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ECON 701 ASSIGNMENT  1

ANSWERS

Problem 1 (50 marks).  Consider a consumer with wealth w who consumes two goods, which we shall call goods 1, 2.  Let the amount of good l that the con- sumer consumes be xe  and the price of good l be pe . Suppose that the consumer’s preferences are described by the utility function
u(x1 , x2 ) = x1 x2
(1) Set up the utility maximisation problem and write down the Lagrangian.
Answer:   The maximisation problem is
max  x1 x2
zl .z2
subject to p1 x1 + p2 x2 - w < 0
x1  > 0, x2  > 0,
and the Lagrangian is
c(x1 , x2 , λ, p1 , p1 , w) = x1 x2 - λ(p1 x1 + p2 x2 - w).

(2) Write down the Karush Kuhn Tucker conditions for a solution to this prob- lem.


Answer:   The KKT conditions are
∂c

∂x1

= x1 - λp2  < 0
∂x2
∂c
∂x2

= -(p1 x1 + p2 x2 - w) > 0
∂λ


Dα芒e : Due 29 April, 2021.
1


2                                                 JOHN HILLAS   UNIVERSITY OF AUCKLAND
(3) Argue as to the nature of the solution and which conditions will be binding.

Answer:    We can see that if either x1  = 0 or x2  = 0 then the objective function is 0 while there are feasible values of (x1 , x2 ) wich give a value for the objective function greater than 0.  Thus the solution will never involve either x1  = 0 or x2  = 0 and so the nonnegativity constraint will not be binding.  Also if -(p1 x1 + p2 x2  - w) > 0 then one could increase both x1 and x2  thereby increasing the value of the objective function. So the budget constraint will be binding, that is, satisfied as an equality.
(4) Solve the Karush Kuhn Tucker conditions to obtain the Marshallian de- mand functions.
Answer:   From the previous answer the conditions we need to satisfy are

x2 - λp1  = 0
x1 - λp2  = 0
-(p1 x1 + p2 x2 - w) = 0

We have solved such equations a number of times (and you should remember what the solution to the utility maximisation problem for Cobb-Douglas utility functions is) so I’ll just give the solutions.

w 
x1 (p, w) =

2p2

(5) Substitute the Marshallian demands back into the utility function to obtain the indirect utility function.

Answer:
w2   
v(p, w) =


(6) State Roy’s Theorem and verify that Roy’s Theorem does indeed give the same Marshallian demands that you found above.
Answer:     Roy’s  Theorem states  that if the indirect  utility function is differentiable at (p, w) then

xe (p, w) = -

In our case ,u(p3w) ,p廷
,u(p3w) . ,w
∂v(p.w)           w2    
=    -
∂p1                   4p1 2p2
∂v(p.w)       2w  
=
∂w         4p1p2 .

So

And

x1 (p, w) = - 
w 
=
2p1 .
similarly
w 
x2 (p, w) =

These are the same functions that we found in part (4).
(7) Invert the indirect utility function to find the expenditure function.
Answer:    When we say  “invert the indirect utility function” we mean to solve the duality relationship
v(p, e(p, u)) = u
to obtain the function e(p, u). In our case this is

(e(p, u))2
4p1p2
or
e(p, u) = 2 ′up1p2 .

(8)  Consider the expenditure minimisation problem
min    p1 x1 + p2 x2
z l .z2
subject to: x1 x2  > u
x1  > 0, x2  > 0.
Write down the Lagrangian for this problem.
Answer:   the Lagrangian is
c(x1 , x2 , λ, p1 , p1 , uw) = p1 x1 + p2 x2 + λ(x1 x2 - u).

(9) Write down the Karush Kuhn Tucker conditions for a solution to this prob- lem.  You can do this either by restating the problem as a maximisation problem or by using conditions appropriate for a minimisation problem.
Answer:    I shall rewrite the problem as a maximisation problem.  Then the problem above becomes

max zl .z2

- (p1 x1 + p2 x2 )

subject to: u - x1 x2  < 0
x1  > 0, x2  > 0.
And the Lagrangian is
c(x1 , x2 , λ, p1 , p1 , u) = -(p1 x1 + p2 x2 ) - λ(u - x1 x2 ).

And the KKT conditions are
= -p1 + λx2  < 0
∂x1
∂c
∂x1

= -p2 + λx1  < 0
∂x2
∂c
∂x2

= -(u - x1 x2 ) > 0
∂λ

(10) Solve to obtain the Hicksian (or compensated) demand functions.  [Again, you should be able to make an argument as to which of the KKT conditions will be binding.]
Answer:   If u = 0 then the solution is obvious, namely, that x1  = x2  = 0. So assume that u > 0.  Then it must be that x1  > 0 and x2  > 0 to satisfy the constraint that x1 x2  > u.  So the nonnegativity constraints are never binding. Also if x1 x2  > u then we can make the objective function greater by decreasing both x1  and x2  a little, so this constraint will be satisfied as an equality.  Then the KKT conditions above become

-p1 + λx2  = 0
-p2 + λx1  = 0
-(u - x1 x2 ) = 0.

As  usual,  we solve  this  by dividing equation  (31)  by equation  (32) and rearranging a bit to obtain

2  =
and substituting this into equation (33) gives
u = x1 2  x 
or
h1 (p, u) = (  .
And, similarly,
h2 (p, u) = (  .

(11) Take the Hicksian demand functions that you just found, which we shall denote he (p1 , p2 , u), and substitute them back into the objective function, p1 x1 + p2 x2 , to obtain the 44智4n441/y4 人/n.14Xn
e(p1 , p2 , u) = p1 h1 (p1 , p2 , u) + p2 h2 (p1 , p2 , u).

and confirm that it is the same as the expenditure function you found above.

Answer:
e(p, u) = p1 (  + p2 (  
= ′up1p2 +′up1p2
= 2 ′up1p2 ,                                                   (37)
which is the same function that we found in part (7).
(12)  Use Shephard’s Lemma to find the Hicksian demand functions from the expenditure function and confirm that are the same functions that you found by solving the expenditure minimisation problem.
Answer:     Shephard’s Lemma says  that if the  expenditure function is

differentiable then
he (p, u) =
In our case this gives

∂e(p, u)
∂pe      .

h1 (p, u) = 2 x  x (up1p2 )_   x (up2 )
= (  .
And, similarly,
h2 (p, u) = (  .
And these are the same functions we found in part (10).

(13)  Substitute the indirect utility function v(p1 , p2 , w) you found in part (5) into the Hicksian demand functions you just found and confirm that you obtain the Marshallian demand functions that you found in part (4).
Answer:   We claim that
x1 (p, w) = h1 (p, v(p, w))
= )  
w
2p1 .
And, similarly,
x2 (p, w) =                                                         (41)

(14)  Solve the equation
e(p1 , p2 , v(p1 , p2 , w)) = w
to find the indirect utility function and confirm that it is the same as the function you found above.
Answer:
e(p1 , p2 , v(p1 , p2 , w)) = 2 (v(p, w)p1p2  = w

or
4v(p, w)p1p2  = w2
or
v(p, w) =                                                            (42)
And this is the same function that we found in part (5).
(15) Write the indirect utility function as a function of the normalised prices q1  = p1 /w and q2  = p2 /w rather than as functions of p1 , p2 , and w.  Call this function (q1 , q2 ) = v(q1 , q2 , 1).
Answer:
(q1 , q2 ) = v(q1 , q2 , 1)
1
= 4q1 q2

(16)  Consider the minimisation problem

min gl .g2

(q1 , q2 )

subject to: q1 x1 + q2 x2  = 1.
with x1 , x2  > 0. Write down the Lagrangian for this problem.
Answer:   And the Lagrangian is

c(q1 , q2 , λ, x1 , x1 ) = (q1 , q2 ) + λ(q1 x1 + q2 x2 - 1)
=  + λ(q1 x1 + q2 x2 - 1).

(17) Write down the first order necessary conditions for an interior minimum.
Answer:    This is a constrained minimisation problem with equality con- straints and we are not explicitly including any nonnegativity constraints so we use classical optimisation conditions reathe than nonlinear program- ming.  The first order conditions are

= -            + λx1  = 0
= -            + λx2  = 0
∂c(q1 , q2 )

(18)  Solve the first order conditions to obtain the inverse demand functions, giving normalised prices as a function of the consumption bundle. Call this function q : R → R .
Answer:    In this problem the variables for which we shall solve are q1 and q2   which  we shall find in  terms of the parameters x1   and x2 .   It is important to keep this clearly in our minds. In words we are finding, for a fixed consumprion bundle (x1 , x2 ) the normalised proces (q1 , q2 ) that make the given consumption bundle the optimal choice.

We divide equation (44) by equation (45) to obtain
q2          x1
=
q1         x2
or
q2 x2  = q1 x1 .
We now substitute equation (47) into equation (46) to obtain

7

q1 x1 + q1 x1  = 1

or
q1 (x1 , x2 ) =   1  
And similarly
q2 (x1 , x2 ) =   1 
or
q(x1 , x2 ) = │  , ← .

(19)  Take these inverse demand functions and substitute them into  to give a utility function u*  giving utility as a function of the consumption bundle. Confirm that this is the same as the utility function that you started with.

Answer:   We have
(q(x1 , x2 )) = 
1
= 4 ╱ ← ╱ ←
= x1 x2 ,  
which is the utility function that we started with.
(20) Write the Marshallian demands you found in part 4 as functions of the normalised prices q1  and q2 , rather than p1 , p2 , and w. Call these functions 1  and 2 , so we have  : R → R. Show that  is the inverse function to q .
Answer:   In part (4) we found the Marshallian demand functions
w 
x1 (p1 , p2 , w) =
w 
x2 (p1 , p2 , w) =
Writing these as functions of q1 and q2 , either by setting w = 1 or by writing the functions explicitly as functions of p1 /w and p2  gives
1 (q1 , q2 ) = 
2 (q1 , q2 ) =   1 
Now, with the Cobb-Douglas utility function the function 1  actually de- pends only on q1   and the function q1   depends only on x1   and similarly for good  2.    Thus  for  this  case  the  functions  1    :  R++   →  R++   and q1  : R++  → R++  are mutually inverses. But in general the function 1  will depend on both q1  and q2  and the function q1  will depend on both x1  and

x2  so I shall write the answer to show that the functions  : R → R that map pairs of strictly positive normalised prices  to pairs of strictly positive consumption bundles and q : R → R that maps strictly posi- tive consumption bundles to pairs of strictly positive normalised prices are mutually inverse, that is, that
(q(x1 , x2 )) = (x1 , x2 )
for all (x1 , x2 ) in R and
q((q1 , q2 )) = (q1 , q2 )
for all (q1 , q2 ) in R .
Now,
(q(x1 , x2 )) = │  , ←
= ﹔  , 2 x1 \
= (x1 , x2 ),
as required. And,
q((q1 , q2 )) = │  , ←
= ﹔  , 2 x1 \
= (q1 , q2 ),
as required.
Problem 2 (50 marks).  Consider a consumer with wealth w who consumes two goods, which we shall call goods  1, 2.   Let the amount of good l that the con- sumer consumes be xe  and the price of good l be pe . Suppose that the consumer’s preferences are described by the utility function
u(x1 , x2 ) = (x1 + 1) (x2 + 1)
(1)  Draw, as neatly and accurately as you can, a diagram showing the indiffer- ence curves through the points (1.1) and (2, 2).
Answer:
You are asked to draw, as neatly and accurately as you can, the required indifference curves.  To do so you should, at a minimum, find where these indifference curves cut the axes, and probably find a few more points on those curves.  You might also find the slopes of the indifference curves at the points where the indifference curves cut the axes and make sure that your diagrams are consistent with those facts.  For the most part this was reasonably well done.
I shall do the construction in sone detail.  The utility at the point (1, 1) is (1 + 1)(1 + 1) = 4 and the utility at the point 2, 2 is (2 + 1)(2 + 1) = 9. If we let the utility for the indifference curve we are talking about be u then the equation for the indifference curve giving x2  as a function of x1  is
u   
x1 + 1

and the slope of the indifference curve as a function of x1  is given by
dx2                    u      
=   -

I’ll make do with 5 points for each indifference curve. For the indifference curve through (1, 1) I get

1 0 1/2 1 5/3 3
2 3 5/3 1 1/2 0
Slope -4 -16/9 -1 -9/16 -1/9
And for the indifference curve through (2, 2)

1 0 1/2 1 5/3 3
2 3 5/3 1 1/2 0
Slope -4 -16/9 -1 -9/16 -1/9
With these we draw the indifference curves given in Figure 1

(2)  Set up the utility maximisation problem and write down the Lagrangian.
Answer:   The maximisation problem is
max   (x1 + 1)(x2 + 1)
zl .z2
subject to p1 x1 + p2 x2 - w < 0
x1  > 0, x2  > 0,
and the Lagrangian is
c(x1 , x2 , λ, p1 , p1 , w) = (x1 + 1)(x2 + 1) - λ(p1 x1 + p2 x2 - w).

(3) Write down the Karush Kuhn Tucker conditions for a solution to this prob- lem.
Answer:   The KKT conditions are
∂c

∂x1

= (x1 + 1) - λp2  < 0                                                       (55) ∂x2
∂c
∂x2

= -(p1 x1 + p2 x2 - w) > 0                                               (58) ∂λ

(4) For what values of (p1 , p2 , w) will the constraint x1  > 0 be binding?  For what values of (p1 , p2 , w) will the constraint x2  > 0 be binding?
Answer:   Suppose that we let z1  = x1 + 1 and z2  = x2 + 1.  Thus we have x1  = z1 - 1 and x2  = z2 - 1. And since we have, from the budget constraint that p1 x1 +p2 x2  < w this implies that p1 z1 +p2 z2  < w +p1 +p2 . And since xe  > 0 we have ze  > 1.  Thus we can solve the problem by solving instead the problem
max  z1 z2
z l .z2
subject to p1 z1 + p2 z2 - w - p1 - p2  < 0
z1  > 1, z2  > 1.
And we saw in Problem 1 that if we can ignore the constraints z1  > 1 and z2  > 1 then the solution is
w + p1 + p2
w + p1 + p2
2p2           ,
which is, in terms of x1  and x2 ,
x1  =                    - 1 =
w + p1 + p2                    w + p1 - p2
2p2                                      2p2           .
Thus if w - p1 + p2  > 0, or equivalently w > p1 - p2  then the nonnegativity constraint on x1  is not binding. Similarly if w +p1 -p2  > 0, or equivalently w > p2  - p1  then the nonnegativity constraint on x2  is not binding.  So if w > |p1 -p2 | then neither nonnegativity constraint is binding. If w < p1 -p2 then the nonnegativity constraint on x1 is binding and the optimal solution

will have x1  = 0. Ifw < p2 - p1  then the nonnegativity constraint on x2  is binding and the optimal solution will have x2  = 0.
(5) By solving the Karush Kuhn Tucker conditions, or by some other method, obtain the Marshallian demand functions.
Answer:    The answer to the previous part gives us that if w > |p1  - p2 | then the solution is given by
w - p1 + p2
w + p1 - p2
2p2           .
If w < p1  - p2   then x1  = 0.  In this case the consumer will spend all of
w on good 2 and x2  = w/p2 .  Similarly if w < p2  - p1  then x2  = 0 and x1  = w/p1 .  Thus we have

,．．╱0,  ←      (x1 (p, w), x2 (p, w)) = ．  ╱ ,
．．．．╱  , 0←

if w < p1 - p2
if w > |p1 - p2 | .
if w < p2 - p1

Notice on the  boundary between the regions  both answrs give the same result so it does not matter in  which region  we include the equality,  or indeed if we include it in both regions.
(6) Substitute the Marshallian demands back into the utility function to obtain the indirect utility function.
Answer:    Simply making the appropriate substitution in the three cases given above gives

,
．
v(p, w) = ．
．

w+p2
p2
(w+pl +p2 )2 4pl p2
w+pl
pl

if w < p1 - p2
if w > |p1 - p2 | .
if w < p2 - p1

(7) State Roy’s Theorem and verify that Roy’s Theorem does indeed give the same Marshallian demands that you found above.
Answer:     Roy’s  Theorem states  that if the indirect  utility function is differentiable at (p, w) then

xe (p, w) = -

,u(p3w) ,p廷
,u(p3w) . ,w

We first consider the case that w < p1 - p2 . In this case v(p, w) =  =  + 1 and
∂v 
∂v          w 
=    -
∂p2             p2 2
∂v       1 
=
∂w     p2
giving x1 (p, w)  =  0 and x2 (p, w)  =  w/p2 2   as required.   The  case  that w < p2 - p1  is similar.

We shall calculate x1 (p, w) for the case that w > |p1 - p2 |.  The calcula- tion for x2 (p, w) is completely symmetric.
In this region we have
(w + p1 + p2 )2
4p1p2            .
Thus
∂v       2 (w + p1 + p2 ) 4p1p2
=
∂w                16p1 2p2 2                .
And
∂v       2 (w + p1 + p2 ) 4p1p2 - (w + p1 + p2 )2 4p2
=
∂p1                                                 16p1 2p2 2
(w + p1 + p2 ) (8p1p2 - 4p2 (w + p1 + p2 ))
=
16p1 2p2 2
=                         (-4p2 (w - p1 + p2 )) .

(-4p2 (w - p1 + p2 ))      w - p1 + p2
8p1p2                                     2p1           ,
as required.
(8) Invert the indirect utility function to find the expenditure function.
Answer:   Recall
,．             if w < p1 - p2
v(p, w) = ．     if w > |p1 - p2 | .
．．            if w < p2 - p1
The duality result says that v(p, e(p, u)) = u.  Thus if e(p, u) < p1 -p2  then
e(p, u) +p2
p2
or
e(p, u) = (u - 1)p2
and so, in this case we have
(u - 1)p2  < p1 - p2
or
u <
p2 .
Similarly if e(p, u) < p2 - p1  then
e(p, u) = (u - 1)p1
and, in that case it must be that
2

Suppose that neither of these cases hold so that e(p, u) > |p1  - p2 |.  Then
we have
(e(p, u) + p1 + p2 )2
4p1p2
or
e(p, u) + p1 + p2  = (4up1p2

or
e(p, u) = 2 ′up1p2 - p1 - p2
Let us consider the case that p1  > p2  so the condition becomes e(p, u) > p1 - p2 .  Thus we have that
2 ′up1p2 - p1 - p2  > p1 - p2
or
2 ′up1p2  > 2p1
or
up1p2  > p1 2
or
u >

p1 .
Thus we have

,
．
e(p, u) = ．
．

(u - 1)p2
2 ′up1p2 - p1 - p2 (u - 1)p1

if u < 
if u > max{ ,  } .
if u < 

(9)  Use Shephard’s Lemma to find the Hicksian demand functions from the expenditure function.
Answer:   Shephard’s Lemma says that if e(p, u) is differentiable then
∂e(p, u
∂pe     .
Thus we obtain
,
． ．
(h1 (p, u), h2 (p, u)) = ．
． ．

(0, u - 1)
╱） - 1, ）  - 1←
(u - 1, 0)

if u < 
if u > max{ ,  } . if u <