Answers to Econ 210A Midterm, October 2010
Question 1.
f
(
x
1
,x
2
) =
p
max
{
x
1
,x
2
}
A.
The function
f
is homogeneous of degree 1/2. To see this, note that for all
t >
0 and all (
x
1
,x
2
)
f
(
tx
1
,x
2
) =
p
max
{
tx
1
,tx
2
}
=
p
t
max
{
x
1
,x
2
}
=
t
1
/
2
p
max
{
x
1
,x
2
}
(1)
=
t
1
/
2
f
(
x
1
,x
2
)
B.
The function
f
is neither concave, nor quasiconcave. One should be able to
see this quickly by drawing an indiﬀerence curve and seeing that the atleast
asgood set is not convex. Therefore the function is not quasiconcave. Since a
concave function must be quasiconcave,
f
cannot be concave either.
One can show that the function is not quasiconcave by means of a single
example. Let
x
= (1
,
0) and
x
0
= (0
,
1). Then
f
(
x
) = 1
≥
f
(
x
0
) = 1. Now
consider the convex combination
tx
+ (1

t
)
x
0
= (
t,t
) where 0
< t <
1. If
f
is
quasiconcave, then it must be that
f
(
tx
+ (1

t
)
x
0
)
≥
f
(
x
0
). But
f
(
tx
+ (1

t
)
x
0
) =
t < f
(
x
0
) = 1. So
f
cannot be quasiconcave. We also note from this
same example that
tf
(
x
) + (1

t
)
f
(
x
0
) = 1
> f
(
tx
+ (1

t
)
x
0
) =
t
. But if
f
is
a concave function
tf
(
x
)+(1

t
)
f
(
x
0
)
≤
f
(
tx
+(1

t
)
x
0
), so
x
is not a concave
function.
C
The function
f
is not a convex function. The square root sign should be a
tipoﬀ. Square root is a strictly concave function. To show that
f
is not a convex
function, consider the two points
x
= (1
,
0) and
x
0
= (4
,
0). Then
f
(
x
) = 1 and
f
(
x
0
) = 2. Therefore
1
2
f
(
x
) +
1
2
f
(
x
0
) = 2
.
5
.
But
f
(
1
2
x
+
1
2
x
0
) =
f
(2
.
5
,
0) =
√
2
.
5
<
1
2
f
(
x
) +
1
2
f
(
x
0
)
The function
f
is a quasiconvex function. A quick way to see that this
must be true is to look at an indiﬀerence curve and see that it looks like the
“worsethan” sets must be convex.
To show this formally, suppose
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 Spring '10
 fallahi
 Derivative, X1, Convex function, 2 m, A. Euler

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