BEEM103 UNIVERSITY OF EXETER BUSINESS School January 2009

BEEM103
UNIVERSITY OF EXETER
BUSINESS School
January 2009
Sample exam
OPTIMIZATION TECHNIQUES
FOR ECONOMISTS
solutions
Duration : TWO HOURS
The paper has 3 parts. Your marks on the …rst part will be rounded down to 55 marks:
Your marks on the second part will be rounded down to 15 marks. Obviously, you cannot
get over 100 marks overall. This examination is closed book and no materials are allowed.
Full work must be shown on your script. Please write legibly.
Explanation to the external examiner: Part A consists of variations of exercises we have
done in class. It tests whether a student has achieved the goal of module, namely whether
she has a basic understanding and knowledge of optimization problems of various levels
and the underlying mathematical techniques, starting with algebra that most of them
screwed up the …rst time around. Part B test whether she can synthesize this knowledge
to solve more complex questions. Part C consists of hard questions checking for excellence.
1
Part A
(You can gain no more than 55 marks on this part.)
Problem 1 (5 marks) Rewrite the …rst term such that it is de…ned for h = 0 and x > 0:
Simplify the second.
p
p
h
i
x+h
x
3
3
2 3
3
ab3
a6 b 6
8a6
=
(8a)6
(x + h) x
Solution 1
p
p
p
p
p
p
p 2
p
2
x+h
x
x+h+ x
x+h
x
x+h
x
p
p
=
=
p
p =
(x + h) x
h x+h+ x
h x+h+ x
1
x+h x
p
= p
p
p
h x+h+ x
x+h+ x
p
p
x+h
x
(Notice that limh!0 (x+h)
= limh!0
x
p
that the derivative of x is p12x .)
h
ab3
3
a6 b 6
2
i3
( a)
3
2a6
9
b
=
27 36 36
a b
p
1 p
x+h+ x
(2a)6
3
26 a6
2a6
3
=
p 1p
x+ x
=
= ( 1)3
= a27 b9
1
p
.
2 x
3 3
ab3
25
3
This calculation proves
a6 b 6
2 3
= 32 768 a3 b
(2a)6
2a6
!3
9
Problem 2 (5 marks) Solve
ln x3 + ln y 2 = ln 2 + ln 36
ln x + ln y = ln 6
Solution 2 We multiply the second equation by the factor 2 and then subtract the second
from the …rst equation.
3 ln x + 2 ln y
ln x + ln y
2 ln x + 2 ln y
ln x
x
ln 2 + ln y
ln y
y
=
=
=
=
=
=
=
=
ln 2 + ln 36
ln 6
2 ln 6 = ln 26
ln 2
2
ln (2 3) = ln 2 + ln 3
ln 3
3
Problem 3 (10 marks) Consider the function
y (x) =
2
1
1 + x2
=
i) Calculate and draw a sign diagram for the …rst derivative. Where is the function
increasing or decreasing? Are there any peaks or troughs? Does the function have a
(global) maximum. Is the function quasi concave?
ii) Calculate and draw a sign diagram for the second derivative. Where is the function
convex or concave. Are there any in‡ection points?
Solution 3
0.8
0.6
y
0.4
0.2
-6
-4
-2
0
2x
4
2
a) y 0 = 2x (1 + x2 ) , unique critical point zero, y 0 positive to the left and negative to
the right; function increasing to the left, decreasing to the right. Thus x = 0 unique maxdh
= (1 u) 2 > 0.
imum. The function y = h (u) = 1 1 u is increasing for positive u since du
The function u = g (x) = x2 is concave because its second derivative, 2, is negative everywhere. As a monotone transformation of a concave function the function
1
y = h (g (x)) = 1+x
2 is quasi concave.
b)
y 00 =
=
2 1 + x2
2
2
2x
(1 + x2 ) 4x2
=
(1 + x2 ) 3
2 1 + x2
2
3
2x
1 3x2
(1 + x2 ) 3
The second derivative is zero at x = p13 ; in between it is negative, outside positive.
x = p12 are hence in‡ection points with the function concave in between the roots and
convex outside.
Problem 4 (5 marks) For the function
1
y (x) = x3
3
x
…nd the (global) maxima and minima a) on the interval [ 2; 0] and b) on the interval
[ 2; 2] :
Solution 4 y 0 = x2 1 = 0 for x = 1. these are the critical points. y 00 = 2x positive
for x > 0 and negative for x < 0. Function is concave for x < 0 and convex for x > 0. It
3
has an in‡ection point at x = 0. The previous calculations also show that x =
local minimum and x = 1 a local maximum. We have
y ( 2) =
2
2
, y ( 1) = , y (0) = 0, y (1) =
3
3
1 is a
2
2
, y (2) =
3
3
a) We see that the maximum of the function on the interval [ 2; 0] is at x = 1 and the
minimum at x = 2. Over the interval [ 2; 2] we have two maxima, at x = 1 and at
x = 2, and two minima, namely x = 2 and at x = 1.
1
0.8
0.6
y
0.4
0.2
-2
-1
0
-0.2
1x
2
-0.4
-0.6
-0.8
-1
Problem 5 (5 marks) Find the equation of the tangent plane of
z (x; y) = x2 y 2
at the point (x ; y ; z ) = (1; 4; z (1; 4)).
Solution 5 z = 16
@z
= 2xy 2 ,
@x
@z
= 2x2 y,
@y
@z
=2
@x jx=1;y=4
@z
=2
@x jx=1;y=4
16 = 32
4=8
The formula for the total di¤erential at x = 1; y = 4
dz =
@z
@z
dx +
dy
@x jx=1;y=4
@y jx=1;y=4
yields the formula for the tangent
(z
16) = 32 (x 1) + 8 (y 4)
z = 32x + 8y 64 + 16 = 32x + 8y
Problem 6 (10 marks) Show that the function
u (x; y) = ln (5x + y)
is concave.
4
5 (x + y)2
48
Solution 6 We have
@u
@x
@u
@y
@2u
@x2
@2u
@x@y
@2u
@y 2
Set a =
p
=
=
=
=
=
5
10 (x + y)
5x + y
1
10 (x + y)
5x + y
25
10 < 0
(5x + y)2
5
10 < 0
(5x + y)2
5
10 < 0
(5x + y)2
10 and b = (5x + y) 1 . The Hessian of u is
H=
a2
a2
25b2
5b2
a2
a2
5b2
b2
and hence
det H = a2 + 25b2 a2 + b2
a2 + 5b2
= a4 + 26a2 b2 + 25b4 a4 10a2 b2
= 16a2 b2 = (2a)2 (2b)2 > 0
2
25b4
We see that the leading principle minors of the Hessian have the right sign and so u is
concave,pas was to be proved. (Remark: One could have worked here with A = 10 instead
of a = 10 and then get det H = 16Ab2 . We used the notation only for comparability
with other exercises.)
Problem 7 (10 marks) A consumer has the utility function
u (x; y) = 5x2 + 6xy + y 2 + 38x + 19y
a) Determine the marginal utility for the two commodities. Is more always better for the
consumer?
b) The consumer has a budget of £ 90. A unit of the …rst commodity costs £ 10 and a
unit of the second £ 20. Write down the budget equation.
c) The consumer wants to maximize his utility subject to his budget constraint. Write
down the Lagrangian for this problem.
d) Calculate the …rst order conditions for a critical point of the Lagrangian.
e) Assume only the budget constraint binds. Derive a linear equation to be satis…ed by
a critical point that does not involve the Lagrange multiplier for the budget constraint.
f) Use the equation from e) and the budget equation to …nd the constrained optimum.
5
Solution 7 a)
@u
= 10x + 6y + 38
@x
@u
= 6x + 2y + 19
@y
yes, more is always better. b)
L =5x2 + 6xy + y 2 + 38x + 19y + (90
20x
10y)
c etc) The FOC for the Lagranngian yield
@L
= 10x + 6y + 38 20 = 0
@x
5x + 3y + 19 10 = 0
@L
= 6x + 2y + 19 10 = 0
@y
5x + 3y + 19 = 10 = 6x + 2y + 19
5x + 3y = 6x + 2y
y = x
Together with the budget equation we obtain
90
x = y
10y = 0
20x
Solution is: [x = 3; y = 3]. (Notice that
be.)
= (5
3+2
3 + 19) =10 = 4 > 0; as it should
Problem 8 (5 marks) Consider a price-taking …rm with total cost function T C (Q; w)
C
where w is the wage rate. Assuming @T
> 0 and an interior equilibrium show that the
@w
pro…ts of the pro…t maximizing …rm decrease if the wage rate increases.
Solution 8 The pro…t function is for the given price p of output
(Q; t) = Qp
and hence
T C (Q; w)
@
@T C
=
<0
@w
@w
By the envelope theorem a marginal increase in w changes pro…ts by @@w , so pro…t decreases. (More precisely, let Q (w) be the optimal quantity at wage rate w determined
by
@
@T C
=p
= p M C (Q) = 0
@Q
@Q
6
and let
(w) = max
Q
(Q; w) =
(Q (w) ; w) :
Then, by the chain rule,
d
@
=
dw
@Q
dQ
@
+
dw
@w
dw
@
=
=
dw
@w
@T C
< 0.
@w
The envelope theorem is just this application of the chain rule to a function obtained by
optimization like here
(w).)
Problem 9 (10 marks) Solve the problem
max
ut 2[0;+1]
T
X
[ ut + ln xt ] , xt+1 = ut xt for t = 0;
;T
1, x0 given.
t=0
Solution 9 The value function takes the form
Vt (x) = (T
t + 1) ln xt + Ct
with some constant Ct . This holds for t = T because uT is maximized over [0; 1] at 0.
Suppose it holds for t + 1 with 0 t < T . Then we must …nd
Vt (xt ) =
=
=
max [Wt (xt ; ut )]
ut 2[0;+1]
max [ ut + ln xt + Vt+1 (xt ut )]
ut 2[0;+1]
max [ ut + ln xt + (T
ut 2[0;+1]
(t + 1) + 1) ln [ut xt ] + Ct+1 ]
The FOC yields
T t+1
@Wt
= 1+
= 0 , ut = T t + 1
@ut
ut
ut = 1=2 is clearly the maximum and the claim holds because
Vt (xt ) = ln xt + (T
t) ln xt + Ct = (T
t + 1) ln xt + Ct
The xt .are inductively de…ned by
x t = x 0 u0 u 1 : : : ut
= (T t + 1)!x0
(Hint: Find the value function gradually for T = 2 and t = 2; 1; 0. I will give almost full
marks in the exam if you solve the case for T = 2:)
Problem 10 (5 marks) Find a solution to the di¤erential equations
dx
= xt, x (0) = 1
dt
d2 x
b) 2 = x, x (0) = 0, x0 (0) = 1
dt
a)
7
Solution 10 a)
dx
= tdt
Z
Z x
dx
=
tdt
x
1 2
ln x =
t +C
2
1 2
x = e 2 t +C
1 2
Since x (0) = 1 we have C = 0. The solution is x (t) = e 2 t :
b) The general solution to x• (t) = x (t) is x (t) = Aet + Be t with arbitrary constants.
(Check and remember for future use! By the way, the general solution to x• (t) = x (t) is
A sin (t) + B cos (t). ) Since we require x (0) we have
x (0) = A + B = 0, so A =
B
Therefore
x_ (t) = Aet Be t = A et + e t
x_ (0) = A (1 + 1) = 1, so A = 1=2 and
et e t
x (t) =
2
Problem 11 (10 marks) Solve the problem
Z
max
1 4x 2u2 dt, x_ = u, x (0) = 0, x (1) free:
Solution 11 The Hamiltonian is
H= 1
4x
2u2 + pu
The maximum of the Hamiltonian with respect to u is given by the FOC 4u + p = 0, so
p = 4u. The solution to p_ = Hx is p_ = 4. Thus p (t) = 4t + C. Because x (1) is free
we must have p (1) = 0 and so p (t) = 4 (1 t). Hence u (t) = 41 p (t) = 1 t. We obtain
x_ (t) = 1 t, x (0) = 0
Z t
x (t) =
(1
) dt = 2t
t2 + C
0
with C = 0.
Part B
(You can gain no more than 15 marks on this part.)
Problem 12 (15 marks) For a consumer with the utility function
u (x; y) = u (x; y) = 5x2 + 6xy + y 2 + 38x + 18y
derive his demand function. What is the marginal increase in utility if the budget of the
consumer is slightly increased?
8
Solution 12 This one took me quite a bit of work to solve completely. for a simpler
example check the lecture handout for week 5. The following estimate turns out to be
essential:
Claim 1 For all ;
0 we have
1
3
2 + 18
9
6 + 18
<
<
6 + 38
18
10 + 38
3
5
To see this we calculate
@ 2 + 18
@
6 + 38
@
6 + 18
@
10 + 38
32
>0
(6 + 38 )2
12
=
<0
(5 + 19 )2
=
Thus the …rst functional expression in the above inequality achieves it minimum 62 at
= 0;is then increasing in
and never reaches its limit 18
at +1. The second is
38
6
decreasing in and hence has its maximum 10
at = 0 and never falls below its limit 18
38
at +1.
Now to the question. Let us write the budget constraint as
px x + py y
b
where we assume px ; py ; b > 0.
The utility function
u (x; y) = 5x2 + 6xy + y 2 + 38x + 18y
has the partial derivatives
@u
= 10x + 6y + 38
@x
@u
= 6x + 2y + 18
@y
which, as we have seen in a previous question, is always positive and so utility is monotonic.
Therefore the budget constraint must be binding in the consumer optimum and there are
only three possibilities. Either none of the non-negativity constraints x 0 and y 0 is
binding or exactly one of them is binding.1
The Lagrangian for the constrained optimization problem is is
L = u (x; y) +
1
(b
px x
1
py y) +
2x
+
3y
Sometimes I use in exercises utility functions with an “ideal consumption bundle” xi ; y i
(0; 0)
which is an absolute macimum of the utility function, so more is not always better. In this case there is
a fourth possibility, namely if the ideal consumption bundle is a¤ordable, then it is the optimum and no
constraint, including the budget constraint, is binding at the optimum.
9
the two FOC are
@L
@u
=
@x
@x
@L
@u
=
@x
@y
1 px
2
=0
1 py
3
=0
As indicated, we must distinguish 3 cases:
(A) The non-negativity constraint x = 0 is binding. Since the budget constraint must
also be binding we get x = 0 and y = b=py > 0 in the optimum. Moreover, we must
have 3 = 0 by the complementarity conditions and so the second FOC yields
1
The …rst FOC and the requirement
2
=
1 px
@u px
@u
=
@x
@y py
=
@u
=py > 0
@y
2
0 yield
@u
@x
0,
py
px
@u
@y
@u
6x + 2y + 18
=
@x
10x + 6y + 38
where x and y must be taken at their optimal values x = 0 and y = b=py > 0, so
2b=py + 18
6b=py + 38
py
px
(A)
The condition expresses that the marginal rate of substitution exceeds the price ratio at
the point on the budget line where x = 0 and hence the latter is optimal. It is useful to
draw a picture of this case with y on the horizontal and x on the vertical axis. The slope
of the budget line in this case is py =px while the slope of the tangent to the indi¤erence
y +18
curve 2b=p
. See also my script for the lecture in week 5, where I wisely chose to look
6b=py +38
at a much simpler example.
(B) The non-negativity constraint y = 0 is binding. Since the budget constraint must
also be binding we get x = b=px and y = 0 in the optimum. Moreover, we must have
2 = 0 by the complementarity conditions and so the …rst FOC yields
1
=
@u
=px > 0
@x
The second FOC and the requirement
3
0 yield
3
=
1 py
@u
@u py
=
@x
@y px
@u
@x
0,
py
px
@u
@y
6x + 2y + 18
@u
=
@x
10x + 6y + 38
where x and y must be taken at their optimal values,so
py
px
6b=px + 18
10b=px + 38
(C) Only the budget constraint binds. In this case
conditions.We rewrite the budget equation as
x+
py
b
b
y=
,x=
px
px
px
10
(B)
2
=
3
py
y
px
= 0 by the complementarity
and get from the two FOC
py
= 1 py = (6x + 2y + 18)
px
b
py
py
b
py
y + 6y + 38
= 6
y + 2y + 18
px px
px
px px
!
2
py
py
b
py
b
10
12 + 2 y =
10 + 38
6
px
px
px
px
px
(10x + 6y + 38)
10
py
px
py
2 5
px
2 5
1
py
px
1
1
py
px
py
px
py
2 5
px
1
py
px
1
1
10
y =
I have to admit that it took me a while to
Similarly
py
(10x + 6y + 38)
px
b
px
py
10x + 6
x + 38
py
py
px
px
py
x
12x + 2
10
px
py
!
2
py
py
10
12 x + 2 x
px
px
2 5
b
+ 38
px
b
6
18
px
1 y =
6
10
b
px
18
b
+ 38
px
py
px
(C)
see the factorization on the left.
=
1 py
= (6x + 2y + 18)
px
x + 18
py
b
py
b
6 + 38
+ 2 + 18
py
px
py
= 6x + 2
=
=
6
b
py
b
b
+ 38 + 2 + 18
py
py
px
py
b
b
px
+ 38 + 2 + 18
py
py
py
b
px
b
2 + 18
6 + 38
py
py
py
1 x =
py
px
py
px
18
6
x =
(D)
Equations (C) and (D) give us the optimum under the assumption that no non-negativity
constraint is binding. However, the assumption may not hold and this would show in
either x or y being negative. So we have to …nd out for which values of px ; py and b the
two equation give non-negative values for x and y. This is a bit too tricky for an exam
question, but we can do it in a sample exam.
Suppose the product in front of x or y in formulae (C) or (D) is negative. Then either
py
1
<
or ppxy > 1. In the …rst case we have, since 15 < 13 ; by the claim shown in the
px
5
beginning of the solution that
py
2 + 18py =b
2b=py + 18
<
=
,
px
6 + 38py =b
6b=py + 38
6
b
+ 38
py
2
b
+ 18
py
px
<0
py
But then equation (D) implies that x is negative in the optimum, we don’t get an admissible solution. Similarly, ppxy > 1 > 53 implies
py
6 + 18px =b
6b=px + 18
>
=
,
px
10 + 38px =b
10b=px + 38
11
6
b
px
18
10
b
+ 38
px
py
<0
px
which leads to the contradiction that y is negative in optimum. We conclude that both
left hand side and right hand side in equations (C) and (D) must be positive for the
solution (x ; y ) to be positive. In addition, the two terms on the right of the equations
must be positive which, as is shown by calculations as above, holds if and only
2b=py + 18
6b=py + 38
By the claim we have then also
x
y
1
5
<
py
px
py
px
2b=py + 18
6b=py + 38
< 1. We see that the formulae
b
py
3 pby + 19
=
5 ppxy
6 pbx
=
(*)
1
px
py
py
px
1
10 pbx + 38
18
5 ppxy
+9
1
1
py
px
py
px
yield an admissible solution if and only if (*) holds. As was to be expected, (*) is just
the algebraic condition expressing that we are not in in the cases (A) or (B).
Concerning the last question in the problem, let u (px ; py ; b) be the optimal utility
achievable for the given budget. Then, by the envelope theorem
@L
@u
=
=
@b
@b
1
so a marginal increase in the budget raises utility by the value of the Lagrange multiplier
for the budget constraint. In our example this yields in case (A)
=
2b=py + 18
@u
=py =
@y
px
=
@u
10b=px + 38
=px =
@x
py
1
in case (B)
1
and in case (C)
1
=
@u
6x + 2y + 18
=py =
@y
pY
where you are cordially invited to substitute the above expressions and, if possible, simplify.
Problem 13 (15 marks) Solve the problem
max
1
X
1
ln (xt
t
3
t=0
ut ) subject to xt+1 = ut , x0 > 0, ut > 0
12
t
t
Solution 13 H = 31t ln (x u) + pu. FOC: @H
= pt x1=3u = 0 and pt 1 = @H
= x1=3u .
@u
@x
t
t
t
t
0 must hold for any admissible
Also the transversality condition limt!1 pt (xt xt )
sequence (xt ) and we must have xt+1 = ut , x0 = x0 . We obtain from the …rst two
conditions pt = pt 1 = p, so, since xt+1 = ut
xt
ut = x t
xt+1 =
xt+1 = xt
1
p
1
3
= x0
1
p
1
1=3t
p
t
1
p
= : : : = x0
1
1 t+1
3
1
3
!
1
3
2
3p
= x0
0
+
1
3
1
3
1
1
+
t+1
+
!
1
3
t
!
One can choose an admissible sequence xt > 0 with lim xt arbitrarily small. Transversality
implies hence (since xt > 0 must hold for all t) that limt!1 pt xt = 0. Now
!!
t
2
1
2
2
pxt = p x0
1
! p x0
= px0
=0
3p
3
3p
3
so p =
2
3x0
and then xt = 1=3t x0 , ut = 1=3t+1 x0 :
Part C
Problem 14 (20 marks) Sketch the graph of the region A carved out by the two inequalities
x2 + y 2
y
x; y
4
x
0
For any point (a; b) with a; b 0 in the plane use the Lagrangian approach to determine
the point closest to (a; b) within or on the boundary of A.
Solution 14 It is easy to see that we can ignore the non-negativity constraints in the
exercise. This leaves four cases must be considered with respect to the other two constraints: No constraint is binding, one of the two is binding, or both are binding. Instead
of minimizing the distance we can maximize the negative of the square of the distance
and so the Lagrangian is
L=
(x
a)2
(y
b)2 +
1
4
x2
y2 +
the FOC are
2 (x
2 (y
a)
b)
2 1x +
2 1y
13
2
2
= 0
= 0
2
(x
y)
A) If no constraint is binding we set 1 = 2 = 0 and obtain x = a, y = b. This is the
solution if (a; b) is in the area C. B) Suppose only the …rst constraint binds (x2 + y 2 = 4
in the optimum) and so 2 = 0. Then
(x
a)
a
b)
b
(y
where
1
= 1x
= (1 +
= 1y
= (1 +
1) x
1) y
must be non-negative. Thus the optimum must be given by
1
1+
1
=
1+
x
=
y
a
1
b
1
We see that (x ; y ) is the vector of length 2 in the direction of the vector (a; b), which
must have has length larger than 2:Since y
x we must also have b
a . C) Suppose
next that only y = x binds. Then 1 = 0 and so
(x
(y
a) =
b) =
2
2
or
a+b
2
a = b 2 a 0 we must have b a. To be admissible we must also have
Since 2 = a+b
2
p
p
2
2
(x ) + (x )
4 and so x
2, which implies a + b 2 2.
p
In the last case both constraints are binding and hence x = y = 2. We have
(x
a) =
(x
p
a
2
2
p
2
2
b
b) , x = y =
p
2 2
p
2 2
1
1
+
2
= 0
2
= 0
Addition yields
p
4 2 + 2 (a + b)
and so a + b
so b
p
2 2. Moreover
p
p
2 a +2 2
2 = 2
p
4 2
p
2 2
1
=
1
1
= 0
= (a + b)
p
2 2
2a + a + b = b
a
0,
a. We …nd four regions for the four di¤erent cases, as indicated in the following
14
graph.
y
5
D
4.5
4
3.5
3
2.5
2
1.5
C
1
B
A
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
x
Region A is the region carved out by the constraints. In areas B and C one and in area D
both constraints are binding in the optimum. Four points, one for each area, are shown.
For the points in areas B,C,D a line connects the point to the closest point in (and on the
boundary of) region A.
15