in a basic Keynesian macroeconomic model where Y = C + I the accounting identity C = 20 +0.6Y the consumption function I = 60 exogenously determined 3. Solve for x and y when 600 = 3x + 0.5y 52 = 1.5y − 0.2x 5.6 Row operations Row operations entail multiplying or dividing all the terms in one equation by whatever number is necessary to get the coefficient of one of the unknowns equal to the coefficient of that same unknown in another equation. Then, by subtraction of one equation from the other, this unknown can be eliminated. Alternatively, if two rows have the same absolute value for the coefficient of an unknown but one coefficient is positive and the other is negative, then this unknown can be eliminated by adding the two rows. Example 5.8 Given the equations below, use row operations to solve for x and y. 10x + 3y = 250 (1) 5x + y = 100 (2) Solution Multiplying (2) by 3 15x + 3y = 300 Subtracting (1) 10x + 3y = 250 Gives 5x = 50 x = 10 © 1993, 2003 Mike Rosser Substituting this value of x back into (1), 10(10) +3y = 250 100 + 3y = 250 3y = 150 y = 50 Example 5.9 A firm makes two goods A and B which require two inputs K and L. One unit of A requires 6 units of K plus 3 units of L and one unit of B requires 4 units of K plus 5 units of L. The firm has 420 units of K and 300 units of L at its disposal. How much of A and B should it produce if it wishes to exhaust its supplies of K and L totally? (NB. This question requires you to use the economic information given to set up a mathe- matical problem in a format that can be used to derive the desired solution. Learning how to set up a problem is just as important as learning how to solve it.) Solution The total requirements of input K are 6 for every unit of A and 4 for each unit of B, which can be written as K = 6A + 4B Similarly, the total requirements of input L can be specified as L = 3A + 5B As we know that K = 420 and L = 300 because all resources are used up, then 420 = 6A +4B (1) and 300 = 3A +5B (2) Multiplying (2) by 2 600 = 6A + 10B Subtracting (1) 420 = 6A +4B gives 180 = 6B 30 = B Substituting this value for B into (1) gives 420 = 6A +4(30) 420 = 6A +120 300 = 6A 50 = A The firm should therefore produce 50 units of A and 30 units of B. © 1993, 2003 Mike Rosser (Note that the method of setting up this problem will be used again when we get to linear programming in the Appendix to this chapter.) Test Yourself, Exercise 5.4 1. Solve for x and y if 420 = 4x + 5y and 600 = 2x + 9y 2. A firm produces the two goods A and B using inputs K and L. Each unit of A requires 2 units of K plus 6 units of L. Each unit of B requires 3 units of K plus 4 units of L. The amounts of K and L available are 120 and 180, respectively. What output levels of A and B will use up all the available K and L? 3. Solve for x and y when 160 = 8x − 2y and 295 = 11x + y 5.7 More than two unknowns With more than two unknowns it is usually best to use the row operations method. The basic idea is to use one pair of equations to eliminate one unknown and then bring in another equation to eliminate the same variable, repeating the process until a single equation in one unknown is obtained. The exact operations necessary will depend on the format of the particular problem. There are several ways in which row operations can be used to solve most problems and you will only learn which is the quickest method to use through practising examples yourself. Example 5.10 Solve for x, y and z, given that x + 12y + 3z = 120 (1) 2x + y + 2z = 80 (2) 4x + 3y + 6z = 219 (3) Solution Multiplying (2) by 2 4x + 2y + 4z = 160 (4) Subtracting (4) from (3) y + 2z = 59 (5) We have now eliminated x from equations (2) and (3) and so the next step is to eliminate x from equation (1) by row operations with one of the other two equations. In this example the © 1993, 2003 Mike Rosser [...]... function (1) equal to demand price ∗ ps = pd (55 + 4q)(1 + t) = 3 75 − 2.5q 55 + 55 t + 4q + 4qt = 3 75 − 2.5q 6.5q + 4qt = 320 − 55 t q(6 .5 + 4t) = 320 − 55 t q= 320 − 55 t 6 .5 + 4t This reduced form equation for equilibrium q is a bit more complicated than the one we derived for the per unit sales tax case However, we can still use it to work out the predicted value of q for a few values of t Normally we would... reduced form equation for equilibrium q is not a simple linear function of t Lastly, we can derive the reduced form equation for equilibrium p by substituting the reduced form for q that we have already found into the demand schedule Thus p = 3 75 − 2.5q = 3 75 − 2 .5 320 − 55 t 6 .5 + 4t 2,437 .5 + 1 ,50 0t − 800 + 137.5t 1,637 .5 + 1,637.5t = 6 .5 + 4t 6 .5 + 4t 1,637 .5( 1 + t) = 6 .5 + 4t = To check this reduced form... lie between 0% and 100%, giving a value of t in decimal format between 0 and 1 © 1993, 2003 Mike Rosser 320 − 55 (0.1) 320 − 5. 5 314 .5 = = = 45. 58 6 .5 + 4(0.1) 6 .5 + 0.4 6.9 320 − 55 (0.2) 320 − 11 309 If t = 20% = 0.2 then q = = = = 42.33 6 .5 + 4(0.2) 6 .5 + 0.8 7.3 320 − 16 .5 303 .5 320 − 55 (0.3) = = = 39.42 If t = 30% = 0.3 then q = 6 .5 + 4(0.3) 6 .5 + 1.2 7.7 If t = 10% = 0.1 then q = These examples show... example Figure 5. 5 shows that MC will cut MR in the section below the kink K The aggregate profit-maximizing output is found where MR = MC Thus using the MC function given in the question and the aggregated marginal revenue function (5) derived above we get 10 − 0.1q = 1. 75 + 0.05q 8. 25 = 0.15q 55 = q © 1993, 2003 Mike Rosser Therefore MR = 10 − 0.1 (55 ) = 10 − 5. 5 = 4 .5 and so MR1 = 4 .5 MR2 = 4 .5 To determine... work out the reduced form equations, a proportional tax needs to be specified in decimal format Thus a sales tax of 17 .5% becomes 0.1 75 in decimal format Example 5. 17 A proportional sales tax t is imposed in a competitive market where demand price = pd = 3 75 − 2.5q and supply price = ps = 55 + 4q © 1993, 2003 Mike Rosser p St S 55 (1+ t) 55 D q 0 Figure 5. 4 Derive reduced form equations for the equilibrium... + 3(28 .5) = 120 x + 24 + 85. 5 = 120 x = 120 − 109 .5 x = 10 .5 Therefore, the solutions are x = 10 .5, y = 2, z = 28 .5 Example 5. 11 Solve for x, y and z in the following set of simultaneous equations: 14.5x + 3y + 45z = 340 (1) 25x − 6y − 32z = 82 (2) 9x + 2y − 3z = 16 (3) © 1993, 2003 Mike Rosser Solution Multiplying (1) by 2 29x + 6y + 90z = 680 Adding (2) 25x − 6y − 32z = 82 (2) Gives 54 x + 58 z = 762... Solve for x, y and z when 12x + 15y + 5z = 158 4x + 3y + 4z = 50 (2) 5x + 20y + 2z = 148 3 (1) (3) Solve for A, B and C when 32A + 14B + 82C = 664 11.5A + 8B + 52 C = 349 (2) 18A + 26.2B − 62C = 56 0.4 4 (1) (3) Find the values of x, y and z when 4.5x + 7y + 3z = 128 .5 6x + 18.2y + 12z = 270.8 3x + 8y + 7z = 139 5 Solve for A, B, C and D when A + 6B + 25C + 17D = 843 3A + 14B + 60C + 21D = 1,286 .5 10A... equation, we can calculate p for some extreme values of t to see if the prices calculated lie in a reasonable range for this demand schedule If t = 0 (i.e no tax) If t = 100% = 1 1,637 .5 + 1,637 .5( 0) 1,637 .5 = = 251 .92 6 .5 + 4(0) 6 .5 1,637 .5 + 1,637 .5 3,2 75 p= = 311.81 = 10 .5 6 .5 + 4 then p = then These values lie in a range that one would expect for this demand schedule The reduced form of a Keynesian macroeconomic... MR2 = 9 − 0.15q2 (1) q2 = 60 − MR2 0. 15 (2) £ £ £ 12 12 9 9 8. 25 K MR = MR1 + MR2 D1 6. 75 MC D2 4 .5 MR2 MR1 0 25 40 q1 80 0 30 60 120 q2 0 55 100 q Figure 5. 5 For profit maximization MR1 = MR2 = MR (3) and by definition q = q 1 + q2 (4) Therefore, substituting (1), (2) and (3) into (4) q = 40 − MR 0.3 + 60 − MR 0. 15 12 − MR + 18 − 2MR 0.3 30 − 3MR = 100 − 10MR = 0.3 and so MR = 10 − 0.1q = (5) This function... increased by one unit In this example the reduced form equation for price (4) tells us that for every one unit increase in t the equilibrium price p increases by 0. 75 This is illustrated below for a few values of t: when t = 4 then p = 6 + 0. 75( 4) = 6 + 3 = 9 when t = 5 then p = 6 + 0. 75( 5) = 6 + 3. 75 = 9. 75 when t = 6 then p = 6 + 0. 75( 6) = 6 + 4 .5 = 10 .5 Note that this method can only be used with linear . Rosser Solution Multiplying(1)by229x+6y+90z=680 Adding(2)25x−6y−32z=82 (2) Gives54x +58 z=762(4) Havingusedequations(1)and(2)toeliminateywenowneedtobringinequation(3)to deriveasecondequationcontainingonlyxandz. Multiplying(3)by327x+6y−9z=48 Adding(2)25x−6y−32z=82 (2) gives52x−41z=130 (5) Multiplying (5) by271,404x−1,107z=3 ,51 0 Multiplying(4)by261,404x+1 ,50 8z =19,812 Subtractinggives−2,615z=−16,302 z=6.234 (Notethatalthoughfinalanswersaremoreneatlyspecifiedtooneortwodecimalplaces, moreaccuracywillbemaintainedifthefullvalueofzaboveisenteredwhensubstitutingto calculateremainingvaluesofunknownvariables.) Substitutingtheabovevalueofzinto (5) gives 52 x−41(6.234)=130 52 x=130+ 255 .59 4 x=7.4 15 Substitutingforbothxandzin(1)gives 14 .5( 7.4 15) +3y+ 45( 6.234)=340 3y=−48. 05 y=−16.02 Thus,solutionsto2decimalplacesare x=7.42y=−16.02z=6.23 Theaboveexamplesshowhowthesolutiontoa3×3setofsimultaneousequationscanbe solvedbyrowoperations.Thesamemethodcanbeusedforlargersetsbutobviouslymore stageswillberequiredtoeliminatetheunknownvariablesonebyoneuntilasingleequation withoneunknownisarrivedat. Itmustbestressedthatitisonlypracticaltousethemethodsofsolutionforlinearequation systemsexplainedherewheretherearearelativelysmallnumberofequationsandunknowns. Forlargesystemsofequationswithmorethanahandfulofunknownsitismoreappropriate tousematrixalgebramethodsandanExcelspreadsheet(seeChapter 15) . ©. 390 46L = 690 L = 15 Substituting this value for L into (1) K = 2( 15) − 20 = 10 and into (2) R = 15 + 50 = 65 Therefore the optimal input combination is K = 10 L = 15 R = 65 Example 5. 13 In a closed. when 12x + 15y + 5z = 158 (1) 4x + 3y + 4z = 50 (2) 5x + 20y + 2z = 148 (3) 3. Solve for A, B and C when 32A + 14B + 82C = 664 (1) 11.5A + 8B + 52 C = 349 (2) 18A + 26.2B − 62C = 56 0.4 (3) 4.