bee1020—basicmathematicaleconomics dieterbalkenborg week14...
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BEE1020 — Basic Mathematical Economics Dieter BalkenborgWeek 14, Lecture Thursday 26.01.2006 Department of EconomicsOptimization University of Exeter
“Since the fabric of the universe is most perfect, and
is the work of a most perfect creator, nothing whatso-
ever takes place in the universe in which some form of
maximum or minimum does not appear.”
Leonhard Euler, 1744
optimization problems:– unconstrained (profit maximization)– constrained (utility maximization with budget constraint)“first order conditions” for optimumyield simultaneous system of equations (→Review)Lagrangian approach for constrained problems
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Unconstrained optimization
Find absolute maximum of function
z = f (x, y)
i.e., find pair (x∗, y∗) such that
f (x∗, y∗) ≥ f (x, y)
for all (x, y).The following must hold: freeze the variable y at optimal valuey∗, vary only x then
f (x, y∗)
Must have maximum at x∗. Obtain first order conditions
∂z
∂x|x=x∗,y=y∗= 0
∂z
∂y |x=x∗,y=y∗= 0
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-40
-30
-20
-10
0
-4 -22 4
y
-4-2
24
x
3
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Example: Production function Q (K,L).r interest ratew wage rateP price of outputprofit
Π (K,L) = PQ (K,L)− rK − rL.
FOC for profit maximum:
∂Π
∂K= P
∂Q
∂K− r = 0 (1)
∂Π
∂L= P
∂Q
∂L− w = 0 (2)
Intuition: Suppose P ∂Q∂K − r > 0...
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Rewrite as∂Q
∂K=r
P(3)
∂Q
∂L=w
P(4)
Division yields:∂Q
∂K
/∂Q
∂L=r
P
/ wP
=r
w(5)
Marginal rate of substitution must equal ratio input prices.
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Example:
Q (K,L) = K16L
12
P = 12, r = 1 and w = 3. Then
∂Q
∂K=
1
6K−
56L
12
∂Q
∂L=
1
2K
16L−
12
∂Q
∂K
/∂Q
∂L=
1
6K−
56L
12
/1
2K
16L−
12 =
1
3K−
56L
12K−
16L
12 =
1
3
L
K
FOC’s:1
6K−
56L
12 =
1
12K−
56L
12 =
1
2(6)
1
2K
16L−
12 =
3
12K
16L−
12 =
1
2(7)
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FOC’s:1
6K−
56L
12 =
1
12K−
56L
12 =
1
2(8)
1
2K
16L−
12 =
3
12K
16L−
12 =
1
2(9)
Division yields1
3
L
K=
1
3L = K
Hence
K−56L
12 = K−
56K
36 = K−
13 =
1
2or
3
√1
K=
1
2⇔
1
K=
1
8⇔ K = L = 8
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10
20
K
0 2 4 612
16 18 20 22
L
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
14
16
18
20
z
8
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Example: A monopolist with total cost function TC (Q) = Q2
sells his product in two different countries. When he sells QA unitsof 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 Bhe obtains the price
PB = 34− 4QB.
How much should the monopolist sell in the two countries in orderto maximize profits?
Solution: Total revenue in country A:
TRA = PAQA = (22− 3QA)QA
Total revenue in country B:
TRB = PBQB = (34− 4QB)QB9
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Total production costs are:
TC = (QA +QB)2
Profit:
Π (QA, QB) = (22− 3QA)QA+(34− 4QB)QB−(QA +QB)2 ,
FOC:∂Π
∂QA= −3QA + (22− 3QA)− 2 (QA +QB) = 22− 8QA − 2QB = 0
∂Π
∂QB= −4QB + (34− 4QB)− 2 (QA +QB) = 34− 2QA − 10QB = 0
or
8QA + 2QB = 22 (10)
2QA + 10QB = 34.
linear simultaneous system
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Simultaneous systems of equations
Linear systems
example
5x + 7y = 50 (11)
4x− 6y = −18
Using the slope-intercept form
7y = 50− 5x
y =50
7−
5
7x
4x− 6y = −18
4x + 18 = 6y
y =2
3x + 3
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2
3
4
5
6
7
8
-2 0 2 4 6x
12
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At intersection point two linear functions must have same y-value.Hence
50
7−
5
7x = y =
2
3x + 3
50
7− 3 =
2
3x +
5
7x | × 3× 7
150− 63 = 14x + 15x
87 = 29x
x =87
29= 3
found value of x. calculate the value of y:
y =2
3× 3 + 3 = 5
solution: x∗ = 3, y∗ = 5
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strongly recommended to check:
5× 3 + 7× 5 = 15 + 35 = 50
4× 3− 6× 5 = 12− 30 = −18
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The method of substitution
solve one of the equations for one of the variables:
4x− 6y = −18
4x + 18 = 6y
y =4
6x + 3 =
2
3x + 3 (12)
replace y in the other equation. (brackets!), obtain equation in onevariable.
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5x + 7y = 50
5x+ 7
(2
3x + 3
)= 50
5x +14
3x + 21 = 50
15
3x +
14
3x + 21 = 50
29
3x = 50− 21 = 29
x =3
29× 29 = 3
use (12) to find y:
y =2
3x + 3 =
2
3× 3 + 3 = 5
solution: x = 3, y = 5.
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The method of elimination
multiply first equation by coefficient of x in second equationmultiply second equation by the coefficient of x in first equation
5x + 7y = 50 | × 4
4x− 6y = −18 | × 5
20x+ 28y = 200
20x− 30y = −90
subtract second equation from first equation
20x + 28y = 20020x− 30y = −90 |−0 + 28y − (−30y) = 200− (−90)58y = 290
y = 29058 = 5
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use one of the original equations to find x
5x + 7y = 50
5x + 35 = 50
5x = 15
x = 3
We could have eliminated y first:
5x + 7y = 50 | × 6
4x− 6y = −18 | × 7
30x+ 42y = 300
28x− 42y = −126 | +
58x = 174
x = 3
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Cramer’s rule
→ linear algebra. Write system of equations as[5 74 −6
] [xy
]=
[50
−18
]
where
[x1x2
]and
[50
−18
]“columns vectors”[5 74 −6
]
is the 2× 2-“matrix of coefficients”.With each 2× 2-matrix
A =
[a bc d
]
associate a new number called the determinant
detA =
∣∣∣∣a bc d
∣∣∣∣ = ad− cb
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vertical lines, not square brackets!!!
For instance,∣∣∣∣5 74 −6
∣∣∣∣ = 5× (−6)− 4× 7 = −30− 28 = −58
Cramer’s rule for system
ax+ by = e
cx+ dy = f
or [a bb d
] [xy
]=
[ef
].
x∗ =
∣∣∣∣e bf d
∣∣∣∣∣∣∣∣a bc d
∣∣∣∣
=ed− bf
ad− bcy∗ =
∣∣∣∣a ec f
∣∣∣∣∣∣∣∣a bc d
∣∣∣∣
=ae− ce
ad− bc
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In our example,
x∗ =
∣∣∣∣50 7
−18 −6
∣∣∣∣∣∣∣∣5 74 −6
∣∣∣∣
=50× (−6)− (−18)× 7
5× (−6)− 4× 7=−300 + 126
−30− 28=−174
−58= 3
y∗ =
∣∣∣∣5 504 −18
∣∣∣∣∣∣∣∣5 74 −6
∣∣∣∣
=5× (−18)− 4× 50
5× (−6)− 4× 7=−90− 200
−58=−290
−58= 5
Exercise:
8QA + 2QB = 22
2QA + 10QB = 34.
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Existence and uniqueness.
linear simultaneous system of equations
ax+ by = e
cx+ dy = f
Rewrite as
y =e
b−a
bx
y =f
d−c
dx
slopes identical when ab = c
d, i.e., when determinant ad − cb iszero. If in addition intercepts are equal, both equations describethe same line.Example:
x + 2y = 3
2x + 4y = 422
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has no solution: The two lines
y =3
2−
1
2x
y = 1−1
2x
are parallel
-1
0
2
-2 -1 1 2 3 4 5x
There is no common solution. Trying to find one yields a contra-
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diction3
2−
1
2x = y = 1−
1
2x | +
1
2x
3
2= 1
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One equation non-linear, one linear
Consider, for instance,
y2 + x− 1 = 0
y +1
2x = 1
In our example it is convenient to solve second equation for x:
1
2x = 1− y
x = 2− 2y
y2 + (2− 2y)− 1 = y2 − 2y + 1 = (y − 1)2 = 0
So the unique solution is y∗ = 1 and x∗ = 2− 2× 1 = 0.Two non-linear equations
no general method. See example above
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Constrained optimization
Example: Utility maximization
preferences of a consumer described by family of indifferencecurves.mathematically convenient as level curves u (x, y) =constant ofutility function.example: u (x, y) = xy. indifference curves for u = 1, u = 2 andu = 3:
0
1
2
3
4
5
1 2 3 4 5x
price of apples is p, price of oranges is q budget b.
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budget constraint
px + qy ≤ b
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maximize utility subject to the budget constraint! constrained
optimization problem
u (x, y) objective function
total expenditure g (x, y) = px + qy is called the constraining
function.Principles lecture budget line must be tangential to the indifferencecurvemarginal rate of substitution must equal relative price:budget line:
px + qy = b or y =b
q−p
qx
−pq = negative of relative price of apples in terms of oranges.slope of the indifference curve is negative of marginal rate of sub-
stitution∂u/∂x∂u/∂y
In optimum:28
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∂u/∂x
∂u/∂y=p
q(13)
For the example u (x, y) = xy: ∂u∂x = y and ∂u∂y = x
y
x=p
q(14)
orqy = px, (15)
budget equationpx + qy = b (16)
To repeat: the constrained optimum (x∗, y∗) is the solution toa simultaneous system of two equations in two unknowns. Thefirst equation expresses that in the optimum the marginal rate ofsubstitution is equal to the relative price. The second one statesthat the consumer spends all his money in the optimum.
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y = pqx. Substituting into budget equation gives
px+ q
(p
qx
)= px + px = 2px = b
x∗ =b
2p
y∗ =p
qx∗ =
p
q
b
2p=b
2q
When the budget is b = 100 and p = 2 and q = 5 then
x∗ =100
2× 2= 25 y =
100
2× 5= 10.
Cost minimization
dual to the utility maximization problemproduction function Q (K,L) = KL.least costly way to produce Q0 units given prices r and w
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minimize total costsrK + wL
subject toQ0 ≤ Q (K,L) .
at cost minimum iso-cost line must be tangential to the isoquant.iso-cost line
rK + wL = constant
has slop − rwslope of isoquant is −
∂Q/∂K∂Q/∂L
In optimum:∂Q/∂K
∂Q/∂L=r
w.
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In our example ∂Q∂K = L and ∂Q∂L = K, so
L
K=r
wor
wL = rK
so in optimum equal amounts are spend on both inputs. This isthe first equation.As the second equation we have that the firm will produce exactlyQ0 units:
Q0 = Q (K,L) .
suppose r = 2, w = 5, Q0 = 250optimum (K∗, L∗) :
L
K=
2
5or L =
2
5K
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and250 = KL.
Substituting in the last equation L = 25K yields
250 = 2× 125 = 2× 53 =2
5K2 54 = K2 K∗ = 52 = 25
L∗ =2
5K∗ = 10.
The optimal input combination is (K∗, L∗) = (25, 10).The general problem
maximize or minimize an objective function
z = f (x, y)
subject to a constraint
g (x, y) ≤ c
where c is a constant.33
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interested in case where constraint is binding in optimum, i.e.,g (x, y) = cThe optimum must then solve the two conditions
∂f/∂x
∂f/∂y=∂g/∂x
∂g/∂yg (x, y) = c
The Lagrangian approach
alternative way to derive these two conditions. The method trans-forms the constrained optimization problem into an unconstrainedoptimization problem in conjunction with a pricing problem.λ ≥ 0 is the so-called Lagrange multiplier.
Lagrangian
L (x, y) = f (x, y)− λ (g (x, y)− c)
— look for an unconstrained maximum or a minimum of this func-tion
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— determine the Lagrange multiplier such that constraint holdswith equalitySuppose found absolute maximum (x∗, y∗) of Lagrangian. Sup-
pose g (x∗, y∗) = c.
L (x∗, y∗) = f (x∗, y∗)− λ (g (x∗, y∗)− c) = f (x∗, y∗) .
consider (x, y) with g (x, y) ≤ c.Then L (x, y) ≤ L (x∗, y∗) because (x∗, y∗) is absolute maximum.−λ (g (x, y)− c) is not positiveHence f (x, y) ≤ L (x, y) ≤ L (x∗, y∗) ≤ f (x∗, y∗)
FOC for maximum of Lagrangian:
∂L
∂x=∂f
∂x− λ
∂g
∂x= 0 or
∂f
∂x= λ
∂g
∂x(17)
∂L
∂y=∂f
∂y− λ
∂g
∂y= 0 or
∂f
∂y= λ
∂g
∂y(18)
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divide:∂f/∂x
∂f/∂y=∂g/∂x
∂g/∂y. (19)
In addition constraint must hold in optimum
g (x, y) = c. (20)
Utility maximization again
Lagrangian
L (x, y) = u (x, y)−λ (px+ qy − b) = u (x, y) +λ (b− px + py)
The savings:s = b− px + py
future utility from saving:λs
λ is the marginal utility of saving a penny.Cost minimization again
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The Lagrangian
L (K,L) = −rK − wL− λ (−Q (K,L) +Q0)
= (λQ (K,L)− rK − wL)− λQ0
Up to a constant the Lagrangian is profit function for output priceλ.
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