John Riley
13 Jan 2015
Maximization with n variables
Consider the following problem.
Max{ f ( x) | x 
x
n

}
Suppose that x 0 is a maximizer. Arguing as in the two variable case, we reduce this to a one
variable problem by considering x  ( x10 ,..., x0j 1 , x j , x0j 1 ,..., xn0 ) . From Module 1, if x 0j  0 , then
the derivative of f ( x10 ,..., x0j 1 , x j , x0j 1 ,..., xn0 ) with respect to x j must be zero at x j  x 0j . And if
x 0j  0 , this derivative cannot be strictly positive.
Thus we have the following result.
Proposition: Necessary conditions for a maximum
If x 0 solves Max{ f ( x) | x 
x
n

} then
f 0
( x )  0 , with equality unless x 0j  0 , j  1,..., n
x j
As in the two variable case, the necessary conditions are also sufficient if f is concave.
Definition 1: Concave Function
The function f is concave on X 
n
if for any x0 , x1  X and any   (0,1)
f ( x )  (1   ) f ( x0 )   f ( x1 )
The vector of all the partial derivatives of a function f is called the gradient vector and is
written as follows:
f
f
f
( x)  ( ( x),...,
( x))
x
x1
xn
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John Riley
13 Jan 2015
Definition 2: Concave function
The differentiable function f is concave on X if for any x0 , x1  X and any   (0,1)
f ( x1 )  f ( x 0 ) 
f 0
( x )  ( x1  x 0 )
x
The proof of the equivalence of these definitions (for differentiable functions) can be found in
the appendix.
Proposition 2: Necessary and sufficient conditions for a maximum
For the problem Max{ f ( x) | x 
x
n

} , if f is a differentiable concave function then the
following conditions are both necessary and sufficient for f to take on its maximum value at
x0 .
f 0
( x )  0 , with equality unless x 0j  0 .
x j
Proof: Appealing to the necessary conditions,
if x 0j  0 then
f 0
f 0 1
( x )  0 and so
( x )( x j  x0j )  0 .
x j
x j
Moreover, if x 0j  0 then
f 0 1
f 0
( x )( x j  x 0j )  0 .
( x )  0 . Also x1j  x0j  0 . Therefore
x j
x j
Summing over j,
n

j 1
f 0 1
f
( x )( x j  x 0j )  ( x 0 )  ( x1  x 0 )  0
x j
x
From Definition 2, for any x1  x0 ,
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John Riley
13 Jan 2015
f ( x1 )  f ( x 0 ) 
f 0
( x )  ( x1  x 0 )
x
We have just argued that the second term on the right side of this inequality is negative.
Therefore f ( x1 )  f ( x0 ) .
QED
Technical remark
When does a function f ( x) have convex upper contour sets and when is it concave?
Here are some key results.

A function f has convex upper contour sets if for some strictly increasing function g ,
h( x)  g ( f ( x)) is concave.


A function is concave if it is the sum of concave functions.
h( x)  g ( f ( x)) is concave if f and g are concave and g is increasing.

If f exhibits constant returns to scale (a mathematician would say homogeneous of
degree 1) and the upper contour sets of f are convex then f is concave.
We have already discussed the first three results in earlier modules. The fourth is proved in the
appendix.
Application 1: f ( x)  x11 x22 . If 1   2  1 then f is concave on
n

.
ln( y) is strictly increasing and ln f ( x)  1 ln x1   2 ln x2 . The terms on the right hand side are
concave thus f ( x) has convex upper contour sets.
As you can quickly check f ( x)   f ( x) so the function exhibits CRS. It follows from the last
bullet that f is concave.
Application 2: f ( x)  x11 x2 2 If 1  2    1 then f is concave on
1
n

.
2
x11 x2 2  ( x1  x2  ) .
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John Riley
13 Jan 2015
1
2
 
Note that 1  2  1 . Appealing to Application 1, x1  x2  is concave. Appealing to the third


bullet, f ( x) is concave.
Exercise: Show that if f ( x)  x11 x2 2 where 1  2    1 , then f is not concave.
HINT: Suppose that x1  x2  z .
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John Riley
13 Jan 2015
Appendix:
Equivalence of Definition 1 and Definition 2
If you have a sufficiently strong background in mathematics you might wish to see if you
can follow each step in the proof that D1 and D2 are equivalent.
We first show that D1 implies D2. Consider any vectors x 0 and x1 . Note that
f ( x )  f ( x0   ( x1  x0 )) . We can therefore think of this as being a function of the parameter
 , that is, we define
g ( )  f ( x )  f ( x0   ( x1  x0 )) .
From the definition of a concave function,
g ( )  (1   ) g (0)   g (1)   (0,1) .
Rearranging this expression,
g ( )  g (0)

 g (1)  g (0),   (0,1) .
Taking the limit as   0 , it follows that
dg
(0)  g (1)  g (0)
d
(1)
Appealing to the Chain Rule, the derivative is
n
dg
f  1
f 
( )  
( x )( x j  x 0j ) 
( x )  ( x1  x 0 ) .
d
x
j 1 x j
(2)
Setting   0 ,
dg
f 0
(0) 
( x )  ( x1  x 0 ) .
d
x
Appealing to (1),
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John Riley
13 Jan 2015
f 0
( x )  ( x1  x0 )  g (1)  g (0)  f ( x1 )  f ( x 0 ) .
x
The proof that D2 implies D1 is almost identical to that for the one variable case.
Appealing to D2, for any x 0 , x1 and convex combination x̂
f ( x0 )  f ( xˆ ) 
f
f
( xˆ )  ( x 0  xˆ ) and f ( x1 )  f ( xˆ )  ( xˆ )  ( x1  xˆ ) .
x
x
Multiplying the first inequality by (1   ) , the second by  and adding,
(1   ) f ( x0 )   f ( x1 )  f ( x )  (1   )
 f ( x ) 
f 
f
( x )  ( x0  x )   ( x )  ( x1  x  )
x
x
f 
( x )  ((1   ) x0   x1  x  )  f ( x  )
x
QED
Proposition: If the upper contour sets of f are convex and f exhibits constant returns to
scale1 (i.e. f ( x)   f ( x) ), then f is concave.
Proof:
We will show that under our assumptions the function f is super additive. That is, for any a
and b ,
f (a  b)  f (a)  f (b)
Choose a  (1   ) x0 and b   x1 . Then
f ((1   ) x0   x1 )  f ((1   ) x0 )  f ( x1 ) .
Appealing to the CRS assumption, f ((1   ) x0 )  (1   ) f ( x0 ) and f ( x1 )   f ( x1 ) .
Hence
f ((1   ) x0   x1 )  (1   ) f ( x0 )   f ( x1 ) .
Super additivity
1
A a mathematician would say that f is homogeneous of degree 1)
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John Riley
13 Jan 2015
Suppose that f (a)  qa and f (b)  qb . Hence
f (a)  f (b)  qa  qb
(3)
Appealing to the CRS assumption
f(
b
a
)  1 and f ( )  1 .
qa
qb
Thus the vectors
a
b
and
are on the boundary of the upper contour set CU  {x | f ( x)  1} .
qa
qb
All convex combinations are in this set. Choose 1   
qb
qa
so that  
.
qa  qb
qa  qb
Then
f(
qa
qb
a
b
ab
( )
( ))  f (
) 1 .
qa  qb qa
aa  qb qb
qa  qb
Appealing to the CRS assumption again
f (a  b)  qa  qb .
Appealing to (3),
f (a  b)  f (a)  f (b)
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