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Occasionally. scientists prefer to study a given function y = f(x) by comparing

In; and Inx. that is; by graphing f on log-log graph paper. See, for example, our

discussion of constant elasticitv demand functions in (7). In this case, they are

working with the change of v&bles

Y = In? and X = Inx.

Since X = In x, x = 6 and dx/dX = $ = x. In XY-coordinates, f becomes

Y = In f(x) = In f(3) = F(x).

102 EXPONENTS AND LOGARITHMS I51

Now, the slope of the graph of Y = F(AвЂ™). that is, of the graph of f in log-log

coordinates, is given by

Wx) _ dWW).-

dx

(by the Chain Rule)

dx dx dx

(22)

= & (In f(x)) g = $g x.

The difference approximation of the last term in (22) is

df(x) L,f.L=af Ax

f(x) f(x) / x:

dx f(x) Ax

the percent change off relative to the percent change of x. This is the quotient we

have been calling the (point) elasticity off with respect to x, especially if f is a

demand function and x represents price or income.

This discussion shows that the slope of the graph off in log-log coordinates

is the (point) elasticity of /:

f вЂ˜(xl x

&=f(X)

In view of this discussion, economists sometimes write this elasticity as

[.5.61 APPLICAT,ONS 103

5.15 The value of a parcel ?f land bought for speculation is increasing according to the

formula V = 2OOOdвЂќ. If the interest rate is IO percent, how long should the parcel

be held to maximize present value?

5.16 Use the logarithmic derivative method to compute the derivative of each of the

following functions: a) J(xвЂ™ + 1)/(x2 + 4), b) (.x2)вЂњ.

5.17 Use the above discussion lo prove that the elasticity of the product of two functions

is the sum of the elasticities.

PART II

Linear

A l g e bra

C H A P T E R 6

Introduction to

Linear Algebra

The analysis of many economic models reduces to the study of systems of qua-

dons. Furthermore, some of the most frequently studied economic models are

linear models. In the next few chapters, will study the simplest possible sys-

WC

tcms of equations- linear rysrcms.

Linear systems have the added advantage that we can often calculate exact

solutions to the equations. By contrast, solutions of nonlinear systems often cannut

be calculated explicitly, and we can only hope to discover indirectly some of the

properties of these solutions. Equally important for linear systems, the precise

relationship between the solution of the linear system and various parameters

determining the system (the uiвЂ™s and h in the equation above) can be easily

described.

Of course, linearity is a simplifying assumption. The real world is nonlinear.

Calculus exploits the managcahility of linear systems to study nonlinear systems.

The fundamental idea of calcuIus is that we can learn much ahout the behavior

of a nonlinear system of equations by studying suitably chosen linear approxi-

mations to the system. For example, the hcst linear approximation 10 the graph

of a nonlincat function at any point on its graph is the tangent line to the graph

at that point. We can lcarn much about the hehwior of a function Neal any point

by examining the slope of the tangent line. Whether the function is increasing

or decreasing can he determined by seeing whcthcr the tangent lint is rising

or falling. The first important cxcrcise in the study of the calculus is to learn

how to calculate this slope-the derivatirv of the function. For a more prosaic

example of the importance of linear approximations, consider that few people

disagree with the proposition that the earth is roughly spherical. and yet in cw-

strutting homes, skyscrapers, and even cities, we assume that the earth is Rat

and obtain some rather impressive results using Euclidean planar gcomctry. Once

again we arc taking advantage of an cffectivs linear approximation to a nonlinear

phenomenon.

Since a primary goal of multivariahle calculus is to provide a mechanism for

approximating complicaled nonlinear systems by simpler lincu ones. it makes

scnsc to hegin hy squccring OUI all the information we can ahwt Iinca systems ˜

the task we take up in the next six chapters.

A final reason for looking at lincar systems first is that some of the mni

frsqucntly studied economic models are linear. We sketch tier such models Ihere.

As we develop our theory of lincx syslems. will often refer back tu thcsc

WC

models and call attention to the insights which linear theory offers. Keferenccs for

further stud!; of thcsc topics can hc found in the notes at the end of Ihe chapter.

6.2 EXAMPLES OF LINEAR MODELS

Example 1: Tax Benefits of Charitable Contributions

of S

A company cilrns before-tax profits 100.000. It has agreed to contrihuir IO

percent of its after-tax profits fo the Red Cross Rclicf Fund. It must pay a stale lax

of 5 percent of its protits (after the Red Cross donation) and a federal tax of 40

pcrccnt of its profits (after the donation and state taxes are paid). How much does

the company pay in state taxes. federal taxes. and Red Cross donation?

[6.2] 109

EXAMPLES OF LINEAR MODELS

Without a model to structure our analysis, this problem is rather difficult be-

cause each of the three payments must take into consideration the other payments.

However, after we write out the (linear) equations which describe the various

dedyctions, we can understand more clearly the relationships between these pay-

ments and then solve in a straightforward manner for the payment amounts.

Let C, S, and F represent the amounts of the charitable contribution, state

tax, and federal tax, respectively. After-tax profits are $lOO,OOO - (S + F); so

C = 0.10 (100,000 - (S + F)). We write this as

c + 0.1s + 0.w = 10,000,

putting all the variables on one side. The statement that the state tax is 5 percent

of the profits net of the donation becomes the equation S = 0.05 (100,000 C),

or

0.05C + S = 5,000.

Federal taxes are 40 percent of the profit after deducting C and S; this relation is

expressed by the equation F = 0.40 [lOO,OOO ˜ (C + S)], or

0.4c + 0.4s + F = 40,000

WC can summarize the payments to be made by the system of linear equations

c + 0. IS + 0.1F = 10,000

= 5,000

0.05c + s (1)

0.4c + 11.4s + F = 40.000.

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