BEE1020 { Basic Mathematical Economics Dieter Balkenborg
Week 12, Lecture Thursday 23.01.03
Department of Economics
Constrained optimization
University of Exeter
1
Objective
We give the \¯rst order conditions" for constrained optimization problems like utility
maximization and costs minimization. The Lagrangean method to obtain these conditions
is introduced and its economic interpretation is discussed.
The relevant reading for this handout is Chapter 7, Sections 4. See also the book by
Begg on the consumer optimum and on cost minimization.
2
2.1
Constrained optimization
Utility maximization
The preferences of a consumer are described by a family of indi®erence curves. A mathematically convenient way to describe a family of indi®erence curves is to describe them
as the level curves of a utility function u (x; y). The utility u (x; y) can be interpreted as
a numerical measure of the satisfaction the consumer has when consuming x units of the
¯rst commodity (say, apples) and y units of the second commodity (say oranges). Along
a level curve utility is constant, so all commodity bundles on a level curve bring equal
satisfaction, they are hence the indi®erence curves. If one bundle has more utility than
another, the consumer will prefer it.
As an example consider the utility function u (x; y) = xy. The indi®erence curves for
u = 1, u = 2 and u = 3 have the form
5
4
3
2
1
0
1
2
x 3
4
5
where consumption on higher indi®erence curves yields higher utility.
Suppose the price of apples is p, the price of oranges is q and the consumer has a
budget b. Then the budget constraint
px + qy · b
must be satis¯ed for the consumption bundle (x; y) which the consumer buys, i.e., his
total expenditure on the two commodities cannot exceed his budget.
The aim of the consumer is to maximize his utility subject to the budget constraint.
This is a constrained optimization problem because the consumer would consume in¯nite
amounts of both commodities if he were not constrained by his budget. u (x; y) is the
objective function for this problem because the objective is to maximize this function.
We call total expenditure g (x; y) = px + qy the constraining function because it appears
in the constraint. We expect the budget constraint to be binding in optimum, i.e., we
expect the budget inquality to hold with equality in the optimum. This is so because by
assumption the consumer has no possibility to save and because no other commodities
are available.
From the Principles lecture you know the essential condition which must hold in an
optimum: The budget line must be tangential to the indi®erence curve. Now the budget
line is given by the equation
px + qy = b
or
y=
b p
¡ x
q q
so it has the slope ¡ pq , which, apart form the sign, is the relative price of apples in terms
of oranges. (To buy an apple more the consumer must give up pq units of oranges.)
The slope of the indi®erence curve is { apart from the sign { the marginal rate of
substitution. In the previous handout we tried to explain that the marginal rate of
substitution is @u=@x
. Thus in optimum the marginal rate of substitution must be equal
@u=@y
to the relative price
@u=@x
p
=
@u=@y
q
(1)
This is the ¯rst equation which characterizes the constrained optimum. For the example
u (x; y) = xy we have @u
= y and @u
= x. The condition becomes
@x
@y
p
y
=
x
q
(2)
qy = px;
(3)
or
i.e., in optimum the consumer spends equal amounts on each commodity. We need a
second equation for the optimum. This second equation is the budget equation
px + qy = b
(4)
To repeat: the constrained optimum (x¤ ; y ¤ ) is the solution of a simultaneous system of
two equations in two unknowns. The ¯rst equation expresses that in the optimum the
marginal rate of substitution is equal to the relative price. The second one states that
the consumer spends all his money in the optimum.
2
In our example condition (1) gives y = pq x. Substituting this into the budget equation
gives
µ ¶
p
px + q
x
= px + px = 2px = b
q
b
x¤ =
2p
b
p ¤ p b
y¤ =
x =
=
q
q 2p
2q
We have found the optimum. For instance, when the budget is b = 100 and the prices are
p = 2 and q = 5 the optimal consumption is
x¤ =
2.2
100
= 25
2£2
y=
100
= 10:
2£5
Cost minimization
The cost minimization problem is in many ways dual to the utility maximization problem
and leads to very similar conditions. Consider a price-taking ¯rm with production function
Q (K; L) = KL. Suppose the ¯rm wants to ¯nd the least costly way to produce Q0 units
of output given the prices r and w for the inputs. Then she wants to minimize total costs
rK + wL
subject to the constraint that she produces at least Q0 units
Q0 · Q (K; L) :
Here expenditure on inputs is the objective function whereas the production function is
the constraining function.
Again from the principles lecture we know that in the cost minimum the iso-cost line
must be tangential to the isoquant. Since the slope of an iso-cost line
rK + wL = constant
is the negative of the relative price of inputs wr it follows that in the optimum the marginal
must equal the relative price
rate of substitution @Q=@K
@Q=@L
r
@Q=@K
= :
@Q=@L
w
In our example
@Q
@K
= L and
@Q
@L
= K, so
L
r
=
K
w
or
wL = rK
3
so in optimum equal amounts are spend on both inputs. This is the ¯rst equation. As
the second equation we have that in the optimum the ¯rm will produce exactly Q0 units
and no more, the cost-minimizing optimum (K ¤ ; L¤ ) must lie on the isoquant for Q0
Q0 = Q (K; L) :
In our example suppose that r = 2, that w = 5 and that the ¯rm wants to produce
250 units of output at the lowest costs. Then the optimum (K ¤ ; L¤ ) must satisfy the
conditions
L
2
=
K
5
or
2
L= K
5
and
250 = KL:
Substituting in the last equation L = 25 K yields
2
250 = 2 £ 125 = 2 £ 53 = K 2
5
2 ¤
¤
L =
K = 10:
5
54 = K 2
K ¤ = 52 = 25
The optimal input combination is (K ¤ ; L¤ ) = (5; 10).
2.3
The general problem
In general we want to maximize or to minimize an objective function
z = f (x; y)
susbject to a constraint
g (x; y) · c
where c is a constant. We are interested in the case where the constraint is binding in
the optimum, i.e., where g (x; y) = c holds at the optimum. (Otherwise the constraint is
always satis¯ed near the optimum and the optimum must be a critical point of f.)
The optimum must then solve the two conditions
@g=@x
@f =@x
=
@f =@y
@g=@y
g (x; y) = c
4
3
The Lagrangian approach
The Lagrangian method is an alternative way to derive these two conditions. The method
transform the constrained optimization problem into an unconstrained optimization problem in conjunction with a pricing problem. We discuss here only the maximization problem. First we form out of the two given functions f and g a new function which depends
on an additional parameter ¸, the so-called Lagrange multiplier. This new function {
called the Lagrangian { is
L (x; y) = f (x; y) ¡ ¸ (g (x; y) ¡ c)
Then we look for an unconstrained maximum or a minimum of this function and determine
the Lagrange multiplier such that the constraint holds with equality in the optimum.
For a maximization problem the interpretation is as follows: Instead of strictly forbidding that the constraint g (x; y) · c ever gets violated, we allow to violate it, but at
a price. This price is ¸: It is a ¯ctituous price in a ¯ctitious problem and hence called
a \shadow price". Suppose the constraint is violated. g (x; y) ¡ c is then the number of
units by which the constraint is violated. When each unit of violation costs ¸, the total
amount ¸ (g (x; y) ¡ c) must then be paid for violation. This amount is subtracted from
the value of the objective function which we want to maximize. (If the constraint is not
violated, there is a reward.)
Now we look for an unconstrained maximum (x¤ ; y ¤ ) of the Lagrangian. Suppose we
have found an absolute maximum (x¤ ; y ¤ ) of this function and suppose and we were able to
choose a positive ¸ such that (x¤ ; y ¤ ) satis¯es the constraint g (x¤ ; y ¤ ) = c. Then (x¤ ; y ¤ )
is automatically an optimum of the constrained optimization problem: We have
L (x¤ ; y ¤ ) = f (x¤ ; y ¤ ) ¡ ¸ (g (x¤ ; y ¤ ) ¡ c) = f (x¤ ; y ¤ ) .
Now consider any pair (x; y) that does not violate the constraint, i.e., that satis¯es
g (x; y) · c. Then L (x; y) · L (x¤ ; y ¤ ) because (x¤ ; y ¤ ) is an absolute maximum. Moreover ¡¸ (g (x; y) ¡ c) is positive since (x; y) satis¯es the constraint. Hence f (x; y) ·
L (x; y) and overall we obtain that f (x; y) · f (x¤ ; y ¤ ) holds for all (x; y) satisfying the
constraint. This means that (x¤ ; y ¤ ) is a constrained maximum.
To ¯nd the maximum of the Lagrangian we solve the ¯rst order conditions
@L
@f
@g
=
¡¸
=0
@x
@x
@x
@L
@f
@g
=
¡¸
=0
@y
@y
@y
or
or
@f
@g
=¸
@x
@x
@f
@g
=¸
@y
@y
(5)
(6)
Notice that if we divide the two equations on the right we get
@g=@x
@f =@x
=
:
@f =@y
@g=@y
(7)
In addition we want the constraint to hold with equality in the optimum, i.e., we impose
the equation
g (x; y) = c:
5
(8)
We then solve the two equations (7) and (8) for x and y. The Lagrangian was just a tool
to derive these equations. If the need arises, we can calculate the value of the Lagrange
multiplier ¸ from equation (5) or (6).
3.1
Utility maximization again
In this case the Lagrangian is
L (x; y) = u (x; y) ¡ ¸ (px + qy ¡ b) = u (x; y) + ¸ (b ¡ px + py)
It can be interpreted as follows: In the original problem saving is not possible. Now
suppose it is possible to save and also to go into debt. The savings are
s = b ¡ px + py
Now suppose there is some future utility from saving because the money can be spend in
the future. For simplicity we assume that this future utility is linear in savings and given
by
¸s
Then ¸ is the marginal utility of saving a penny. One also speaks of the marginal utility
of money. The Lagrangian can then be interpreted as the total utility from consumption
today and in the future. When we solve the constrained optimization problem with the
Lagrangian approach we e®ectively ask: What would the marginal utility of saving have
to be such that it is optimal not to save and not to go into debt.
3.2
Cost minimization again
The ¯rm wants to minimize the costs rK + wL subject to the constraint Q0 · Q (K; L).
First we transorm the problem such that it ¯ts with our general framework above. Minimizing rK + wL is the same as maximizing ¡rK ¡ wL. Now we have a maximization
problem. Next we had the constraint in the form g (x; y) · c with the constant on
the right. Therefore we rewrite Q0 · Q (K; L) as ¡Q (K; L) · ¡Q0 . The Lagrangian
becomes:
L (K; L) = ¡rK ¡ wL ¡ ¸ (¡Q (K; L) + Q0 )
= (¸Q (K; L) ¡ rK ¡ wL) ¡ ¸Q0
We see that { up to a constant which will drop out in the di®erentiation { the Lagrangian
is just the pro¯t function of the ¯rm when the output price is ¸. When we solve the cost
minimization problem with the Lagrangian method we therefore ask: What would the
output price have to be such that it is optimal to produce exactly Q0 units of output.
6
4
Objective
Two types of optimization problems frequently occur in economics: unconstrained optimization problems like pro¯t maximization problems and constrained optimization problems like the utility maximization problem of consumer constrained by a budget. In
this handout we discuss the main step to solve an unconstrained optimization problem namely to ¯nd the two \¯rst order conditions" and how to solve them. The solutions for
this system of two equations in two unknowns can be local maxima, minima or saddle
points. How to distinguish between these types will be discussed in a later handout, after
we have discussed constrained optimization problems.
The relevant reading for the previous handout is Chapter 7, Sections 1 and 2 in the
textbook of Ho®mann and Bradley. The relevant reading for this handout is Chapter 7,
Sections 3.
5
5.1
Optimization problems
Unconstrained optimization
Suppose we want to ¯nd the absolute maximum of a function
z = f (x; y)
in two variables, i.e., we want to ¯nd the pair of numbers (x¤ ; y ¤ ) such that
f (x¤ ; y ¤ ) ¸ f (x; y)
holds for all (x; y). If (x¤ ; y ¤ ) is such a maximum the following must hold: If we freeze
the variable y at the optimal value y ¤ and vary only the variable x then the function
f (x; y ¤ )
{ which is now just a function in just one variable { has a maximum at x¤ . Typically the
maximum of this function in one variable will be a critical point. Hence we expect the
@z
partial derivative @x
to be zero at the optimum. Similarly we expect the partial derivative
@z
to
be
zero
at
the
optimum.
We are thus led to the ¯rst order conditions
@y
@z
= 0
@x jx=x¤ ;y=y¤
@z
= 0
@y jx=x¤ ;y=y¤
which must typically be satis¯ed. We speak of ¯rst order conditions because only the
¯rst derivatives are involved. The conditions do not tell us whether we have a maximum,
a minimum or a saddle point. To ¯nd out the latter we will have to look at the second
derivatives and the so-called \second order conditions", as will be discussed in a later
lecture.
7
Example 1 Consider a price-taking ¯rm with the production function Q (K; L). Let r
be the interest rate (the price of capital K), let w be the wage rate (the price of labour L)
and let P be the price of the output the ¯rm is producing. When the ¯rm uses K units of
capital and L units labour she can produce Q (K; L) units of the output and hence make
a revenue of
P £ Q (K; L)
by selling each unit of output at the market price P . Her production costs will then be
her total expenditure on the inputs capital and labour
rK + wL:
Her pro¯t will be the di®erence
¦ (K; L) = P Q (K; L) ¡ rK ¡ rL:
The ¯rm tries to ¯nd the input combination (K ¤ ; L¤ ) which maximizes her pro¯t. To ¯nd
it we want to solve the ¯rst order conditions
@¦
@Q
= P
¡r =0
@K
@K
@Q
@¦
= P
¡w = 0
@L
@L
(9)
(10)
@Q
@Q
These conditions are intuitive: Suppose for instance P @K
¡ r > 0: @K
is the marginal
product of capital, i.e., it tells us how much more can be produced by using one more
@Q
is the marginal increase in revenue when one more unit of capital is
unit of capital. P @K
used and the additional output is sold on the market. r is the marginal cost of using one
@Q
more unit of capital. P @K
¡ r > 0 means that the marginal pro¯t from using one more
unit of capital is positive, i.t., it increases pro¯ts to produce more by using more capital
@Q
and therefore we cannot be in a pro¯t optimum. Correspondingly, P @K
¡ r < 0 would
mean that pro¯t can be increased by producing less and using less capital. So (9) should
hold in the optimum. Similarly (10) should hold.
It is useful to rewrite the two ¯rst order conditions as
r
@Q
=
@K
P
w
@Q
=
@L
P
(11)
(12)
Thus, for both inputs it must be the case that the marginal product is equal to the price
ratio of input- to output price. When we divide here the two left-hand sides and equate
them with the quotient of the right hand sides we obtain as a consequence
Á
@Q @Q
r.w
r
=
=
(13)
@K
@L
P P
w
so the marginal rate of substitution must be equal to the price ratio of the input price
(which is the relative price of capital in terms of labour).
8
Example 2 Let us be more speci¯c and assume that the production function is
1
1
Q (K; L) = K 6 L 2
and that the output- and input prices are P = 12, r = 1 and w = 3. Then
1 ¡5 1
1 1 1
@Q
@Q
=
K 6 L2
= K 6 L¡ 2
6
2
Á @K
Á @L
5
1
1
1
1 ¡5 1 1 1 ¡1
@Q @Q
1
1L
=
K 6 L2
K 6 L 2 = K¡ 6 L 2 K¡ 6 L 2 =
@K
@L
6
2
3
3K
The ¯rst order conditions (11) and (12) become
1 ¡5 1
1
K 6L2 =
6
12
1 1 ¡1
3
K6L 2 =
:
2
12
(14)
(15)
This is a simultaneous system of two equations in two unknowns. The implied condition
(13) becomes
1L
1
=
3K
3
which simpli¯es to
L=K
Example 3 A monopolist with total cost function T C (Q) = Q2 sells his product in two
di®erent countries. When he sells QA units of the good in country A he will obtain the
price
PA = 22 ¡ 3QA
for each unit. When he sells QB units of the good in country B he obtains the price
PB = 34 ¡ 4QB :
How much should the monopolist sell in the two countries in order to maximize pro¯ts?
To solve this problem we calculate ¯rst the pro¯t function as the sum of the revenues
in each country minus the production costs. Total revenue in country A is
T RA = PA QA = (22 ¡ 3QA ) QA
Total revenue in country B is
T RB = PB QB = (34 ¡ 4QB ) QB
Total production costs are
T C = (QA + QB )2
9
Therefore the pro¯t is
¦ (QA ; QB ) = (22 ¡ 3QA ) QA + (34 ¡ 4QB ) QB ¡ (QA + QB )2 ;
a quadratic function in QA and QB . In order to ¯nd the pro¯t maximum we must ¯nd the
critical points, i.e., we must solve the ¯rst order conditions
@¦
= ¡3QA + (22 ¡ 3QA ) ¡ 2 (QA + QB ) = 22 ¡ 8QA ¡ 2QB = 0
@QA
@¦
= ¡4QB + (34 ¡ 4QB ) ¡ 2 (QA + QB ) = 34 ¡ 2QA ¡ 10QB = 0
@QB
or
8QA + 2QB = 22
2QA + 10QB = 34:
(16)
Thus we have to solve a linear simultaneous system of two equations in two unknowns.
6
6.1
Simultaneous systems of equations
Linear systems
We illustrate four methods to solve a linear system of equations using the example
5x + 7y = 50
4x ¡ 6y = ¡18
6.1.1
(17)
Using the slope-intercept form
We rewrite both linear equations in slope-intercept form
7y = 50 ¡ 5x
50 5
¡ x
y =
7
7
4x ¡ 6y = ¡18
4x + 18 = 6y
2
y =
x+3
3
8
7
6
5
4
3
2
-2
0
2
10
x 4
6
The solutions to each equation form a line, which we have now described as the graph of
linear functions. At the intersection point of the two lines the two linear functions must
have the same y-value. Hence
50 5
¡ x
7
7
50
¡3
7
150 ¡ 63
87
2
= y = x+3
3
2
5
=
x+
j£3£7
3
7
= 14x + 15x
= 29x
87
x =
=3
29
We have found the value of x in a solution. To calculate the value of y we use one of the
linear functions in slope-intercept form
2
£3=2
3
The solution to the system of equations is x¤ = 3; y ¤ = 2
It is strongly recommended to check the result for the original system of equations.
y=
5 £ 3 + 7 £ 5 = 15 + 35 = 50
4 £ 3 ¡ 6 £ 5 = 12 ¡ 30 = ¡18
6.1.2
The method of substitution
First we solve one of the equations in (17), say the second, for one of the variables, say y:
4x ¡ 6y = ¡18
4x + 18 = 6y
4
2
y =
x+3 = x+3
6
3
(18)
Then we replace y in the other equation by the result. (Do not forget to place brackets
around the expression.) We obtain an equation in only one variable, x, which we solve
for x.
5x + 7y
¶
2
5x + 7
x+3
3
14
5x + x + 21
3
14
15
x + x + 21
3
3
29
x
3
µ
= 50
= 50
= 50
= 50
= 50 ¡ 21 = 29
x =
11
3
£ 29 = 3
29
We have found x and can now use (18) to ¯nd y:
2
2
y = x+3= £3+3 =5
3
3
Hence the solution is x = 3; y = 5.
6.1.3
The method of elimination
To eliminate x we proceed in two stages: First we multiply the ¯rst equation by the coef¯cient of x in the second equation and we multiply the second equation by the coe±cient
of x in the ¯rst equation
5x + 7y
4x ¡ 6y
20x + 28y
20x ¡ 30y
=
=
=
=
j£4
j£5
50
¡18
200
¡90
Then we subtract the second equation from the ¯rst equation
20x + 28y = 200
20x ¡ 30y = ¡90
j¡
0 + 28y ¡ (¡30y) = 200 ¡ (¡90)
58y = 290
y = 290
=5
58
Having found y we use one of the original equations to ¯nd x
5x + 7y
5x + 35
5x
x
=
=
=
=
50
50
15
3
Instead of ¯rst eliminating x we could have ¯rst eliminated y:
5x + 7y
4x ¡ 6y
30x + 42y
28x ¡ 42y
58x
x
6.1.4
=
=
=
=
=
=
50
j£6
¡18
j£7
300
¡126
j+
174
3
Cramer's rule
This is a little preview on linear algebra. In linear algebra the system of equations is
written as
·
¸· ¸ ·
¸
5
7
x
50
=
4 ¡6
y
¡18
12
·
x1
x2
¸
50
¡18
¸
are so-called columns vectors, simply two numbers written
·
¸
5
7
below each other and surrounded by square brackets.
is the 2 £ 2-matrix
4 ¡6
of coe±cients. A 2 £ 2-matrix is a system of four numbers arranged in a square and
surrounded by square brackets.
With each 2 £ 2-matrix
·
¸
a b
A=
c d
where
and
·
we associate a new number called the determinant
¯
¯
¯ a b ¯
¯ = ad ¡ cb
det A = ¯¯
c d ¯
Notice that the determinant is indicated by vertical lines in contrast to square brackets.
For instance,
¯
¯
¯ 5
¯
7
¯
¯
¯ 4 ¡6 ¯ = 5 £ (¡6) ¡ 4 £ 7 = ¡30 ¡ 28 = 58
Cramer's rule uses determinants to give an explicit formula for the solution of a linear
simultaneous system of equations of the form
ax + by = e
cx + dy = f
or
·
a b
b d
¸·
x
y
¸
=
·
e
f
¸
:
Namely,
¯
¯
¯
¯
x¤ = ¯¯
¯
¯
e
f
a
c
b
d
b
d
¯
¯
¯
¯
ed ¡ bf
¯ =
¯
ad ¡ bc
¯
¯
¯
¯
¯
¯
y ¤ = ¯¯
¯
¯
a
c
a
c
e
f
b
d
¯
¯
¯
¯
ae ¡ ce
¯ =
¯
ad ¡ bc
¯
¯
Thus x¤ and y ¤ are calculated as the quotients of two determinants. In both cases we
divide by the determinant of the matric of the coe±cients. For x¤ the determinant in
the numerator is the determinant of the matrix obtained by replacing in the matrix of
coe±cients the coe±cients a and c of x by the constant terms e and f, i.e., the ¯rst row
is replaced by the row vector with the constant terms. Similarly, for y ¤ the determinant
in the numerator is the determinant of the matrix obtained by replacing in the matrix of
coe±cients the coe±cients b and d of y by the constant terms e and f , i.e., the second
row is replaced by the row vector with the constant terms.
The method works only if the determinant in the denominator is not zero.
13
In our example,
¯
¯
¯ 50
¯
7
¯
¯
¯ ¡18 ¡6 ¯
50 £ (¡6) ¡ (¡18) £ 7
¡300 + 126
¡174
¤
¯
¯ =
x =
=
=
=3
¯ 5
¯
5
£
(¡6)
¡
4
£
7
¡30
¡
28
¡58
7
¯
¯
¯ 4 ¡6 ¯
¯
¯
¯ 5
¯
50
¯
¯
¯ 4 ¡18 ¯
¡90 ¡ 200
¡290
5 £ (¡18) ¡ 4 £ 50
¤
¯ =
y = ¯¯
=
=
=5
¯
5 £ (¡6) ¡ 4 £ 7
¡58
¡58
7 ¯
¯ 5
¯ 4 ¡6 ¯
Exercise 1 Use the above methods to ¯nd the optimum in Example 16
6.1.5
Existence and uniqueness.
A linear simultaneous system of equations
ax + by = e
cx + dy = f
can have zero, exactly one or in¯nitely many solution. To see that only these cases can
arise, bring both equations into slope-intercept form1
e a
¡ x
y =
f
b
f
c
y =
¡ x
d d
If the slopes of these two linear functions are the same, they describe identical or parallel
lines. The slopes are identical when ab = dc , i.e., when the determinant of the matrix of
coe±cients ad ¡ cb is zero. If in addition the intercepts are equal, both equations describe
the same line. In this case all points (x; y) on the line are solutions. If the intercepts are
di®erent the two equations describe parallel lines. These do not intersect and hence there
is no solution.
Example 4
x + 2y = 3
2x + 4y = 4
has no solution: The two lines
3 1
¡ x
2 2
1
y = 1¡ x
2
y =
1
We assume here for simplicity that the coe±cients b and d are not zero. The arguments can be easily
extended to these cases, except when all four coe±cients are zero. The latter case is trivial - either there
is no solution or all pairs of numbers (x; y) are solutions.
14
are parallel
2
-2
-1
0
1
2
x
3
4
5
-1
There is no common solution. Trying to ¯nd one yields a contradiction
3 1
1
¡ x = y =1¡ x
2 2
2
3
= 1
2
6.2
1
j+ x
2
One equation non-linear, one linear
Consider, for instance,
y2 + x ¡ 1 = 0
1
y+ x = 1
2
In this case we solve the linear equation for one of the variables and substitute the result
into the non-linear equation. As a result we obtain one non-linear equation in a single
unknown. This makes the problem simpler, but admittedly some luck is needed to solve
the non-linear equation in one variable. Solving the linear equation for x can sometimes
give a simpler non-linear equation than solving for y and vice versa. One may have to try
both possibilities. In our example it is convenient to solve for x:
1
x = 1¡y
2
x = 2 ¡ 2y
2
y + (2 ¡ 2y) ¡ 1 = y 2 ¡ 2y + 1 = (y ¡ 1)2 = 0
So the unique solution is y ¤ = 1 and x¤ = 2 ¡ 2 £ 1 = 0.
6.3
Two non-linear equations
There is no general method. One needs to memorize some tricks which work in special
cases.
15
In the example of pro¯t-maximization with a Cobb-Douglas production function we
were led to the system
1 ¡5 1
1
K 6L2 =
6
12
1 1 ¡1
3
K6L 2 =
:
2
12
This is a simultaneous system of two equations in two unknowns which looks hard to
solve. However, as we have seen, division of the two equations shows that L = K must
hold in a critical point.
We can substitute this into, say, the ¯rst equation and obtain
5
1
5 3
2
1
1
1 ¡5 1
1
1
1
1
=
K 6 L 2 = K¡ 6 K 2 = K¡ 6 +6 = K¡ 6 = K¡3
12
6
6
6
6
6
1
1
K¡ 3 =
2
1
1
p
=
3
2
K
p
3
K = 2
K = 23 = 8
Thus the solution to the ¯rst order conditions (and, in fact, the optimum) is K ¤ = L¤ = 8.
Remark 1 This method works with any Cobb-Douglas production function Q (K; L) =
K a Lb where the indices a and b are positive and have a sum less than 1.
Remark 2 The ¯rst order conditions (14) and (15) form a \hidden" linear system of
equations. Namely, if you take the logarithms (introduced later!) you get the system
1 5
¡ ln K +
6 6
1 1
ln + ln K ¡
2 5
ln
1
1
ln L = ln
2
12
1
3
ln L = ln
2
12
which is linear in ln K and ln L. One can solve this system for ln K and ln L, which then
gives us the values for L and K.
16