X and the City: Modeling Aspects of Urban Life (10 page)

We shall use equations (3.15) and (3.17) to eliminate
Q
,
L
, and
M
to obtain

where
C
(
α, λ
) is a constant. With two additional assumptions this will enable us to derive a valuable result to be used in
Chapter 17
(The axiomatic city). If the cost of materials is independent of location, and land costs increase with population density
ρ
according to a power law, that is,
l
∝
ρ
^κ
(
κ
> 0), then

p
=
Aρ
^γ
,

where
γ
=
κλ
and
A
is another constant.

Exercise:
Derive equation (3.18)

We conclude this section with a numerical application of equation (3.15) in a related context. It is recommended that the reader consult
Appendix 5
to brush up on the method of
Lagrange multipliers
if necessary.

Suppose that

L
now being the number of labor “units” and
M
the number of capital “units” to produce
Q
units of a product (such as sections of custom-made fencing for a new housing development). If labor costs per unit are
l
= $50, capital costs per unit are
m
= $100, and a total of $500,000 has been budgeted, how should this be allocated between labor and capital in order to maximize production, and what is the maximum number of fence sections that can be produced? (I hope the reader appreciates the “nice” numbers chosen here.)

To solve this problem, note that the total cost is given by

The simplified equation will be the constraint used below. The problem to be solved is therefore to maximize
Q
in equation (3.19) subject to the constraint (3.20). Using the method of Lagrange multipliers we form the function

The critical points are found by setting the respective partial derivatives to zero, thus:

From equations (3.22a) and (3.22b) we eliminate
λ
to find that
M
= 3
L
/2. From (3.22c) it follows that
M
= 2500, and hence that
L
= 3750. Finally, from (3.22a),
λ
= −4(2500)
^−3/4
(3750)
^3/4
≈ −5.4216, so the unique critical point of
F
is (2500, 3750, −5.4216). The maximum value is
Q
(
L
,
M
) = 16(2500)
^1/4
(3750)
^3/4
≈ 54,200 sections of fencing.

Question:
Can we be certain that this is the
maximum
value of
Q
?

Exercise:
Show that had we not simplified the constraint equation (3.20), the value for
λ
would have been
λ
≈ −0.1084 but the maximum value of
Q
would remain the same.

The home has been built; now it’s time to start paying for it. It’s been said that if you think no one cares whether you’re alive or not, try missing a couple of house payments. Anyway, let’s set up the relevant
difference equation
and its solution. It is called a difference equation (as opposed to a
differential
equation) because it describes discrete payments (as opposed to continuous ones).

Suppose that you have borrowed (or currently owe) an amount of money
A
₀
, and the annual interest (assumed constant) is 100
I
% per year (e.g., if
I
= 0.06, the annual interest is 6%), compounded
m
times per year. If you pay off an amount
b
each compounding period (or due date), the governing first-order nonhomogeneous difference equation is readily seen to be

The solution is

Let’s see how to construct this solution using the idea of a fixed point (or equilibrium value). Simply put, a fixed point in this context is one for which
A
_n
+1
=
A
_n
. From equation (3.23) such a fixed point (call it
L
) certainly exists:
L
=
Bm
/
I
. If we now define
a
_n
=
A
_n
−
L
, then it follows from equation (3.24) that

that is

Thus we have reduced the nonhomogeneous difference equation (3.23) to a homogeneous one. Since
a
₁
=
λa
₀
,
a
₂
=
λa
₁
=
λ
²
a
₀
,
a
₃
=
λa
₂
=
λ
³
a
₀
, etc., it is clear that equation (3.25) has a solution of the form
a
_n
=
λ
ⁿ
a
₀
. On reverting to the original variable
A
_n
and substituting for
L
and
λ
the solution (3.24) is recovered immediately.