Basic Mathematics for Economists - Rosser - Chapter 6 pps

The mathematical problem is therefore to solve the quadratic equation 200= 85q − 2q2 3All three solution methods require like terms to be brought together on one side of the equalitysign

6 Quadratic equations

Learning objectives

After completing this chapter students should be able to:

• Use factorization to solve quadratic equations with one unknown variable

• Use the quadratic equation solution formula

• Identify quadratic equations that cannot be solved

• Set up and solve economic problems that involve quadratic functions

• Construct a spreadsheet to plot quadratic and higher order polynomial functions

6.1 Solving quadratic equations

A quadratic equation is one that can be written in the form

ax2+ bx + c = 0

where x is an unknown variable and a, b and c are constant parameters with a

example,

6x2+ 2.5x + 7 = 0

A quadratic equation that includes terms in both x and x2cannot be rearranged to get a single

term in x, so we cannot use the method used to solve linear equations.

There are three possible methods one might try to use to solve for the unknown in aquadratic equation:

(i) by plotting a graph

(ii) by factorization

(iii) using the quadratic ‘formula’

In the next three sections we shall see how each can be used to tackle the following question

If a monopoly can face the linear demand schedule

at what output will total revenue be 200?

It is not immediately obvious that this question involves a quadratic equation We first need

to use economic analysis to set up the mathematical problem to be solved By definition weknow that total revenue will be

So, substituting the function for p from (1) into (2), we get

TR= (85 − 2q)q = 85q − 2q2

This is a quadratic function that cannot be ‘solved’ as it stands It just tells us the value of

TR for any given output What the question asks is ‘at what value of q will this function be

equal to 200’? The mathematical problem is therefore to solve the quadratic equation

200= 85q − 2q2

(3)All three solution methods require like terms to be brought together on one side of the equalitysign, leaving a zero on the other side It is also necessary to put the terms in the order given

in the above definition of a quadratic equation, i.e

unknown squared (q2), unknown (q), constant

Thus (3) above can be rewritten as

2q2− 85q + 200 = 0

It is this quadratic equation that each of the three methods explained in the following sectionswill be used to solve

Before we run through these methods, however, you should note that an equation involving

terms in x2and a constant, but not x, can usually be solved by a simpler method For example,

a solution whilst others have two solutions Only a rough sketch diagram is necessary for thispurpose.

and so the value of y rises again and must cut the q axis a second time.

These values indicate that the graph is a U-shape, as shown inFigure 6.1 This cuts the

horizontal axis twice and so there are two values of q for which y is zero, which means that

there are two solutions to the question The precise values of these solutions, 2.5 and 40, can

be found by the other two methods explained in the following sections or by computation of

y for different values of q (See spreadsheet solution method below.)

If we slightly change the problem in Example 6.1 we can see why there may not always

be a solution to a quadratic equation

Example 6.2

Find out if there is an output level at which total revenue is 1,500 for the function

TR= 85q − 2q2

and calculate a few values, we can see that it falls and then rises again but never cuts the

q axis, asFigure 6.2shows

on the graphics package that you use.

Plotting quadratic functions with Excel

An Excel spreadsheet for calculating different values of the function y in Example 6.1 above

can be constructed by following the instructions inTable 6.1 Rather than building in formulaethat are specific to this example, this spreadsheet is constructed in a format that can be used

to plot any function in the form y = aq2+ bq + c once the parameters a, b and c are entered

in the relevant cells The range for q has been chosen to ensure that it includes the values when y is zero, which is what we are interested in finding.

If you construct this spreadsheet you should get the series of values shown inTable 6.2

The q values which correspond to a y value of zero can now be read off, giving the solutions

2.5 and 40

You may also use the Excel spreadsheet you have created to plot a graph of the function

y = 2q2− 85q + 200 Assuming that you have q and y in single columns, then you just use the Chart Wizard command to obtain a plot with q measured on the X axis and y as variable

Aon the vertical axis (It you don’t know how to use this chart command, refer back toExample 4.17.) To make the chart clearer to read, enlarge it a bit by dragging the corner The

legend box for y can also be cut out to allow the chart area to be enlarged This should give

you a plot similar toFigure 6.3, which clearly shows how this function cuts the horizontalaxis twice

Table 6.1

A1 Ex.6.2 Label to remind you what example this is

A6 =A5+0.5 Calculates a 0.5 unit increment in q

function corresponding to the value of q in

cell A5 and the parameter values in cellsC2, E2 and G2

Note that the $ sign is used so that these cell references do not change when this

function is copied down the y column

This spreadsheet can easily be amended to calculate values and plot graphs of other

quadratic functions by entering different values for the parameters a, b and c in cells C2,

E2 and G2 For example, to calculate values for the function from Example 6.2

y = 2q2− 85q + 1,500

the value in cell G2 should be changed to 1,500 A computer plot of this function shouldproduce the shape shown inFigure 6.2above, confirming again that this function will not cutthe horizontal axis and that there is no solution to the quadratic equation

If a quadratic function which has been rearranged to equal zero can be factorized in this way

then one or the other of the two factors must equal zero (Remember that if A × B = 0 then either A or B, or both, must be zero.)

Example 6.3

Solve by factorization the quadratic equation

2q2− 85q + 200 = 0

As expected, these are the same solutions as those found by the graphical method.

It may be the case that mathematically a quadratic equation has one or more solutions with

a negative value that will not apply in an economic problem One cannot have a negativeoutput, for example

If x represented output, then x= 5 would be the only answer we would use

If a quadratic equation cannot be factorized then the formula method in Section 6.5 belowmust be used The formula method can also be used, however, when an equation can befactorized Therefore, if you cannot quickly see a way of factorizing then you should use theformula method Factorization is only useful as a short-cut way of solving certain quadraticequations It defeats the object of the exercise if you spend half an hour trying to find a way

of factorizing an expression when it would be quicker to use the formula

It should also go without saying that quadratic equations for which no solutions existcannot be factorized For example, it is not possible to factorize the equation

2q2− 85q + 1,500 = 0

which we have already shown to have no solution

Test Yourself, Exercise 6.1

1 Solve for x in the equation x2− 5x + 6 = 0.

2 Find the output at which total revenue is £600 if a firm’s demand schedule is

p = 70 − q

3 A firm faces the average cost function

AC= 40x−1+ 10x

where x is output When will average cost be 40?

4 Is there a positive solution for x when

0= 12x2+ 90x − 48?

5 A firm faces the total cost schedule

TC= 6 − 2q + 2q2

when q > 2 At what output level will TC= £150?

6.4 The quadratic formula

Any quadratic equation expressed in the form

ax2+ bx + c = 0

where a, b and c are given parameters and for which a solution exists can be solved for x by

using the quadratic formula

x =−b ±

√

b2− 4ac 2a

(The sign ± means + or −.) There is no need for you to understand how the formula isderived You just need to know that it works

Example 6.5

Use the quadratic formula to solve the quadratic equation

2q2− 85q + 200 = 0

In the quadratic formula applied to this example a = 2, b = −85 and c = 200 (and, of course, x = q) Note that the minus signs for any negative coefficients must be included.

One also needs to take special care to remember to use the rules for arithmetic

opera-tions using negative numbers Substituting these values for a, b and c into the formula

Test Yourself, Exercise 6.2

(Use the quadratic formula to try to solve these problems.)

1 Solve for x if 0 = x2+ 2.5x − 125.

2 A firm faces the demand schedule q = 400 − 2p − p2 What price does it need

to charge to sell 100 units?

3 If a firm’s demand function is p = 100 − q, what quantities need to be sold to

bring in a total revenue of

( a) £100 ( b) £1,000 ( c) £10,000?

(Give answers to 2 decimal places, where they exist.)

4 Make up your own quadratic equation and then find whether a solution exists

6.5Quadratic simultaneous equations

If one or more equations in a simultaneous equation system are quadratic then it may bepossible to eliminate all but one unknown and to reduce the problem to a single quadraticequation If this can be solved then the other unknowns can be found by substitution

We can ignore the second solution as negative quantities cannot exist Thus the equilibriumquantity is 5.

Substituting this value into the supply function gives equilibrium price

p= 30 + 2 × 5 = 40

You should now be able to link the different mathematical techniques you have learned so far

to tackle more complex problems If you have covered the theory of perfect competition inyour economics course, then you should be able to follow the analysis in the example below

p = 70 − 0.08Q

where Q is industry output (and q is an individual firm’s output).

(i) What will be the short-run price, industry output and profit for each firm?

(ii) What will happen to price, industry output and the number of firms in the long run?(Assume new entrants have the same cost structure.)

= −0.08 ±

√

0.04 0.00024

= −0.08 + 0.2

0.00024 or

−0.08 − 0.2 0.00024

= 0.12

0.00024 or

−0.28 0.00024

= 500 (ignoring the negative answer)

Substituting this value of Q into the demand schedule (3) gives

until each firm is only just breaking even, when price equals the lowest value on the firm’sU-shaped average cost schedule.

How do we find when AC is at its minimum point? The MC and AC functions are given

in the question From cost theory you should know that MC always cuts AC at its minimumpoint Therefore

Therefore p = £16.29 (to the nearest penny)

The old supply schedule does not now apply because of the increased number of firms in theindustry Therefore, substituting this price into the demand schedule (3) to get total outputgives

Test Yourself, Exercise 6.3

3 A monopoly faces the marginal cost function MC= 0.5q

and the marginal revenue function MR= 200 − 4q

What output will maximize profits?

4 A price-discriminating monopoly sells in two markets whose demand schedulesare

5 A firm’s marginal cost schedule is MC= 2.3 + 0.00012q2and it sells its output

in two separate markets with demand schedules

where n is any non-negative integer Linear equations contain polynomials where n = 1

Quadratic equations contain polynomials where n= 2

When n is greater than 2 the solution of a polynomial equation by algebraic means becomes

complex and time-consuming For practical purposes, however, you may use a spreadsheet

to find a solution by the iterative method This means calculating values of a function fordifferent values of the unknown variable until a solution or a good approximation to it isfound As a spreadsheet can quickly perform the necessary calculations, it is an ideal tool forthe calculation of polynomial solutions

The format of the spreadsheet will depend on the problem tackled Below are someexamples of how problems can be approached

A spreadsheet needs to be constructed that will calculate TC for different values of q You

can then experiment with different ranges of q until the solution is found

The method for constructing the spreadsheet is basically the same as that used for quadraticequations as set out inTable 6.1earlier This time the spreadsheet calculates the cubic TCfunction that corresponds to the parameters entered The instructions for doing this are setout in Table 6.3

Although AC and MC may initially fall as a firm’s output increases, its TC function shouldnever fall It would therefore be useful to have a check that this cubic TC function always

increases as q increases and MC is never negative To do this the spreadsheet also includes

a third column where values of MC are calculated (See Section 8.4 for further analysis ofcubic functions with this property.)

Table 6.3

A1 Ex.6.9 Label to remind you what example this is

SOLUTION TO

B3 TC =a + bq + cq^2 + dq^3

Title of spreadsheet (Note that this is not an actual Excelformula.)

Right justify these cells

These are the actual parameter values for

a,b,c and d, respectively, for this example.

Column heading labels

A5 =A4+1 Calculates a one unit increment in q

F$7*A4^3

Formula to calculate value of TC corresponding to value of q in cell A4 and

parameter values in cells F4, F5, F6 and F7

Note the $ sign used to anchor row references for when this formula is copieddown row B

B5 to

B45

Copy formula from cell B4

down column B

Calculates values for TC in each row

corresponding to values of q in column A.

C5 =B5-B4 Calculates values MC as the change in TC

from a one unit increment in q

C6 to

C45

Copy formula from cell C5

down column C

Calculates MC of a unit of q corresponding

to increment in TC shown in column B

B4 to

C45

Highlight these columns and

format to 2 decimal places

TC and MC are both monetary valuesmeasured in £ so use numerical format 0.00

Your spreadsheet should now look like Table 6.4, which shows that when q is 40, TC will

be 42,920 Thus, if output is constrained to whole units, 40 is the maximum output that the

firm’s management can produce for a budget of £43,000.

This spreadsheet also confirms that MC declines in value then increases, but is nevernegative This is what we would expect Save your spreadsheet for use with other examples

This example was constructed for a range of values of q that contained the answer we

were seeking If you had no idea where the solution to this cubic polynomial lay then youcould get a ‘ball park’ estimate by producing a range of values in jumps of 10 in the column

headed q by entering the formula= A4 + 10 in cell A5 and then copying it down the column

for a few dozen rows This would tell you that when q = 31, TC = £19,254.05, and when

q = 41, TC = £46,383.05 Therefore, TC = £43,000 must lie somewhere between these values of q Once you have a rough idea of where the solution value for q will lie, you can change the q column so that values increase in only one unit increments, or smaller units if

necessary, until the actual solution is pinpointed

To solve other cubic polynomials, one simply enters the corresponding parameters into thespreadsheet set up for Example 6.9 above and adjusts the range of the independent variable

(q) until the solution is found.