Part 1: mathematics 1 linear algebra and equation solving text: Chapter 1 in core textbook
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Part 1 Mathematics study guide
- Bu sahifa navigatsiya:
- 1.1.4 Subtraction and negative values
- 1.1.5 Division, reciprocals and fractions
- 1.1.6 Powers (exponents)
- 1.2.3 Simultaneous linear equations with two unknown variables
- 1.3.1 Economics application: Demand, revenue, cost and profit equations for a monopolist
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PART 1: MATHEMATICS 1 LINEAR ALGEBRA and EQUATION SOLVING TEXT: Chapter 1 in core textbook: Curwin, J., Slater, R. and Eadson, D. (2015). Quantitative methods for business decisions, seventh edition. Cengage Learning. 1. Basic Algebra 1.1 Rules of linear algebra This section is revision material. It is worthwhile reading through all of Section 1.1 to make sure you are comfortable with the material. We will not go through this material in detail during the lecture. Although we will use all of these skills during linear and non-linear algebra examples. If you are not comfortable with this material, then you can go online to look for examples, or I would encourage the use of a textbook such as Chapter 1 in the core textbook for this course: Curwin, J., Slater, R. and Eadson, D. (2015). Quantitative methods for business decisions, seventh edition. Cengage Learning. 1.1.1 Addition
a+b = b+a a+b+c = (a+b)+c = a+(b+c) [order of addition does not matter]
ab = ba abc = (ab)c = a(bc)
[order of multiplication does not matter] 1.1.3 Factorisation and the use of brackets
a(b+c) = ab +ac [expansion of brackets]
ab+ac = a(b+c)
[factorisation] (a+b) (c+d) = (a+b)c+(a+b)d = ac+bc+ad+bd
When expanding brackets, always perform the operation inside the brackets first. If there are brackets within brackets, start from the inside and work outwards.
e.g. 3{5a+2(a+b)} = 3{5a+2a+2b}
= 3{7a+2b}
= 21a+6b
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1.1.4 Subtraction and negative values
The formal definition of a negative number is as follows. For any number, there exists a corresponding ‘additive inverse’ number, such that the sum of the number and its additive inverse number is zero.
e.g.
the additive inverse of 3 is –3, so 3+(–3) = 0.
In general, the additive inverse of any positive number is the corresponding negative number. Subtraction of a number is the same as addition of that number’s additive inverse.
a+(b) = a+b a– (b) = a–b
a+(–b) = a–b a– (–b) = a+b
Rules
(a)
Two ‘minuses’ make ‘plus’ (b)
A minus sign outside a bracket changes the signs of all the terms inside the bracket
e.g. a – (x+2y–3z) = a–x–2y+3z (c) ab is positive: either if ‘a’ and ‘b’ are both positive
or if ‘a’ and ‘b’ are both negative (d)
ab is negative if one of ‘a’ and ‘b’ is positive, and the other is negative
The formal definition of a reciprocal is as follows: For any number except zero, there exists a corresponding ‘multiplicative inverse’ number, or ‘reciprocal’ number, such that the product of the number and its reciprocal is 1.
e.g.
the reciprocal of 2 is 1/2, also written as 2 –1
2 1/2 = 1, or 2 (2 –1 ) = 1
Division by a number is the same as multiplication by the reciprocal of that number. a b or a/b = a (1/b) or a (b –1 )
The number 0 can be used in the numerator of a fraction, but not in the denominator. 0/a = 0
a/0 is undefined
0/0 is undefined
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Rules
Multiplication and division: bd ac d c b a ; bc ad d c b a d c b a
Cancellation of common factors: c b
ab
Addition and subtraction: To add/subtract two fractions, put them over a ‘common denominator’; then add or subtract the numerators; then (if possible) simplify by cancelling common factors.
bd bc ad bd bc bd ad d c b a
bd bc ad bd bc bd ad d c b a
1.1.6 Powers (exponents) Let b be any number, and let n be any positive integer (whole number). The expression b n
b 1 = b b 2 = b b b 3 = b b b : b n = b
b .... b [n times]
Similar expressions can be defined for values of n that are not positive integers: b 0 = b/b = 1 b –1 = b/(b b) = 1/b b –2 = b/(b b b) = 1/b 2
: b –n = b/(b b ...
b) = 1/b
n
[n+1 times]
Expressions can also be defined for values of n which are fractional or decimal, provided b is not negative:
4 b 1/2 = square root of b ( b) = the number c such that c 2 = b
b 1/3
= cube root of b ( 3 b) = the number d such that d 3 = b : b 1/n = n’th root of b ( n b) = the number k such that k n = b The following rules and examples may be useful when manipulating expressions containing powers.
b 3/2 = (b
3 ) 1/2 or (b 1/2
) 3
b 0.75
= b 3/4
= (b 3 ) 1/4 or (b
1/4 ) 3 b m
b n = b m+n
b m
b n = b m–n
(b m ) n = b
mn
(ab) n = a
n b n (a + b)
2 = (a + b)(a + b) = a 2 + ba +ab + b 2 = a
2 + 2ab + b 2
2 = (a – b)(a – b) = a 2 – ba – ab + b 2 = a
2 – 2ab + b 2 (a + b)(a – b) = a 2 + ba – ab – b 2 = a
2 – b
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1.1.7 Manipulation of equations
Rules for carrying terms across the ‘equals sign’: If a + b = c then a = c – b If a – b = c then a = c + b
Rules for cross-multiplication: If ab = cd then b cd
If d c b a then d bc a or ad = bc
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1.1.8 Inequalities
a b means ‘a is greater than or equal to b’ a>b means ‘a is greater than b’ a b means ‘a is less than or equal to b’ a
If a>b then a+c>b+c If a>b then ac>bc if c>0 ac
Similar rules apply for the other inequalities: <, and
.
If you are struggling with Section 1.1, then please be sure to revise it thoroughly, as it will benefit you for the rest of the material in this module as well as others.
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1.2 Linear algebra 1.2.1 Linear equations with one unknown variable
Often we need to use the rules of algebra to solve an equation which contains one unknown variable. A linear equation is one which has terms in the unknown variable itself, e.g. x, but no terms in x 2 , x
3 , x
1/2 etc.
Question You see a pack of six apples in the supermarket for 60 pence. What is the price of one apple? Well this is simple as its 10 pence per apple. If you can work this out then you have just used algebra!
We can write the problem down formally, let a stand for the cost of an apple, and then we have:
6a = 60
a = 10
so cost of one apple is ten pence.
Important Rule: Whatever you do to one side of the equation, you must also do to the other side. (look back to Section 1.1.7.) Examples
Method
1.
If there are fractions, eliminate them by putting everything over a common denominator, and then multiplying through by the denominator.
2.
If there are brackets, multiply them out.
3. Collect all the terms in x on one side of the equation, and all the numerical terms on the other side.
4.
Divide both sides through by the coefficient on x, to obtain the numerical solution for x.
5. Substitute the solution back into the original equation and evaluate both sides, to check
whether the solution is correct.
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Examples
Solve the following equations for x: (i)
3x + 4 = 10 (ii)
3(x – 3) = 2(x – 5) + 7
Solutions
(i) 3x + 4 = 10 3x = 10 – 4
3x = 6
x = 6/3 x = 2 [Check : 3(2) + 4 = 10 OK]
(ii)
3(x – 3) = 2(x – 5) + 7 3x – 9 = 2x – 10 + 7
3x – 2x = –10 + 7 + 9
– 5) + 7 OK]
1.2.2 Linear Algebra with two variables
Linear algebra with only one variable such as in Section 1.2.1 have little relevance to ‘real world’ problems. What we usually have is problems where there are more than one variable of interest. Such as the price and the quantity sold of a certain product.
A shop owner has worked out an equation that links the quantity (q) of t-shirts they sell and the price (p) it’s sold at. The equation is as follows:
q
= 250 – 5p By substituting different values for the price, p, we can work out how much quantity of t-shirts sold as follows:
When p = 5, q = 250 – 5*(5) = 225
Or when p = 10, q = 250 – 5*(10) = 200
Price (£) Quantity sold 0 250 5 225
10 200
15 175
20 150
25 125
30 100
35 75 40 50 45 25 50 0
8 The data in the above table graphed in Excel: Note the ‘quantity sold’ is on the y-axis (or vertical axis) and the ‘price’ is on the x-axis (or horizontal axis)
The data draws a straight-line graph. This means that price and quantity sold have a ‘linear’ relationship. As price is increased then quantity sold goes decreases. There is an ‘inverse’ relationship here.
This is a very simple example, because we have not factored in any costs involved. Because if we follow the line left, beyond the £0 price, then the quantity sold would keep going up, but this would obviously yield a loss for the shop owner!
We can draw this type of graph very easily in Excel. Straight-line graphs can always be represented by the equation:
y = mx + c or y = c + mx
c is the intercept, i.e. the value of y at which the line cuts the y-axis m is the slope, where m>0
m=0 the line is flat m<0 the line is downward sloping. The ‘coordinates’ of any point on a graph can be represented in the form (x-value, y-value).
e.g. (5,8) means ‘x = 5 and y = 8’ In our example the slope, m = -5, hence the downward sloping line, and the intercept, c = 250.
9 Some examples
0 10 20 30 40 50 60 70 80 0 2 4 6 8 10 12 14 y- ax is x-axis
y = 5x + 10 -6 -4 -2 0 2 4 6 8 10 -3 -2 -1 0 1 2 3 4 5 y- ax is x-axis
y = 4-2x 10
1.2.3 Simultaneous linear equations with two unknown variables
A slightly more complicated situation arises when we have more than one equation, and there is more than one unknown variable to solve for. In general, we can always solve provided there are as many equations as there are unknown variables; i.e. 2 equations in 2 unknown variables (x and y); 3 equations in 3 unknown variables (x, y and z); and so on.
There are two methods of solution: algebraic and graphical. Algebraic method
1.
Eliminate any fractions or brackets (as in section 1.2.1). Then rearrange both equations so that the terms in x and y are on the left hand side, and the numerical terms are on the right.
Multiply each side of one (or both) of the equations through by a number (or numbers) so that the coefficient on one of the variables (x or y) has the same numerical value in both equations. The signs of the equalised coefficients can be the same or different.
3. Eliminate the chosen variable by subtracting one equation from the other (if the signs of the equalised coefficients are the same) or adding the two equations (if the signs are different).
4. Solve the resulting equation for the remaining variable.
5. Substitute the solution back into one of the original equations, and solve for the other variable.
6.
Check your solution by substituting both values back into the other original equation.
Examples Solve algebraically the following pairs of simultaneous equations:
(i)
x + 2y = 5
x – y = 2
(ii) 2y + 14 = 6x
5x – 3y = 1 (iii)
4x + 2y = 14
2x – y = 1
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Solutions
(i) x + 2y = 5
[1]
x – y = 2 [2]
[1] – [2] 3y = 3
y = 1
substitute into [1] x + 2 1 = 5
x = 5 – 2
check in [2]
3 – 1 = 2
OK
(ii) rearranging
–6x + 2y = –14 [1]
5x – 3y = 1 [2]
[1] 3 –18x + 6y = –42 [3]
[2] 2 10x – 6y = 2
[4]
[3] + [4] –8x = –40
x = –40/–8
x = 5
substitute into [2] 5 5 – 3y = 1
24 = 3y
y = 24/3
check in [1] –6 5 + 2
8 = –14 OK
(iii) 4x + 2y = 14 [1]
2x – y = 1 [2]
[2] 2
4x – 2y = 2 [3]
[1] + [3]
8x = 16
x = 16/8
x = 2 substitute into [1] 4 2 + 2y = 14
2y = 14 – 8 2y = 6 y = 3 12
check in [2] 2 2 – 3 = 1
OK
Graphical method
1.
Rearrange both equations so that y appears by itself on the left hand side of both (i.e. get both equations in the form y = c+mx, where c and m are numbers).
2.
For the first equation, find the numerical value of y when x = 0, and plot the point on the graph. Next find the numerical value of y for a second, arbitrarily chosen, value of x, and plot on the graph. Join the two points with a straight line.
3. Repeat step 2 for the second equation.
4. Locate the solution at the point of intersection of the two lines. If the point of intersection is off the graph, try again with different scales on the axes.
For any linear equation of the form y = c+mx or y = mx+c, c is the intercept, i.e. the value of y at which the line cuts the y-axis m is the slope, where m>0 the line is upward sloping m=0 the line is flat m<0 the line is downward sloping. The ‘coordinates’ of any point on a graph can be represented in the form (x-value, y-value).
e.g. (5,8) means ‘x = 5 and y = 8’
Please look at the other notes available in Blackboard for examples how to draw line graphs in Excel.
Examples
Solve graphically the following pairs of simultaneous equations (taken from the previous example):
(i) x + 2y = 5
(iii) 4x + 2y = 14
x – y = 2
2x – y = 1
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Solutions (to locate two points on each line)
(i) x + 2y = 5 2y = 5 – x
2 x 2 5 y
[1]
x – y = 2 –y = 2 – x y = x – 2 [2]
[1] when x = 0, y = 5/2 = 2.5 when x = 5, y = 5/2 – 5/2 = 0
[2] when x = 0, y = –2 when x = 5, y = 5 – 2 = 3
(iii) 2y + 4x = 14 2y = 14 – 4x y = 7 – 2x [1] 2x – y = 1 –y = 1 – 2x y = 2x – 1 [2]
[1] when x = 0, y = 7
when x = 3, y = 7–6 = 1 [2] when x = 0, y = –1 X X X X 3 5 1 2.5
3 2 x 2 5 y y = x 2 x y 2 (3,1) 14
when x = 3, y = 6–1 = 5
Example The quantity demanded of a certain product (q d , measured in kilos per week) depends on the market price (p per kilo, in £’s), while the quantity supplied (q s , measured in kilos per week) depends on the price producers receive (p * per kilo, in £’s), as follows:- q d = 40–4p
q s = 2p * –8
The market price is the price which producers receive, plus any tax, minus any government subsidy, so: p = p * +t, where t>0 tax
t<0 subsidy X X X X 3 7 1 1 5 y=7 2x y=2x 1 2 3 y x (2,3) 15
Using algebraic methods, find the equilibrium values of p, p * and q d (=q
s )
(i) When there are no taxes or subsidies, (ii) When a tax of £3 per kilo is imposed, (iii) When a subsidy of £1.50 per kilo is awarded.
Solution
If p = p * +t
p
* =
p – t q s = 2(p–t) – 8
(i) When t=0, q d = 40 – 4p q s = 2p – 8
For market clearing, let q d = q
s = q
40 – 4p = 2p – 8
48 = 6p
p = 48/6
p = 8
From demand equation, q = 40 – 4 8
q = 40 – 32
q = 8
Check in supply equation 2
8 – 8 = 8
OK
(ii) When t=3, q d = 40 – 4p q s = 2(p – 3) – 8 q
s = 2p – 14
For market clearing, let q d = q
s = q
40 – 4p = 2p – 14
54 = 6p
p = 54/6
p = 9
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From demand equation, q = 40 – 4 9
q = 40 – 36
q = 4
Check in supply equation 2
9 – 14 = 4
OK
(iii) When t=–1.5, q d = 40 – 4p q s = 2(p + 1.5) – 8 q s = 2p – 5
For market clearing, let q d = q
s = q
40 – 4p = 2p – 5
45 = 6p
p = 45/6
p = 7.5
From demand equation, q = 40 – 4 7.5
q = 40 – 30
q = 10
Check in supply equation 2
7.5 – 5 = 10
OK
In order to show the graphical solution, note that economists’ graphs of demand and supply curves are constructed differently to the normal mathematician’s graphs of the relationship between two variables, y and x. Essentially, the x-axis and y-axis are swapped!
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q d =40 4p q s =2p 8 8 8 p q (8,8) q d =40 4p q s =2p 8 8 8 p q (8,8) q s =2p 14 9 4 Effect of tax of £3 per unit y x p q y=3 2x x=1 y=1 q d =40 4p p=6 q=16
read anti-clockwise read clockwise 18
Note: We will cover how to draw these types of graphs in Excel during the lecture. And we will see how we can draw an ‘economic’ style graph by swapping over the x and y-axis. All this information will be made available in Blackboard.
This graph has been drawn in Excel:
q d =40 4p q s =2p 8 8 8 p q (8,8) q s =2p 5 7.5 10 Effect of a subsidy of £1.50 per unit 19
1.3 Quadratic equations
Quadratic equations are equations in one variable (x), which take the form: ax 2 +bx+c = 0 , where a, b, c are constants (b or c, but not a, could be equal to zero)
Normally there are two solutions for x (but in some cases there can be one solution or no solutions).
Two methods for solving are:- (a) Factorisation. This is easy if you can spot how to do it, but sometimes you can’t! Also, the method of factorisation will not help identify a case for which there is no solution.
(b) Formula method. This takes longer and can be tedious, but it is guaranteed to work if you do it right, and it will always identify cases for which there is no solution.
(a) Method of factorisation
Rewrite the equation as the product of two factors: ax 2 +bx+c = (mx+p)(nx+q) = 0 where mn = a
mq+np = b
pq = c
Then, if ax 2 +bx+c = 0, either mx+p = 0 x = –p/m or nx+q = 0 x = –q/n These expressions give the two solutions. The two factors (i.e. the numerical values of m, n and p) have to be determined by trial and error.
If it helps, the original equation can be ‘multiplied or divided through’ by any number to simplify the factorisation.
Examples Solve the following for x:
(i) x 2 +4x+3 = 0
(vi) x 2 +6x+9 = 0 (ii) x 2 –5x+4 = 0
(vii) 2x 2 –x–3 = 0 (iii) x 2 +2x–8 = 0
(viii) 6x 2 +x–12 = 0 (iv) 2x 2 +2x–40 = 0
(ix) x 2 +8x+4 = 0 (v) x 2 –9 = 0
(x) x 2 –2x+7 = 0
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Solutions
(i) x 2 +4x+3 = 0
x = –3 or x = –1
(ii) x 2 –5x+4 = 0 (x–4)(x–1) = 0 x = 4 or x = 1
(iii) x 2 +2x–8 = 0 (x+4)(x–2) = 0 x = –4 or x = 2
(iv) 2x 2 +2x–40 = 0 x 2 +x–20 = 0 (x+5)(x–4)=0 x = –5 or x = 4 (v) x 2 –9 = 0 (x+3)(x–3) = 0 x = –3 or x = 3 [(a 2 –b 2 ) = (a+b)(a–b)]
(vi) x 2 +6x+9 = 0 (x+3)(x+3) = 0 x = –3 [only one solution] (vii) 2x 2 –x–3 = 0 (2x–3)(x+1) = 0 x = 3/2 or x = –1 (viii) 6x 2 +x–12 = 0 (2x+3)(3x–4) = 0 x = –3/2 or x = 4/3 (ix) x 2 +8x+4 = 0 There is no obvious factorisation which will solve this. In fact, there is a solution, but the formula method is needed to find it (see below).
(x+0.5359)(x+7.4641) = 0 x = –0.5359, x = –7.4641 (x) x 2 –2x+7 = 0 Again, there is no obvious factorisation. In this case, application of the formula method will show that there is no solution (see below).
(b) Formula method For any quadratic equation of the form:
ax
+bx+c = 0
the solution can be obtained by substituting the numerical values of the coefficients a, b and c into the following formula:
a 2 ac 4 b b x 2
A solution exists only if b 2 –4ac
0, because the square root of a negative number is undefined.
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Examples
Solve the following for x: (i) x
2 +4x+3 = 0 (ii) x 2 –5x+4 = 0 (v) x 2 –9 = 0 (ix) x 2 +8x+4 = 0 (x) x 2 –2x+7 = 0 Solutions
(i) x
2 +4x+3 = 0 a = 1, b = 4, c = 3
1 2 3 1 4 4 4 x 2 = 2 12 16 4 = 2 4 4 = 2 2 4
x = –6/2 or x = –2/2 x = –3 or x = –1
(ii) x 2 –5x+4 = 0 a = 1, b = –5, c = 4
1 2 4 1 4 5 5 x 2 = 2 16 25 5 = 2 9 5 = 2 3 5
x = 8/2 or x = 2/2 x = 4 or x = 1
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(v) x 2 –9 = 0 a = 1, b = 0, c = –9
2 6 0 2 36 0 1 2 9 1 4 0 0 x 2
x = 3 or x = –3
(ix) x 2 +8x+4 = 0 a = 1, b = 8, c = 4
2 9282
. 6 8 2 48 8 2 16 64 8 1 2 4 1 4 8 8 x 2
x = –7.4641 or x = –0.5359 (x) x 2 –2x+7 = 0 a = 1, b = –2, c = 7 2 24 2 2 28 4 2 1 2 7 1 4 2 2 x 2
There is no solution because the square root of a negative number [ –24] is undefined. Graphical representation of quadratic equations
Although not recommended as a method for solving quadratic equations, it is sometimes useful to draw a sketch-diagram to get a visual representation of the solution. For the equation ax
2 +bx+c = 0, we write: y = ax 2 +bx+c i.e. ‘let the expression ax 2 +bx+c equal y’ To obtain a graphical representation, the steps are as follows:
1.
Find y when x = 0, and plot the point.
2. Find the values of x at which y = 0 [by solving the original quadratic equation] and plot the points.
3.
Join the plotted points with a curve.
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Example
Produce a graphical representation of the solution to the following quadratic equation: (ii) x 2 –5x+4 = 0 Solution
Let y = x 2 –5x+4
When x=0, y = 0 2 – 5
0 + 4 = 4
When y=0, x=4 or x=1 (solutions obtained above, by factorisation or by the formula method).
This can be quite hard to draw by hand! But Excel allows you to do it easily. Please find Excel material in Blackboard.
You will not be asked to draw a detailed curved graph in the exam for this module. But you may be asked to draw a rough graph.
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1.3.1 Economics application: Demand, revenue, cost and profit equations for a monopolist
Consider a monopolist with a linear demand equation, q = 16–p, where q = quantity demanded and p = price.
To obtain an expression for the monopolist’s total revenue, TR, in terms of the quantity of output which it produces and sells, start by rearranging the demand equation:
q = 16 – p p = 16 – q
By definition, total revenue = price quantity
q = 16q–q 2
2 +bq+c,
with a = –1, b = 16 and c = 0.
Suppose now the firm’s total cost, TC, in terms of the quantity of output which it produces is the following: TC = 4q+20
By definition profit, , is the difference between total revenue, TR, and total cost, TC. Therefore we can write: = TR – TC = (16q – q 2 ) – (4q + 20) = –q
2 + 12q – 20
Profit is a quadratic expression in q, where a = –1, b = 12 and c = –20. To find the firm’s break-even levels of production, i.e. the values of q at which = 0, we should solve the following quadratic equation for q:
–q 2 + 12q – 20 = 0
Simplify by multiplying through by –1:
q 2 – 12q + 20 = 0
(q – 2)(q – 10) = 0 [using the method of factorisation] q = 2 and q = 10 are the break-even levels of output at which profit, = 0. Alternatively, using the formula method with a =1, b = –12, c = 20:
25
2 8 12 2 64 12 2 80 144 12 1 2 20 1 4 12 12 q 2
q = 2 or q = 10 as above.
q TR = 16q – q 2
AR = 16 – q TC = 4q + 20 = –q 2 + 12q – 20 0 0
20 –20
1 15
15 24
–9 2 28 14 28
0 3 39 13 32
7 4 48 12 36
12 5 55 11 40
15 6 60 10 44
16 7 63 9 48
15 8 64 8 52
12 9 63 7 56
7 10
60 6 60 0 11
55 5 64 –9
The shaded areas represent the break-even levels of production.
q TR, TC, = q 2 +12q 20 10 2 TR=16q q 2 TC=4q+20 26
Example
A monopolist is faced with the demand equation, q = 100 – p, where q = quantity demanded and p = price.
The total cost equation is: TC = q 2 + 36q +120, where q = quantity supplied.
(i)
By manipulating the demand equation, show that the total revenue equation is:
TR = 100q – q 2
(ii) Write down an expression for the firm’s profit ( = TR–TC) in terms of q, and find the firm’s two break-even levels of production, i.e. the values of q at which = 0. Solution
(i)
By definition, TR = price quantity Rearranging the demand equation, q = 100 – p
p = 100 – q
TR = pq = (100 – q) q
TR = 100q – q 2
(ii) By definition,
= TR – TC
= (100q – q 2 ) – (q
2 + 36q + 120)
= –2q
2 + 64q – 120
At break-even levels of production, = 0
–2q 2 + 64q – 120 = 0
q 2 – 32q + 60 = 0 [dividing through by –2]
(q – 2)(q – 30) = 0
Therefore the break-even levels of production are q = 2 and q = 30.
27
q TR, TC, = 2q 2 +64q
120
30 2 TR=100q q 2 TC=q 2
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