Chapter 3 (pp. 90-99): Existence of Solutions.

1. As a counterexample consider f : (0, 1) R such that f (x) = x1 :

sup f (x) = limx0
(1,0)

1
= +.
x

2. It can be shown (e.g. by construction) that in R (in any completely ordered set)
sup and inf of any finite set coincide respectively with max and min.
If D Rn is finite also f (D) = {x R | x D s.t. f (x) = x} is.
The result is implied by the Weierstrass theorem because every finite set A is compact, i.e. every sequence in A have a constant (hence converging) subsequence.
3. a) Consider x
= max{D}, which exists because D is compact subset of R,
f (
x) is the desired maximum, because x D such that x x
,
= x D such that f (x) f (
x);
b) consider D {(x, y) R2+ | x + y = 1} R2 ,
D is the closed segment from (0, 1) to (1, 0), hence it is compact,
there are no two points in D which are ordered by the partial ordering,
hence every function D R is nondecreasing,
consider:
f (x, y) =
max f on D is limx0 f (x, y) = limx0

1
x

if x = 0
0 if x = 0

1
x

= +.

4. A finite set is compact, because every sequence have a constant (hence converging) subsequence.
Every function from a finite set is continuous, because we are dealing with the
discrete topology, and every subset is open (see exercise 11 of Appendix C for the
correct statement of Corollary C.23 (page 340)).
We can take a subset A of cardinality k in Rn and construct a one-to-one function
assigning to every element of A a different element in R.
A compact and convex subset A Rn cannot have a finite number k 2 of
elements (because otherwise aA ak is different from every a A but should be
in A).
Suppose A is convex and a, b A such that f (a) = f (b), then f[a,b] () =
f (a + (1 )b) is a restriction of f : A R but can be also considered as a
function f[a,b] : [0, 1] R.
By continuity every element of [f (a), f (b)] R must be in the codomain of f[a,b]

(the rigorous proof is long, see Th. 1.69, page 60), which is then not finite, so that
neither the codomain of f is.
5. Consider sup(f ), it cannot be less than 1, if it is 1, then max(f ) = f (0).
Suppose sup(f ) > 1, then, by continuity, we can construct a sequence {xn
x s.t. x >
R+ | f (xn ) = sup(f ) sup(fn )1 , moreover, since limx f (x) = 0,
x
, f (x) < 1.
{xn } is then limited in the compact set [0, x], hence it must converge to x
. f (
x) is
however sup(f ) and then f (
x) = max(f ).
f (x) = ex , restricted to R+ , satisfies the conditions but has no minimum.
6. Continuity (see exercise 11 of Appendix C for the correct statement of Corollary
C.23 (page 340)) is invariant under composition, i.e. the composition of continuous
functions is still continuous.
Suppose g : V1 V2 and f : V2 V3 are continuous, if A V3 is open then, by
continuity of f , also f 1 (A) V2 is, and, by continuity of g, also g1 (f 1 (A)) V1
is.
g1 f 1 : V3 V1 is the inverse function of f g : V1 V3 , which is then also
continuous.
We are in the conditions of the Weierstrass theorem.
7.

1 x 0
g(x) =
x
x (0, 1)

1
x1

f (x) = e|x|

arg max(f ) = 0, max(f ) = e0 = 1 and also sup(f g) = 1, which is however never

attained.
8.
min p x subject to x D {y Rn+ | u(y) u
}
s.t. u : Rn+ R is continuous, p

a) p x is a linear, and hence continuous function of x,

there are two possibilities: (i) u is nondecreasing in x, (ii) u is not;
(i) D is not compact,
we can however consider a utility U and define WU {y Rn+ | u(y) U },
by continuity of u, WU is compact,
for U large enough D WU is not empty and the minimum can be found in it;

(ii) for the components of x where u is decreasing,

maximal utility may be at 0,
so D may be compact, if not we can proceed as in (i);
b) if u is not continuous we could have no maximum even if it is nondecreasing
(see exercise 3);
if pi < 0, and u is nondecreasing in xi , the problem minimizes for xi +.
9.
max p g(x) w x s.t. x Rn+
with p > 0, g continuous, w
a solution is not guaranteed because Rn+ is not compact.

10.
F (p, w) = {(x, l) Rn+1
| p x w(H l), l H}
+
F (p, w) compact = p

0 (by absurd):

if pi 0, the constrain could be satisfied also at the limit xi +,

F would not be compact;
p

0 = F (p, w) compact:

F is closed because inequalities are not strict,

l is limited in [0, H],
i {1, . . . n}, xi is surely limited in [0, w(Hl)
].
pi

11.
max U (y()) subject to { RN | p 0, y() 0}

S
where ys () = s + N
i=1 i zis , U : R+ R is continuous and strictly increasing
(y
y in RS+ if yi yi i {1, . . . , S} and at least one inequality is strict).

Arbitrage: RN such that p 0 and Z

0.

A solution exists no arbitrage

(=) (by absurd: (A = B) (B = A) )
suppose RN such that p 0 and Z t
0, then any multiple k of
has the same property,

y(k ) increases linearly with k and then also U (y(k )) increases with k,
max U (y(k )) is not bounded above;
(=) RN such that p 0 there are two possibilities:
either N
i=1 i zis = 0 s {1, . . . S},
or s {1, . . . S} such that N
i=1 i zis < 0;
in the first case the function of k U (y(k )) : R R is constant,
then a maximum exists,
in the second case y() 0 impose a convex constraint on the feasible set,
by Weierstrass theorem a solution exists.
12.
max W u1 (x1 , h(x)), . . . un (xn , h(x))
n
n+1

with h(x) 0, (x, x) [0, w]

, wx=

xi
i=1

sufficient condition for continuity is that W , u and h are continuous,

the problem is moreover well defined only if x [0, w] such that h(x) 0;
the simplex w {(x, x) [0, w]n+1 | x + ni=1 xi = w} is closed and bounded,
if h(x) is continuous H = {(x, x) [0, w]n+1 | h(x) 0} is closed because the
inequality is not strict,
then the feasible set w H is closed and bounded because intersection of closed
and bounded sets.
13.
max (x) = max xp(x) c(x)
xR+

with p : R+ R+ and c : R+ R+ continuous, c(0) = 0 and p() decreasing;

a) x > 0 such that p(x ) = 0, then, x > x , (x) (0) = 0,
the solution can then be found in the compact set [0, x ] R+ ;
b) now x > x , (x) = (0) = 0, as before;
c) consider c(x) = 2p x, now (x) =
not bounded above.

p
2

x, so that the maximization problem is

14. a)
max v(c(1), c(2)) subject to c(1)

0, c(2)

10

0, p(1) c(1) +

p(2)
c(2) W0
1+r

b) we can consider, in R2n , the arrays c = (c(1), c(2)) and p = (p(1), p(2)
1+r ), the
conditions
c(1)

0, c(2)

0, p(1) c(1) +

p(2)
c(2) W0
1+r

become
c

0, p c W0

we are in the conditions of example 3.6 (pages 92-93), and p

p(1)
0 and p(2)
0.

0 if and only if

15.

max (x1 ) + (x2 ) + (x3 )

0 x1 y 1
0 x2 f (y1 x1 )
subject to

0 x3 f (f (y1 x1 ) x2 )

lets call A the subset of R3+ that satisfies the conditions, A is compact if it is closed
and bounded (Th. 1.21, page 23).
It is bounded because
A [0, y1 ] [0, max f ([0, y1 ])] [0, max f ([0, max f ([0, y1 ])])] R3+
and maxima exist because of continuity.
Consider (
x1 , x
2 , x
3 ) Ac in R3+ , then (if we reasonably suppose that f (0) = 0),
becasue of continuity, at least one of the following holds:
x
1 > y1 , x
2 > f (y1 x
1 ) or x
3 > f (f (y1 x
1 ) x
2 ).
If we take r = max{
x1 y 1 , x
2 f (y1 x
1 ), x3 f (f (y1 x
1 ) + x
2 )}, r > 0 and
c
B((
x1 , x
2 , x
3 ), r) A .
Ac is open = A is closed.

11

Chapter 4 (pp. 100-111): Unconstrained Optima.

1. Consider D = [0, 1] and f : D R, f (x) = x:
arg maxxD f (x) = 1, but Df (1) = 1 = 0.
2.
2

f (x) = 1 2x 3x = 0 for x =

1+3
=
3

1
1
3

f (1) = 4
f ( 13 )
= 4

f (x) = 2 6x ,

1 is a local minimum and 13 a local maximum, they are not global because
limx = + and limx+ = .
3. The proof is analogous to the proof at page 107, reverting the inequalities.

4. a)
fx = 6x2 + y 2 + 10x
fy = 2xy + 2y
D2 f (x, y) =

= y = 0 x = 0

12x + 10
2y
2y
2x + 2

(0, 0) is a minimum;
b)
fx = e2x + 2e2x (x + y 2 + 2y) = e2x (1 + 2(x + y 2 + 2y))
fy = e2x (2y + 2)

= y = 1

1
1
1
= f ( , 1) = e
2
2
2
since limx f (x, 1) = 0 and limx+ f (x, 1) = +, the only critical point
( 21 , 1) is a global minimum;
1 + 2(x + 1 2) = 0 = x =

c) the function is symmetric, limits for x or y to are +, a R:

fx = ay 2xy y 2
fy = ax 2xy x2

1
1
2
2
= (0, 0) (0, a) (a, 0) ( a, a) ( a, a)
5
5
5
5

D2 f (x, y) =

2y
a 2x 2y
a 2x 2y
2x

12

( 52 a, 51 a) and ( 15 a, 52 a) are local minima a R;

d) when sin y = 0, limits for x are ,
fx = sin y
fy = x cos y

= (0, ky) k Z = {. . . , 2, 1, 0, 1, . . .}

f is null for critical points, adding any ( , ) to them ( < 2 ), f is positive,

adding ( , ) to them, f is negative,
critical points are saddles;
e) limits for x or y to are +, a R,
fx = 4x3 + 2xy 2
fy = 2x2 y 1

= y =

1
1
= 4x3 = = impossible
2
2x
x

there are no critical points;

f) limits for x or y to are +,
fx = 4x3 3x2
fy = 4y 3

3
= (0, 0) ( , 0)
4

since f (0, 0) = 0 and f ( 43 , 0) < 0, ( 34 , 0) is a minimum and (0, 0) a saddle;

g) limits for x or y to are 0,
fx =
fy =
since f (1, 0) =
minimum;

1
2

1x2 +y 2
(1+x2 +y 2 )2
2xy
(1+x2 +y 2 )2

= (1, 0) (1, 0)

and f (1, 0) = 21 , the previous is a maximum, the latter a

h) limits for x to are +,

limits for y to depends on the sign of (x2 1),
3

fx = x8 + 2xy 2 1
fy = 2y(x2 1)

D 2 f (x, y) =

3 2
8x

y = 0 = x = 2

x = 1 = y = 47

x = 1 = impossible

+ 2y 2
4xy
4xy
2y(x2 1)

13

D2 f (2, 0)
D2 f (1,

3
2

0
0 0

7
4 )

D2 f (1,

= saddle,

5
7

= saddle.
0
7

7
4 )

4
7

7
0

= saddle,

5. The unconstrained function is given by the substitution y = 9 x:

f (x) = 2 + 2x + 2(9 x) x2 (9 x)2 = 2x2 + 18x 61
which is a parabola with a global maximum in x =

9
2

= y = 92 .

6. a) x is a local maximum of f if R+ such that y B(x , ), f (y) f (x ),

considering then that:
lim inf
f (x ) f (y) =

yx

lim inf

yB(x , )x

f (x ) f (y) 0

y < x , x y > 0 y > x , x y < 0

1 lim inf = lim sup
we have the result;
b) a limit may not exist, while lim inf and lim sup always exist if the function
is defined on all R;
c) if x is a strict local maximum the inequality for lim inf yx is strict, so also
the two inequalities to prove are.
7. Consider f : R R:
f (x) =

1 xQ
0 otherwise

where Q R are the rational number; x = 0 is a local not strict maximum, f is

not constant in x = 0.
8. f is not constant null, otherwise also f would be constant null,
since limy f (y) = 0, f has a global maximum x
, with f (
x) > 0 and f (
x)
x is the only point for which f (x) = 0 = x = x
.
9. a) > 0:
d f
df
= (f )
,
dxi
dxi

df
d f
= 0 =
=0
dxi
dxi

Df (x ) = 0 = D f (x ) = 0

14

b)
df
df df
d
d2 f
d2 f
(f )
= (f )
=
+ (f )
dxi dxj
dxj
dxi
dxi dxj
dxi dxj
df
d2 f
d2 f
= 0 =
= (f )
dxi
dxi dxj
dxi dxj
Df (x ) = 0 = D 2 f (x ) = (f (x )) D 2 f (x )
a negative definite matrix is still so if multiplied by a constant.

10. Let us check conditions on principal minors (see e.g. Hal R. Varian, 1992,
Microeconomic analysis, Norton, pp.500-501).

f x1 x1 . . . fx1 xn

..
..
D2 f (x) = ...
.
.
f x1 xn . . . fxn xn

is positive definite if:

1) fx1 x1 > 0,
2) fx1 x1 fx2 x2 fx21 x2 > 0,
...
n) |D 2 f (x)| > 0;

fx1 x1 . . . fx1 xn
gx1 x1 . . . gx1 xn

..
..
.. = D 2 f (x) =
..
..
D2 g(x) = ...

.
.
.
.
.
fx1 xn . . . fxn xn
gx1 xn . . . gxn xn

is negative definite if:

1) fx1x1 < 0,
2) (fx1 x1 ) (fx2 x2 ) (fx1 x2 )2 > 0,
...
n) |D 2 f (x)| < 0 if n is odd, |D 2 f (x)| > 0 if n is even;

the ith principal minor is a polinimial where all the elements have order i,
it is easy to check that odd principal minors mantain the sign of its arguments,
while even ones are always positive,
hence D 2 f (x) positive definite = D 2 f (x) negative definite.

15

Chapter 5 (pp. 112-144): Equality Constraints.

1. a)
L(x, y, ) = x2 y 2 + (x2 + y 2 1)

=0

Lx = 2x + 2x = = 1 x = 0
=0

Ly = 2y + 2y = = 1 y = 0
=0

L = x2 + y 2 1 =

x = 0 y = 1 = 1
y = 0 x = 1 = 1

when = 1 we have a minimum, when = 1 a maximum;

b)
f (x) = x2 (1 x2 ) = 2x2 1 =

min for x = 0
max for x =

the solution is different from (a) because the right substitution is y

admissible only for x [1, 1].

1 x2 ,

2. a) Substituting y 1 x:
f (x) = x3 (1 x)3 = 3x2 3x + 1 = max for x =

b)

L(x, y, ) = x3 + y 3 + (x + y 1)
=0

Lx = 3x2 + = = 3x2
=0

Ly = 3y 2 + = = 3y 2

= x = y
x=y

L = x + y 1 = x = y =
x=y=

1
2

is the unique local minimum, as can be checked in (a).

1
2

3. a)
L(x, y, ) = xy + (x2 + y 2 2a2 )

=0

Lx = y + 2x = (y = 0 = 0) y = 2x
=0

Ly = x + 2y = (x = 0 = 0) x = 2y

x = 0 y = 2a

=0
L = x2 + y 2 2a2 =
y = 0 x = 2a

y = 2x x = 2y

16

(0, 2a) and ( 2a, 0) are saddle points, because f (x, y) can be positive or negative for any ball around them;
y = 2x x = 2y imply:
= 12 , x = y, and x = a,
the sign of x does not matter, when x = y we have a maximum, when x = y a
minimum.
b) substitute x
x1 , y

1
y

and a

1
a

L(
x, y, ) = x
+ y + (
x2 + y2 a
2 )

1
=0
Lx = 1 + 2
x = x
=
2
1
=0

Ly = 1 + 2
y = y = = x
2
2
=0
a
L = x
2 + y2 a
2 = x
= y =
2

for x
and y negative we have a minimum, otherwise a maximum.
c)
L(x, y, x, ) = x + y + z + (x1 + y 1 + z 1 1)

=0
Lx = 1 x2 = x =

=0
. . . = y =

=0
. . . = z =

we have a minimum when they are all negative (x = y = z = 3) and a maximum

when all positive (x = y = z = 3).
d) substitute xy = 8 (x + y)z = 8 (5 z)z:
f (z) = z(8 (5 z)z) = z 3 5z 2 + 8z
2

=0

fz = 3z 10z + 8 = z =

25 24
=
3

as z , f (z) ,
fzz = 6z 10 is negative for z = 34 (local maximum)
and positive for z = 2 (local minimum),

17

2
4
3

when z = 34 , x + y = 5 z = 11
3 and xy = 8 (x + y)z =
= one is also 43 and the other 73 ,

28
9

when z = 2, x + y = 5 z = 3 and xy = 8 (x + y)z = 2

= one is also 2 and the other is 1;
e) substituting y

16
x:

=0

= fx = 1 x162 = x = 4
f (x) = x + 16
x
when x = 4 and y = 41 we have a minimum, maxima are unbounded for limx0
and limx ;
f) substituting z 6 x and y 2x:
f (x) = x2 + 4x (6 x)2 = 16x 36 ,
f (x) is a linear function with no maxima nor minima.

4. Actually a lemniscate is (x2 +y 2 )2 = x2 y 2 , while (x2 y 2 )2 = x2 +y 2 identifies

only the point (0, 0).

In the lemniscate x + y maximizes, by simmetry, in the positive quadrant,

in the point where the tangent of the explicit function y = f (x) is 1;
for x and y positive the explicit function becomes:
(x2 + y 2 )2 = x2 y 2

x4 + 2x2 y 2 + y 4 x2 + y 2 = 0

y 4 + (2x2 + 1)y 2 + x4 x2 = 0
y

y = f (x) =

2x2 1 +

4x4 + 4x2 + 1 4x4 + 4x2

2

2x2 + 1 + 8x2 + 1
2

computing f (x) and finding where it is 1 we find the arg max = (x , y ),

(x , y ) will be the arg min.

18

5. a) (x 1)3 = y 2 implies x 1,
f (x) = x3 2x2 + 3x 1 = fx = 3x2 4x + 3 which is always positive for
x 1,
since f (x) is increasing, min f (x) = f (1) = 1 (y = 0);
b) the derivative of the constraint D((x 1)3 y 2 ) is (3x2 6x + 3, 2y) ,
which is (0, 0) in (1, 0) and in any other point satisfying the constraint,
hence the rank condition in the Theorem of Lagrange is violated.
6. a)
1
L(x, ) = c x + x Dx + (Ax b)
2
n
n
n
1
xj Dji xi +
ci xi +
=
2
i=1

i=1

Lxi = ci + Dii xi +

1
2

j=1
n

xj Dji +
j=1,j=i
m

i
i=1
n

j=1

Aij xj bi
m

j Aji

xj Dij +
j=1

j=1,j=i

j Aji

xj Dij +

= ci +

1
2

j=1

j=1
n

Li

=
j=1

Aij xj bi

b). . .
7. The constraint is the normalixation |x| = 1, the system is:
f (x) = x Ax
n

i=1
n

j=1

xj Aji xi

=
f xi = 2

xj Aji
j=1
n

j=1,j=i xj Aji

xi =

Aii

x =

A21
A11

..
.

0
..
.

...
...
..
.

A1n
Ann

A2n
Ann

...

0
A12
A22

such that |x | = 1

19

An1
A11
An2
A22

..
.
0

x B x

for any eigenvectors of the square matrix B, its two normalization (one opposite
of the other) are critical points of the problem;

A11 . . . An1

.. , x is constant,
..
since D2 f (x) = 2 ...
.
.

A1n . . . Ann
all critical points are maxima, minima or saddles, according wether A is positivedefinite, negative-definite or neither.
8. a)

dt
The function s(t) quantifying the stock is periodic of period T = x dI
,
RT

s(t)

x
in this period the average stock is 0 T = xT
2T = 2 ,
in the long run this will be the total average;

b)
L(x, n, )
Lx = C2h + n
Ln = C0 + x

x
+ C0 n + (nx A)
2
C0
Ch
=0
x=
= n =
2

Ch

=0

Ch C0
= A = =
22

L = nx A =

x and n negative have no meaning so must be negative.

Ch C0
2A

9. The condition for equality constraint to suffice is that u(x1 , x2 ) is nondecreasing,

this happens for 1 1;
if otherwise one of them , say is less than 1,
the problem is unbounded at limx1 0 x1 + x2 = +.
10.

20

min w1 x1 + w2 x2 s.t. (x1 , x2 ) X {(x1 , x2 ) R2+ | x21 + x22 1}

a)

if w1 < w2 it is clear from the graph that (1, 0) is the cheapest point in X,
similarly, if w2 < w1 , the cheapest point is 0, 1,
if w1 = w2 they both cost w1 = w2 ;
b) if nonnegativity constraints are ignored:
min w1 x1 + w2 x2 =

x21 +x22 1

lim

(x1 ,x2 )(,)

w1 x1 + w2 x2 =

similarly for (x1 , x2 ) (+, +) if w1 and w2 are positive.

11. [x 2 is not defined if x < 0. . . ]

1

max x 2 + y 2 s.t. px + y = 1

L(x, y, )
Lx =
Ln =

1 12
2x
1 12
2y

+ p
+

x 2 + y 2 + (px + y 1)

=0

= x = (2p)2 y = (2)2
=0

L = px + y 1 = p(2p)2 + (2)2 = 1

p
1
p2 + p
+
=
= 1 = =
42 p2 42
42 p2

p2 + p
2p

the sign of does not matter to identify x = p21+p and y =

the unique critical point, with both positive components,
1

x 2 + y 2 = 1+p
=
2
p +p

1+p
p

is greater e.g. than 0 2 + 1 2 = 1,

being the unique critical point it is a maximum.

21

p2
,
p2 +p

Chapter 6 (pp. 145-171): Inequality Constraints.

1. Since the function is increasing in both variables, we can search maxima on the
boundary,

substituting y = 1 x2 (which means x [0, 1] = y [0, 1]):

f (x) = ln x+ln

1 1 2x
1 2x2
1
=
1 x2 = ln x+ ln(1x2 ) = fx =
2
x 2 1 x2
(1 x2 )x

2
2
= x =
= y =
2
2
which is the argument of the maximum.
=0

2. The function is increasing, we search maxima on the boundary,

we can substitute to polar coordinates y x sin , z x cos :

f () =

x(py sin + pz cos ) = f =

=0

py
sin
=
cos
pz

x(py cos pz sin )

y
py
=
z
pz

economically speaking, marginal rate of substitution equals the price ratio,

sin =

p2y

py
+ p2z

cos =

p2y

pz
+ p2z

y= x

p2y

py
+ p2z

z= x

p2y

pz
+ p2z

py pz
py
pz
x p2 +p2 ) is a maximum and ( x p2 +p
( x p2 +p
x p2 +p2 ) is a minimum.
2,
2,
y

3. a) We can search maxima on the boundary determined by I:

max x1 x2 x3 s.t. x1 [0, 1], x2 [2, 4], x3 = 4 x1 x2
max f (x1 , x2 ) = 4x1 x2 x21 x2 x1 x22 s.t. x1 [0, 1], x2 [2, 4]
consider only x1 ,
4x2 x22
= 2 21 x2 ,
f (x1 ) = 4x1 x2 x21 x2 x1 x22 has a maximum for x1 = 2x
2
by simmetry a maximum is for x1 = x2 = 34 , which is however not feasible,
nevertheless, for x1 [0, 1] the maximum value for x2 [2, 4] is 2,
we get then x1 = 1 and x3 = 1;
b)

22

max x1 x2 x3 s.t. x1 [0, 1], x2 [2, 4], x3 =

6 x1 2x2
3

1
2
max f (x1 , x2 ) = 2x1 x2 x21 x2 x1 x22 s.t. x1 [0, 1], x2 [2, 4]
3
3
2x2 32 x22
2
x
3 2

f (x1 ) has a maximum when x1 =

f (x2 ) when x2 =

2x1 31 x22
4
x
3 2

3
2

= 3 x2 ,

14 x1 ,

the two together give x1 = 3 32 + 41 x1 = x1 = 2 and x2 = 1, not feasible,

nevertheless, for x1 [0, 1] the maximum value for x2 [2, 4] is again 2,
hence x1 is again 1 and x3 = 1.
4. The argument of square-root must be positive, the function is increasing in all
variables, so we can use Lagrangean method:
T

L(x, ) =

t=1

t=1

Lx i =

2t xt +

1
2i1 xi 2

xt 1

if = 0 all xi are 0 (minimum);

otherwise, for i < j, i, j {1, . . . T }:
12

2i1 xi

21

= 2j1 xj

1
2

xi
xj

xi
2j+1
= 22(ji)
= 2ji =
2i+1
xj

the system becomes x1 = x1 , x2 = 14 x1 , . . . xT =

since they must sum to 1:
x1 = 1/

T 1
i=0

1
4i , x2 = /
4

T 1

1
x ,
22(T 1) 1

4i , . . . xT =

i=0

as T increases the sum quickly converges to

1
1 14

4
3.

/
4(T 1)

T 1
i=0

5. a)
min w1 x1 + w2 x2 s.t. x1 x2 = y2 , x1 1, x2 > 0
the feasible set is closed but unbounded as x1 +
b) substituting x2 =

y2
x1

we get:

23

4i

f (x1 ) = w1 x1 + w2 xy1
2

=0

f (x1 ) = w1 w2 xy2

= x1 = +

f (x1 ) = 2w2 xy3 > 0

1

if

w2
w1 y

w2
y
w1

since x1 > 1

since x1 > 1
w1
w2 y,

1, this is the solution, and x2 =

otherwise 1 is, and x2 = y 2 ,

in both cases satisfying Kuhn-Tucker conditions:
L(x1 , x2 , 1 , 2 )
for

w2
y,
w1

w1 x1 + w2 x2 + 1 (
y 2 x1 x2 ) + 2 (x1 1)

w1
y
w2
w2
y

=0

Lx2 = w2 1 x1 = 1 =

w1
w2

=0

Lx1 = w1 1 x2 + 2 = 2 = 0
for (1, y 2 )
=0

Lx2 = w2 1 x1 = 1 = w2
=0

Lx1 = w1 1 x2 + 2 = 2 = w2 y 2 w1 .
6.
1

max

(x1 ,x2 )R2+

f (x1 , x2 ) =

max

(x1 ,x2 )R2+

p1 x12 + p2 x12 x23 w1 x1 w2 x2

substitute x x12 and y x23 , the problem becomes

max f (x, y) = max x(p1 + p2 y) w1 x2 w2 y 3

(x,y)R2+

(x,y)R2+

=0

fx = p1 + p2 y 2w1 x = x =

p2

p22 y

p2 p1
=0
fy = xp2 3w2 y =
+
3w2 y 2 = y =
2w1
2w1
2

p1 + p2 y
2w1

only the positive value will be admissible;

for p1 = p2 = 1 and w1 = w2 = 2 we have:

24

2w21

p22
2w1

3w2

+ 6 w2wp12 p1

y=

41 +

1
16

+6

97 1
97
= x =
24
96
4
1
1 12y

it is a maximum, because D 2 f (x, y) =

is negative semidefinite for

every y 0,
it is a global one for positive values because it is the only critical point.

7. a) Substituting x x12 and y x2 , considering that utility is increasing in both

variables, the problem is:
max f (x, y) = x + x2 y such that 4x2 + 5y = 100

(x,y)R2+

substituting y = 20 54 x2 the problem becomes:

4
max f (x) = x + 20x2 x4
5

xR+

16 3
x
5
with numerical methods it is possible to calculate that the only positive x for which
fx = 0 is x 3.5480,
where f (x) 128.5418,
48
2
since fxx = 40 48
5 x < 40 5 5 < 0, it is a maximum,
we obtain x1 12.5883 and x2 9.9294;
fx = 1 + 40x

b) buying the coupon the problem is:

max fa (x, y) = x + x2 y such that 3x2 + 5y = 100 a

(x,y)R2+

which becomes:
max fa (x) = x +

xR+

100 a 2 3 4
x x
5
5

12 3
x
5
the value of a that makes the choice indifferent is when fa (x)
without coupon) and fa = 0,
for lower values of a the coupon is clearly desirable.
fa = 1 + (200 2a)x

8. a)
max u(f, e, l) s.t. l [0, H], uf > 0, ue > 0, ul > 0

25

128.5418 (as

if is the amount of income spent in food, f =

max u

wl (1 )wl
,
,l
p
q

wl
p

and e =

(1)wl
:
q

s.t. l [0, H], [0, 1], uf > 0, ue > 0, ul < 0 ;

b)

L(l, , ) = u
Ll =
L =

wl (1 )wl
,
, l + 1 l + 2 (H l) + 3 + 4 (1 )
p
q

du
+ 1 2
dl
du
+ 3 4
d

and the constraints;

c) the problem is
max

l[0,16], [0,1]

f (l, ) = (3l) 3 ((1 )3l) 3 l2

1

= () 3 (1 ) 3 (3l) 3 l2
we can decompose the two variables function:
2
1
1
f (l, ) = g()(3l) 3 l2 , where g() () 3 (1 ) 3
2
since (3l) 3 is always positive, for l [0, 16],
and g() is always positive, for [0, 1],
g() maximizes alone for = 12 , where f ( 21 ) = 64;
now we have:
1

f (l) 64(3l) 3 l2 = fl = 128 (3l) 3 2l

1

=0

= l = 64 3 3 l 3
4

= l 3

34

fll = 128 (3l)

64 0.6934 = l

17.1938

always negative

l is a maximum but is not feasible,

however, since the function is concave, f (l) is increasing in [0, 16],
the agent maximizes working 16 hours (the model does not include the time for
leisure in the utility) and splitting the resources on the two commodities.
9.
1

max x13 + min{x2 , x3 } s.t. p1 x1 + p2 x2 + p3 x3 I

26

in principle Weierstrass theorem applies but not Kuhn-Tucker because min is continuous but not differentiable.
The cheapest way to maximize min{x2 , x3 } is however when x2 = x3 , the problem
becomes:
1

max x13 + x2 s.t. p1 x1 + (p2 + p3 )x2 I

now also Kuhn-Tucker applies.

10. a)
max p f (L + l) w1 L w2 l s.t. l 0, f C 1 is concave = fl < 0
b)
L(l, ) = p f (L + l) w1 L w2 l + l
Ll = p fl (L + l) w2 +

L = l

l = 0 and = w2 p fl (L + l);
c) pf (L +l)w1 L w2 l maximizes once (by concavity) for pfl (L +l) = w2 ,
when this happens for l 0, the maximum is (by chance) the Lagrangean point,
when instead this happens for l > 0, the maximum is not on the boundary.
11. a)
1

max py x14 x24 p1 (x1 K1 ) p2 (x2 K2 ) s.t. x1 K1 , x2 K2

1

L(x1 , x2 , 1 , 2 ) = py x14 x24 p1 (x1 K1 ) p2 (x2 K2 ) + 1 (K1 + x1 ) + 2 (K2 + x2 )

1
3 1
Lx 1 =
py x1 4 x24 p1 x1 + 1 x1
4
1
1
3
py x14 x2 4 p2 x2 + 2 x2
Lx 2 =
4
L1 = K1 + x1
L2

= K2 + x2
1

b) the unbounded maximum of f (x1 , x2 ) = x14 x24 x1 x2 + 4 is for:

27

43
1
x
x
4 1 2

3
4

1
4

fx1 = 14 x1 4 x24 1 = 0 = x14 = 41 x24

f x2 =

1
4

= x1 = x2 =

1 = 0 = x2 = 41 x1

bounds are respected,

the firm sells most of x1 and buys only

1
16

units of x2 to produce

c) is analogous to (b) since the problem is symmetric.

1
4

1
4

1
16

units of y;

12. a)
max py (x1 (x2 + x3 )) w x
L(x, ) = py (x1 (x2 + x3 )) w x + x
Lx i

Lx1 = py (x2 + x3 ) w1 + 1

i={2,3}

= py x1 wi + i

Li = xi

x1 =

w3 3
w2 2
=
py
py
x2 + x3

= 3 2 = w3 w2
=

w1 1
py

b) [ w4 ??? ] x R3+ and R3 are not defined by the equations;

c) the problem can be solved considering the cheapest between x2 and x3 ,
suppose it is x2 , the problem becomes:
max py x1 x2 w1 x1 w2 x2
which has critical point ( wpy2 , wpy1 ) but its Hessian

0 py
py 0

is not negative semidef-

inite.
The problem maximizes at (+, +) for any choice of py R+ and w R3+ .

28