Questions and Answers from Econ 210A Final: Fall 2008
I have gone to some trouble to explain the answers to all of these questions,
because I think that there is much to be learned by working through them.
Please let me know if you find mistakes or inadequate explanations.
Ted Bergstrom
Question 1)
Firm A has production function
f
(
x
1
, x
2
) =
x
1
/
2
1
+
x
1
/
2
2
2
k
where
k >
0.
a) For what values of
k
is
f
a quasiconcave function? For what values of
k
is
f
a concave function? Explain your answers.
Firm B has production function
g
(
x
1
, x
2
) =
(
x
2
1
+
x
2
2
)
k/
2
where
k >
0.
b) For what values of
k
is
g
a quasiconcave function? For what values of
k
is
g
a concave function? Explain your answers.
Answers to Question 1
Answer to 1a:
The easiest way to check for quasiconcavity of
f
is to remem
ber that a function is quasiconcave if and only if every monotonic increasing
transformation of that function is quasiconcave. In particular, the function
f
(
x
1
, x
2
) =
x
1
/
2
1
+
x
1
/
2
2
2
k
is a monotonic increasing transformation of the function
g
(
x
1
, x
2
) =
x
1
/
2
1
+
x
1
/
2
2
,
But it is easy to verify that
g
is quasiconcave. You could do this by checking
the bordered Hessian conditions, which are now pretty simple because there are
no offdiagonal terms in the regular Hessian part of the bordered Hessian. You
could make the problem even easier by checking that
g
is a concave function
and then using the fact that a concave function must be quasiconcave. Or you
could alternatively just note that the isoquants for
g
have diminishing marginal
rate of substitution.
When is the function
f
concave?
We could check this by writing out the
Hessian matrix and making sure that it is negative semidefinite. But there is
an easier way. We see that the function
f
is homogeneous of degree
k
. We know
that it is quasiconcave, and our textbook tells us that if a function is quasi
concave and homogeneous of degree 1 it is concave. Suppose that
k <
1. Define
1
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the function
h
(
x
) =
f
(
x
)
1
/k
. We know that
h
(
x
) is homogeneous of degree one
and quasiconcave, so it is concave. But now
f
(
x
) =
h
(
x
)
k
where
k <
1. This
means that
f
is a concave function of a concave function and hence must be
concave.
What if
f
is homogeneous of degree
k >
1? Then it will not be a concave
function. There are increasing returns to scale. It is easy to check that
f
is not
a concave function. One could prove this by showing a single example of two
points that violate the condition for concavity For example let
x
= (0
,
0) and
x
0
= (1
,
0). Then
f
(
x
) = 0,
f
(
x
0
) = 1. Now for any
t
∈
(0
,
1),
tx
0
+ (1

t
)
x
=
(
t,
0). We have (1

t
)
f
(
x
) +
tf
(
x
0
) =
t
. But
f
((1

t
)
x
+
tx
0
) =
f
(
t,
0) =
t
k
.
If
k >
1 and 0
< t <
1, we see that
t
k
< t
.
Therefore
f
((1

t
)
x
+
tx
0
)
<
(1

t
)
f
(
x
) +
tf
(
x
0
) =
t
, which means that
f
is not a concave function.
Answer to 1b:
This one should be really easy if you think about it. The isoquants for this
production function have the equation
x
2
1
+
x
2
2
= Constant. What do they look
like? Can this function be quasiconcave? No! This should be obvious once you
look at an isoquant. To be a little more formal about this: Much as we remarked
in Part 1a) the function
f
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