Mathematics for Economists
[Section A – Part 2 / 4]
Instructor: Annika M. Mueller, Ph.D., The Wang
Yanan Institute for Studies in Economics
Main Reference: C. P. Simon and L. Blume
(1994), Mathematics for Economists
Further Reading: W. Nicholson and C. Snyder
(2011), Microeconomic Theory
Constrained Optimization

Suppose that not all values for the x’s
are feasible
– the values of a specific x may all have to be
positive
– a consumer’s choices may be limited by the
budget she has available

One method used to solve constrained
optimization 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
Lagrangian Multiplier Method

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

Note that when the constraint holds, L
= f because g(x1, x2,…, xn) = 0
Lagrangian Multiplier Method

First-Order Conditions
L/x1 = f1 + g1 = 0
L/x2 = f2 + g2 = 0
.
.
.
L/xn = fn + gn = 0
L/ = g(x1, x2,…, xn) = 0
Lagrangian Multiplier Method


The first-order conditions can generally
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
Constrained Maximization




Suppose a farmer had a certain length of
fence in yards (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
Constrained Maximization

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
Constrained Maximization

Since y/2 = x/2 = , x must be equal to y
– the field should be square

Since x = y and y = 2, we can use the
constraint to show that
x = y = P/4
 = P/8
Interpretation of λ
In the example, we maximze A=xy
subject to 2x+2y=P.
Ask yourself, how does the maximized
A (A*) change for a small exogenous
change in P? (Suppose the farmer
suddenly got a bit more fence dP so
that he now has a length of fence
P+dP).
Interpretation of λ
Differentiating A* totally gives us
dA*=x*.dy + y*.dx
Since from the FOC we have y*/2 = x*/2 =  this
implies
dA*=2(dy+dx)= (2.dy+2.dx)=dP
This means: dA*/ dP =
Basically, the Lagrange multiplier shows us how
much extra value (in this case area) we can get if
our constraint (in this case length of fence) were
increased (or relaxed) by a tiny bit.
Interpretation of λ
Other (standard) example:
“How much does maximized utility increase when
we have a bit more income in a utility
maximization problem?”
Here, the Lagrange multiplier is the marginal utility
of income.
Envelope Theorem &
Constrained Maximization

Suppose that we are interested in dy*/da for
y = f(x1,…,xn;a)
subject to the constraint
g(x1,…,xn;a) = 0

One way to solve would be to set up the
Lagrangian expression and solve the firstorder conditions, plug those into y, then
take the derivative wrt a.
Envelope Theorem &
Constrained Maximization

Alternatively, it can be shown that
dy*/da = L/a(x1*,…,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
An Example of the Envelope Theorem
for Constrained Optimization
Verify this for Specific Production
Function
Inequality Constraints


In some economic problems the
constraints need not hold exactly
For example, suppose we seek to
maximize y = f(x1,x2) subject to
g(x1,x2)  0,
x1  0, and
x2  0
Inequality Constraints


One way to solve this problem is to
introduce three new variables (a, b, and
c) that convert the inequalities into
equalities
To ensure that the inequalities continue
to hold, we will square these new
variables to ensure that their values are
positive
Inequality Constraints
g(x1,x2) - a2 = 0;
x1 - b2 = 0; and
x2 - c2 = 0

Any solution that obeys these three
equality constraints will also obey the
inequality constraints
Inequality Constraints

We can set up the Lagrangian
L = f(x1,x2) + 1[g(x1,x2) - a2] + 2[x1 - b2] +
3[x2 - c2]

This will lead to eight first-order
conditions
Inequality Constraints
L/x1 = f1 + 1g1 + 2 = 0
L/x2 = f1 + 1g2 + 3 = 0
L/a = -2a1 = 0
L/b = -2b2 = 0
L/c = -2c3 = 0
L/1 = g(x1,x2) - a2 = 0
L/2 = x1 - b2 = 0
L/3 = x2 - c2 = 0
Inequality Constraints

According to the third condition, either a
or 1 = 0
– if a = 0, the constraint g(x1,x2) holds exactly
– if 1 = 0, the availability of some slackness
of the constraint implies that its value to the
objective function is 0

Similar
complementary
slackness
relationships also hold for x1 and x2
Inequality Constraints

These results are sometimes called
Kuhn-Tucker conditions
– they show that solutions to optimization
problems involving inequality constraints
will differ from similar problems involving
equality constraints in rather simple ways
– we cannot go wrong by working primarily
with constraints involving equalities