Basic Mathematics for Economists phần 2 ppsx

per week were worked, then gross pay = (40 hours @ £8) +(10 hours @ £12) = £320 +£120 = £440 tax payable = (£440 −£80)0.25 = (£360)0.25 = £90 Therefore net pay = £440 −£90 = £350 UsingtheexpressionderivedinExample3.14,ifH=50then net pay = 9H − 100 = 9(50) − 100 = 450 − 100 = £350 This checks out with the answer above and so we know our expression works. It is rather more complicated to multiply pairs of brackets together. One method that can be used is rather like the long multiplication that you probably learned at school, but instead of keeping all units, tens, hundreds etc. in the same column it is the same algebraic terms that are kept in the same column during the multiplying process so that they can be added together. Example 3.15 Simplify (6 +2x)(4 − 2x). Solution Writing this as a long multiplication problem: 6 +2x × 4 −2x Multiplying (6 +2x)by−2x −12x − 4x 2 Multiplying (6 +2x)by4 24+ 8x Adding together gives the answer 24 −4x − 4x 2 One does not have to use the long multiplication format for multiplying out sets of brackets. The basic principle is that each term in one set of brackets must be multiplied by each term in the other set. Like terms can then be collected together to simplify the resulting expression. Example 3.16 Simplify (3x + 4y)(5x − 2y). © 1993, 2003 Mike Rosser Solution Multiplying the terms in the second set of brackets by 3x gives: 15x 2 − 6xy (1) Multiplying the terms in the second set of brackets by 4y gives: 20xy − 8y 2 (2) Therefore, adding (1) and (2) the whole expression is 15x 2 − 6xy + 20xy − 8y 2 = 15x 2 + 14xy − 8y 2 Example 3.17 Simplify (x + y) 2 . Solution (x + y) 2 = (x + y)(x + y) = x 2 + xy + yx + y 2 = x 2 + 2xy + y 2 The above answer can be checked by referring to Figure 3.1. The area enclosed in the square with sides of length x +y can be calculated by squaring the lengths of the sides, i.e. finding (x + y) 2 . One can also see that this square is made up of the four rectangles A, B, C and D whose areas are x 2 ,xy,xy and y 2 respectively – in other words, x 2 + 2xy + y 2 , which is the answer obtained above. x y A D B C xy Figure 3.1 © 1993, 2003 Mike Rosser Example 3.18 Simplify (6 −5x)(10 − 2x + 3y). Solution Multiplying out gives 60 −12x + 18y − 50x + 10x 2 − 15xy = 60 − 62x + 18y +10x 2 − 15xy Some expressions may be best left with the brackets still in. Example 3.19 If a sum of £x is invested at an interest rate of r% write an expression for the value of the investment at the end of 2 years. Solution After 1 year the investment’s value (in £) is x  1 + r 100  After 2 years the investment’s value (in £) is x  1 + r 100  2 One could multiply out but in this particular case the expression is probably clearer, and also easier to evaluate, if the brackets are left in. The next section explains how in some cases, some expressions may be ‘simplified’ by reformatted into two expressions in brackets multiplied together. This is called ‘factorization’. Test Yourself, Exercise 3.4 Simplify the following expressions: 1. 6x(x − 4) 2. (x + 3) 2 − 2x 3. (2x + y)(x + 3) 4. (6x + 2y)(7x − 8y) + 4y + 2y 5. (4x − y + 7)(2y − 3) + (9x −3y)(5 + 6y) 6. (12 − x + 3y + 4z)(10 + x + 2y) 7. A good costs a basic £180 a unit but if an order is made for more than 10 units this price is reduced by a discount of £2 for every one unit increase in the size of an order (up to a maximum of 60 units purchased), i.e. if the order size is 11, price is £178; if it is 12, price is £176 etc. Write an expression for the total cost of an order in terms of order size and simplify it. Assume order size is between 10 and 60 units. © 1993, 2003 Mike Rosser 8. A holiday excursion costs £8 per person for transport plus £5 per adult and £3 per child for meals. Write an expression for the total cost of an excursion for x adults and y children and simplify it. 9. A firm is building a car park for its employees. Assume that a car park to accom- modate x cars must have a length (in metres) of 4x +10 and a width of 2x +10. If 24 square metres will be specifically allocated for visitors’ cars, write an expres- sion for the amount of space available for the cars of the workforce in terms of x if x is the planned capacity of the car park. 10. A firm buys a raw material that costs £220 a tonne for the first 40 tonnes, £180 a tonne for the next 40 tonnes and £150 for any further quantities. Write an expression for the firm’s total expenditure on this input in terms of the total amount used (which can be assumed to be greater than 80 tonnes), and simplify. 3.5 Simplification: factorizing Forsomepurposes(see,e.g.Section3.6belowandChapter6onquadraticequations)itmay be helpful if an algebraic expression can be simplified into a format of two sets of brackets multiplied together. For example x 2 + 4x +4 = (x + 2)(x + 2) This is rather like the arithmetical process of factorizing a number, which means finding all the prime numbers which when multiplied together equal that number, e.g. 126 = 2 ×3 ×3 ×7 If an expression has only one unknown variable, x, and it is possible to factorize it into two sets of brackets that do not contain terms in x to a power other than 1, the expression must be in the form ax 2 + bx + c However, not all expressions in this form can be factorized into sets of brackets that only involve integers, i.e. whole numbers. There are no set rules for working out if and how an expression may be factorized. However, if the term in x 2 does not have a number in front of it (i.e. a = 1) then the expression can be factorized if there are two numbers which (i) give c when multiplied together, and (ii) give b when added together. Example 3.20 Attempt to factorize the expression x 2 + 6x +9. © 1993, 2003 Mike Rosser Solution In this example a = 1,b = 6 and c = 9. Since 3 ×3 = 9 and 3 +3 = 6, it can be factorized, as follows x 2 + 6x +9 = (x + 3)(x + 3) This can be checked as x + 3 × x + 3 3x + 9 x 2 + 3x x 2 + 6x +9 Example 3.21 Attempt to factorize the expression x 2 − 2x − 80. Solution Since (−10) ×8 =−80 and (−10) + 8 =−2 then the expression can be factorized and x 2 − 2x − 80 = (x − 10)(x + 8) Check this answer yourself by multiplying out. Example 3.22 Attempt to factorize the expression x 2 + 3x +11. Solution There are no two numbers which when multiplied together give 11 and when added together give 3. Therefore this expression cannot be factorized. It is sometimes possible to simplify an expression before factorizing if all the terms are divisible by the same number. Example 3.23 Attempt to factorize the expression 2x 2 − 10x + 12. © 1993, 2003 Mike Rosser Solution 2x 2 − 10x + 12 = 2(x 2 − 5x +6) The term in brackets can be simplified as x 2 − 5x +6 = (x − 3)(x − 2) Therefore 2x 2 − 10x + 12 = 2(x − 3)(x − 2) In expressions in the format ax 2 +bx +c where a is not equal to 1, then one still has to find two numbers which multiply together to give c. However, one also has to find two numbers for the coefficients of the two terms in x within the two sets of brackets that when multiplied together equal a, and allow the coefficient b to be derived when multiplying out. Example 3.24 Attempt to factorize the expression 30x 2 + 52x + 14. Solution If we use the results that 6 ×5 = 30 and 2 × 7 = 14 we can try multiplying 5x + 7 × 6x + 2 10x + 14 30x 2 + 42x This gives 30x 2 + 52x + 14 Thus 30x 2 + 52x + 14 = (5x + 7)(6x + 2) Similar rules apply when one attempts to factorize an expression with two unknown variables, x and y. This may be in the format ax 2 + bxy + cy 2 where a, b and c are specified parameters. Example 3.25 Attempt to factorize the expression x 2 − y 2 . © 1993, 2003 Mike Rosser Solution Inthisexamplea=1,b=0andc=−1.Thetwonumbers−1and1give−1when multipliedtogetherand0whenadded.Thus x 2 −y 2 =(x−y)(x+y) Tocheckthis,multiplyout: (x−y)(x−y)=x 2 −xy+y 2 −yx=x 2 −y 2 Example3.26 Attempttofactorizetheexpression3x 2 +8x+23. Solution As23isapositiveprimenumber,theonlypairsofpositiveintegersthatcouldpossiblybe multipliedtogethertogive23are1and23.Thus,whateverpermutationsofcombinations withtermsinxthatwetry,theterminxwhenbracketsaremultipliedoutwillbeatleast 24x,e.g.(3x+23)(x+1)=3x 2 +26x+23,whereasthegivenexpressioncontainsthe term8x.Itisthereforenotpossibletofactorizethisexpression. Unfortunately,itisnotalwayssoobviouswhetherornotanexpressioncanbe factorized. Example3.27 Attempttofactorizetheexpression3x 2 +24+16. Solution Althoughthenumberslookpromising,ifyoutryvariouspermutationsyouwillfindthatthis expressiondoesnotfactorize. Thereisnoeasywayoffactorizingexpressionsanditisjustamatteroftrialanderror.Do notdespairthough!Asyouwillseelateron,factorizingmayhelpyoutouseshort-cut methodsofsolvingcertainproblems.Ifyouspendagestryingtofactorizeanexpres- sionthenthiswilldefeattheobjectofusingtheshort-cutmethod.Ifitisnotobvious howanexpressioncanbefactorizedafterafewminutesofthoughtandexperimentation withsomepotentialpossiblesolutionsthenitisusuallymoreefficienttoforgetfactoriza- tionandusesomeothermethodofsolvingtheproblem.Weshallreturntothistopicin Chapter6. © 1993, 2003 Mike Rosser Test Yourself, Exercise 3.5 Attempt to factorize the following expressions: 1. x 2 + 8x +16 2. x 2 − 6xy + 9y 2 3. x 2 + 7x +22 4. 8x 2 − 10x + 33 5. Make up your own expression in the format ax 2 +bx+c and attempt to factorize it. Check your answer by multiplying out. 3.6 Simplification: division To divide an algebraic expression by a number one divides every term in the expression by the number, cancelling where appropriate. Example 3.28 15x 2 + 2xy + 90 3 = 5x 2 + 2 3 xy + 30 To divide by an unknown variable the same rule is used although, of course, where the numerator of a fraction does not contain that variable it cannot be simplified any further. Example 3.29 2x 2 x = 2x Example 3.30 4x 3 − 2x 2 + 10x x = x(4x 2 − 2x + 10) x = 4x 2 − 2x + 10 Example 3.31 16x + 120 x = 16 + 120 x © 1993, 2003 Mike Rosser Example 3.32 A firm’s total costs are 25x + 2x 2 , where x is output. Write an expression for average cost. Solution Average cost is total cost divided by output. Therefore AC = 25x + 2x 2 x = 25 +2x If one expression is divided by another expression with more than one term in it then terms can only be cancelled top and bottom if the numerator and denominator are both multiples of the same factor. Example 3.33 x 2 + 2x x + 2 = x(x + 2) x + 2 = x Example 3.34 x 2 + 5x +6 x + 3 = (x + 3)(x + 2) x + 3 = x + 2 Example 3.35 x 2 + 5x +6 x 2 + x − 2 = (x + 2)(x + 3) (x + 2)(x − 1) = x + 3 x − 1 Test Yourself, Exercise 3.6 1. Simplify 6x 2 + 14x −40 2x 2. Simplify x 2 + 12x + 27 x + 3 © 1993, 2003 Mike Rosser 3. Simplify 8xy + 2x 2 + 24x 2x 4. A firm has to pay fixed costs of £200 and then £16 labour plus £5 raw materi- als for each unit produced of good X. Write an expression for average cost and simplify. 5. A firm sells 40% of its output at £200 a unit, 30% at £180 and 30% at £150. Write an expression for the average revenue received on each unit sold and then simplify it. 6. You have all come across this sort of party trick: Think of a number. Add 3. Double it. Add 4. Take away the number you first thought of. Take away 3. Take away the number you thought of again. Add 2. Your answer is 9. Show how this answer can be derived by algebraic simplification by letting x equal the number first thought of. 7. Make up your own ‘think of a number’ trick, writing down the different steps in the form of an algebraic expression that checks out the answer. 3.7 Solving simple equations We have seen that evaluating an expression means calculating its value when one is given specific values for unknown variables. This section explains how it is possible to work backwards to discover the value of an unknown variable when the total value of the expression is given. When an algebraic expression is known to equal a number, or another algebraic expression, we can write an equation, i.e. the two concepts are written on either side of an equality sign. For example 45 = 24 +3x In this chapter we have already written some equations when simplifying algebraic expres- sions. However, the ones we have come across so far have usually not been in a format where the value of the unknown variables can be worked out. Take, for example, the simplification exercise 3x + 14x − 5x = 12x The expressions on either side of the equality sign are equal, but x cannot be calculated from the information given. Some equations are what are known as ‘identities’, which means that they must always be true. For example, a firm’s total costs (TC) can be split into the two components total fixed costs (TFC) and total variable costs (TVC). It must therefore always be the case that TC = TFC +TVC Identities are sometimes written with the three bar equality sign ‘≡’ instead of ‘=’, but usually only when it is necessary to distinguish them from other forms of equations, such as functions. © 1993, 2003 Mike Rosser [...]... price per share be x Therefore, working in pence, 69 ,20 0 = 500x − 2, 000 Adding 2, 000 to both sides 71 ,20 0 = 500x Dividing both sides by 500 gives the solution 1 42. 4 = x Thus the share price is 1 42. 4p Test Yourself, Exercise 3.7 1 Solve for x when 16x = 2x + 56 2 Solve for x when 14 = 6 + 4x 5x Solve for x when 45 = 24 + 3x Solve for x if 5x 2 + 20 = 1,000 If q = 560 − 3p solve for p when q = 314 You get... 98, i.e by 22 .5%, which is greater than Table 3 .2 Percentage of expenditure (xi ) Durable goods Food Alcohol and tobacco Footwear and clothing Energy Other goods Rent, rates, water Other services Prices, year 0 (pi0 ) Prices, year 1 (pi1 ) 9 17 11 7 8 11 12 25 20 0 80 70 120 26 5 62 94 52 216 98 92 130 27 0 71 98 60 100 Note All prices are in £ © 1993, 20 03 Mike Rosser the inflation rate Adjusted for inflation,... However, spreadsheets can be used to do tedious calculations and in some cases short-cut formulae may be used (see Chapter 7) © 1993, 20 03 Mike Rosser Example 3.44 Evaluate n (20 + 3i) for n = 6 i=3 Solution Note that in this example i starts at 3 Thus 6 (20 + 3i) = (20 + 9) + (20 + 12) + (20 + 15) + (20 + 18) i=3 = 29 + 32 + 35 + 38 = 134 The second way in which the summation sign can be used requires a... solution x = 24 Example 3.40 Solve for x if 6x 2 + 12 = 1 62 Solution Subtracting 12 from both sides 6x 2 = 150 Dividing through by 6 x 2 = 25 Taking square roots gives the solution x = 5 or − 5 Example 3.41 A firm has to pay fixed costs of £1,500 plus another £60 for each unit produced How much can it produce for a budget of £4,800? Solution budget = total expenditure on production Therefore if x is... inequality sign between your possible overall mark for the course and the percentage figures below (a) 18% ? overall mark (c) 88% ? overall mark 2 If x ≥ 1, insert the appropriate inequality sign between: (a) (x + 2) 2 and 3 (c) (x + 2) 2 and 3x 3 (b) (x + 2) 2 and 9 (d) (x + 2) 2 and 6x If Q1 and Q2 represent positive production levels of a good and the equality Q2 = Z n Q1 always holds where Z > 1, what can... then P = 120 − 0 .2( 400) = 120 − 80 = 40 If Q = 500 then P = 120 − 0 .2( 500) = 120 − 100 = 20 These are the values of P originally specified and so we are satisfied that the line that passes through points A and B in Figure 4.6 is the linear function P = 120 − 0.2Q © 1993, 20 03 Mike Rosser The inverse of this function will be Q = 600 −5P Precise values of Q can now be derived for given values of P For example,... vertical axis at 20 0 However, when C = 0, then 0 = 20 0 + 0.6Y −0.6Y = 20 0 Y =− 20 0 0.6 As negative values of Y are unacceptable, just choose another pair of values, e.g when Y = 500 then C = 20 0 + 0.6(500) = 20 0 + 300 = 500 This graph is shown in Figure 4.4 C C = 20 0 + 0.6y 500 20 0 0 Figure 4.4 © 1993, 20 03 Mike Rosser 500 y P 20 0 Demand function Q = 800 – 4P 0 800 Q Figure 4.5 In mathematics the usual... functions Sales are not determined by ‘time’ (£’000 s) Sales revenue 20 0 180 160 140 120 100 80 Sales 60 40 20 0 1997 Figure 4.1 © 1993, 20 03 Mike Rosser 1998 1999 20 00 20 01 20 02 y y = 5 + 0.6x A 17 11 5 –x 0 10 20 x –y Figure 4 .2 Mathematical functions are mapped out on what is known as a set of ‘Cartesian axes’, as shown in Figure 4 .2 Variable x is measured by equal increments on the horizontal axis... produce and then see what price they can get for this output, i.e P = f(Q) Example 4.3 Given the demand function Q = 20 0 − 4P , derive the inverse demand function Solution Q = 20 0 − 4P 4P + Q = 20 0 4P = 20 0 − Q P = 50 − 0 .25 Q Test Yourself, Exercise 4 .2 1 To convert temperature from degrees Fahrenheit to degrees Celsius one uses the formula ◦ 2 C= 5 ◦ ( F − 32) 9 What is the inverse of this function?... figure Write an expression for the firm’s total sales over n years and evaluate for n = 3 Observations of a firm’s sales revenue (in £’000) per month are as follows: Month 1 2 3 4 5 6 7 8 9 10 Revenue 4.5 4 .2 4.6 4.4 5.0 5.3 5 .2 4.9 4.7 5.4 11 5.3 12 5.8 (a) Write an expression for average monthly sales revenue for the first n months and evaluate for n = 4 (b) Write an expression for average monthly sales . 3 .29 2x 2 x = 2x Example 3.30 4x 3 − 2x 2 + 10x x = x(4x 2 − 2x + 10) x = 4x 2 − 2x + 10 Example 3.31 16x + 120 x = 16 + 120 x © 1993, 20 03 Mike Rosser Example 3. 32 A firm’s total costs are 25 x. factor. Example 3.33 x 2 + 2x x + 2 = x(x + 2) x + 2 = x Example 3.34 x 2 + 5x +6 x + 3 = (x + 3)(x + 2) x + 3 = x + 2 Example 3.35 x 2 + 5x +6 x 2 + x − 2 = (x + 2) (x + 3) (x + 2) (x − 1) = x + 3 x. Simplify 6x 2 + 14x −40 2x 2. Simplify x 2 + 12x + 27 x + 3 © 1993, 20 03 Mike Rosser 3. Simplify 8xy + 2x 2 + 24 x 2x 4. A firm has to pay fixed costs of 20 0 and then £16 labour plus £5 raw materi- als for