Constrained Maximization

Constrained Maximization
• What if all values for the x’s are not
feasible?
– The values of x may all have to be positive
– A consumer’s choices are limited by the
amount of purchasing power available
• One method used to solve constrained
maximization problems is the Lagrangian
multiplier method

Lagrangian Multiplier Method
• Suppose that we wish to find the values
of x1, x2,…, xn that maximize
y = f(x1, x2,…, xn)
subject to a constraint that permits only
certain values of the x’s to be used
g(x1, x2,…, xn) = 0

• The Lagrangian multiplier method starts
with setting up the expression
L = f(x1, x2,…, xn ) + g(x1, x2,…, xn)
where  is an additional variable called
a Lagrangian multiplier
• When the constraint holds, L = f
because g(x1, x2,…, xn) = 0

• First-Order Conditions
L/x1 = f1 + g1 = 0
L/x2 = f2 + g2 = 0
.
L/xn = fn + gn = 0
.
.
L/ = g(x1, x2,…, xn) = 0

• The first-order conditions can be solved
for x1, x2,…, xn and 
• The solution will have two properties:
– The x’s will obey the constraint
– These x’s will make the value of L (and
therefore f) as large as possible

• The Lagrangian multiplier () has an
important economic interpretation
• The first-order conditions imply that
f1/-g1 = f2/-g2 =…= fn/-gn = 
– The numerators above measure the
marginal benefit that one more unit of xi will
have for the function f
– The denominators reflect the added burden
on the constraint of using more xi

• At the optimal choices for the x’s, the
ratio of the marginal benefit of increasing
xi to the marginal cost of increasing xi
should be the same for every x
•  is the common cost-benefit ratio for all
of the x’s
i
i
x
x
ofcostmarginal
ofbenefitmarginal


• If the constraint was relaxed slightly, it
would not matter which x is changed
• The Lagrangian multiplier provides a
measure of how the relaxation in the
constraint will affect the value of y
•  provides a “shadow price” to the
constraint

• A high value of  indicates that y could
be increased substantially by relaxing
the constraint
– each x has a high cost-benefit ratio
• A low value of  indicates that there is
not much to be gained by relaxing the
constraint
• =0 implies that the constraint is not
binding

Duality
• Any constrained maximization problem
has associated with it a dual problem in
constrained minimization that focuses
attention on the constraints in the
original problem

Duality
• Individuals maximize utility subject to a
budget constraint
– Dual problem: individuals minimize the
expenditure needed to achieve a given level
of utility
• Firms minimize cost of inputs to produce
a given level of output
– Dual problem: firms maximize output for a
given cost of inputs purchased

• Suppose a farmer had a certain length of
fence (P) and wished to enclose the
largest possible rectangular shape
• Let x be the length of one side
• Let y be the length of the other side
• Problem: choose x and y so as to
maximize the area (A = x·y) subject to the
constraint that the perimeter is fixed at P =
2x + 2y

• Setting up the Lagrangian multiplier
L = x·y + (P - 2x - 2y)
• The first-order conditions for a maximum
are
L/x = y - 2 = 0
L/y = x - 2 = 0
L/ = P - 2x - 2y = 0

• Since y/2 = x/2 = , x must be equal to y
– The field should be square
– x and y should be chosen so that the ratio of
marginal benefits to marginal costs should be
the same
• Since x = y and y = 2, we can use the
constraint to show that
x = y = P/4
 = P/8

• Interpretation of the Lagrangian multiplier:
– If the farmer was interested in knowing how
much more field could be fenced by adding an
extra yard of fence,  suggests that he could
find out by dividing the present perimeter (P)
by 8
– The Lagrangian multiplier provides
information about the implicit value of the
constraint

• Dual problem: choose x and y to minimize
the amount of fence required to surround a
field of a given size
minimize P = 2x + 2y subject to A = x·y
• Setting up the Lagrangian:
LD = 2x + 2y + D(A - x - y)

• First-order conditions:
LD/x = 2 - D·y = 0
LD/y = 2 - D·x = 0
LD/D = A - x ·y = 0
• Solving, we get
x = y = A1/2
• The Lagrangian multiplier (D) = 2A-1/2

Envelope Theorem &
• Suppose that we want to maximize
y = f(x1, x2,…, xn)
subject to the constraint
g(x1, x2,…, xn; a) = 0
• One way to solve would be to set up the
Lagrangian expression and solve the first-
order conditions

Envelope Theorem &
• Alternatively, it can be shown that
dy*/da = L/a(x1*, x2*,…, xn*;a)
• The change in the maximal value of y that
results when a changes can be found by
partially differentiating L and evaluating
the partial derivative at the optimal point

• Suppose we want to choose x1 and x2
to maximize
y = f(x1, x2)
• subject to the linear constraint
c - b1x1 - b2x2 = 0
• We can set up the Lagrangian
L = f(x1, x2) - (c - b1x1 - b2x2)

• The first-order conditions are
f1 - b1 = 0
f2 - b2 = 0
c - b1x1 - b2x2 = 0
• To ensure we have a maximum, we
must use the “second” total differential
d 2y = f11dx1
2 + 2f12dx2dx1 + f22dx2
2

• Only the values of x1 and x2 that satisfy
the constraint can be considered valid
alternatives to the critical point
• Thus, we must calculate the total
differential of the constraint
-b1 dx1 - b2 dx2 = 0
dx2 = -(b1/b2)dx1
• These are the allowable relative changes
in x1 and x2

• Because the first-order conditions imply
that f1/f2 = b1/b2, we can substitute and
get
dx2 = -(f1/f2) dx1
• Since
d 2y = f11dx1
2 + 2f12dx2dx1 + f22dx2
2
we can substitute for dx2 and get
d 2y = f11dx1
2 - 2f12(f1/f2)dx1 + f22(f1
2/f2
2)dx1
2

• Combining terms and rearranging
d 2y = f11 f2
2 - 2f12f1f2 + f22f1
2 [dx1
2/ f2
2]
• Therefore, for d 2y < 0, it must be true
that
f11 f2
2 - 2f12f1f2 + f22f1
2 < 0
• This equation characterizes a set of
functions termed quasi-concave functions
– Any two points within the set can be joined
by a line contained completely in the set

• Recall the fence problem: Maximize A
= f(x,y) = xy subject to the constraint P -
2x - 2y = 0
• Setting up the Lagrangian [L = x·y + (P
- 2x - 2y)] yields the following first-order
conditions:
L/x = y - 2 = 0
L/y = x - 2 = 0
L/ = P - 2x - 2y = 0

• Solving for the optimal values of x, y,
and  yields
x = y = P/4 and  = P/8
• To examine the second-order
conditions, we compute
f1 = fx = y f2 = fy = x
f11 = fxx = 0 f12 = fxy = 1
f22 = fyy = 0

• Substituting into
f11 f2
2 - 2f12f1f2 + f22f1
2
we get
0 ·x2 - 2 ·1 ·y ·x + 0 ·y2 = -2xy
• Since x and y are both positive in this
problem, the second-order conditions
are satisfied

Constrained Maximization

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