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

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

f (x) = lim x01x

= + .

2. It can be shown ( e.g. by construction) that in R (in any completely ordered set)sup and inf of any nite set coincide respectively with max and min .If DR n is nite also f (D) = {xR | xD s.t. f ( x) = x} is.The result is implied by the Weierstrass theorem because every nite set A is com-pact, 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 xD such that x x,= xD such that f (x) f (x);b) consider D {(x, y )R 2+ | x + y = 1}R 2,

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 ) = 1

x if x = 00 if x = 0

max f on D is limx0 f (x, y ) = lim x0 1x = + . 4. A nite set is compact, because every sequence have a constant (hence converg-ing) subsequence.Every function from a nite set is continuous, because we are dealing with thediscrete topology, and every subset is open (see exercise 11 of Appendix C for thecorrect statement of Corollary C.23 (page 340)).

We can take a subset A of cardinality k in Rn and construct a one-to-one functionassigning to every element of A a different element in R .

A compact and convex subset A Rn cannot have a nite number k 2 of elements (because otherwise aAak is different from every a A but should bein 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 afunction 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 ]7

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(the rigorous proof is long, see Th. 1.69, page 60), which is then not nite, so thatneither 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 R + | f (xn ) = sup( f )

sup (f )1n , moreover, since limxf (x) = 0 , x s.t. x >x, f (x) < 1.{xn } is then limited in the compact set [0, x], hence it must converge to x. f (x) ishowever sup( f ) and then f (x) = max( f ).f (x) = ex , restricted to R+ , satises the conditions but has no minimum.

6. Continuity (see exercise 11 of Appendix C for the correct statement of CorollaryC.23 (page 340)) is invariant under composition, i.e. the composition of continuousfunctions is still continuous.Suppose g : V 1 V 2 and f : V 2 V 3 are continuous, if A V 3 is open then, bycontinuity of f , also f 1(A)V 2 is, and, by continuity of g, also g1(f 1(A))V 1is.g1 f 1 : V 3 V 1 is the inverse function of f g : V 1 V 3, which is then alsocontinuous.We are in the conditions of the Weierstrass theorem.

7.

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

, f (x) = e|x |

arg max( f ) = 0 , max( f ) = e0 = 1 and also sup( f g) = 1 , which is however neverattained.

8.

min p x subject to xD {y R n+ | u( y) u}s.t. u : R n+ R is continuous , p 0

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 dene W U {y R n+ | u( y) U },by continuity of u, W U is compact,for U large enough D W U is not empty and the minimum can be found in it;

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(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 x i , the problem minimizes for x i + . 9.

max p g( x) w x s.t. xR n+

with p > 0, g continuous , w 0

a solution is not guaranteed because Rn+ is not compact.

10.

F ( p, w) = {( x, l)R n +1+ | p x w(H l), l H }F ( p, w) compact = p 0 (by absurd):

if pi 0, the constrain could be satised 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 }, x i is surely limited in [0, w(H l) pi ].

11.

max U ( y( )) subject to { R N | p 0, y( ) 0}where y

s( ) =

s + N

i=1

iz

is, U : RS

+ R is continuous and strictly increasing

( y y in RS + if yi yi i {1, . . . , S } and at least one inequality is strict).Arbitrage: R N such that p 0 and Z 0.A solution exists no arbitrage

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

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y(k

) increases linearly with k and then also U ( y(k

)) increases with k,

max U ( y(k )) is not bounded above;(=) R N 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 rst 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)) , . . . u n (xn , h(x))

with h(x) 0, (x, x)[0, w]n +1 , w x =n

i=1

x i

sufficient condition for continuity is that W , u and h are continuous,the problem is moreover well dened only if x[0, w] such that h(x) 0;the simplex w {( x, x)[0, w]n +1 | x + ni=1 x i = w} is closed and bounded,if h(x) is continuous H = {( x, x) [0, w]n +1 | h(x) 0} is closed because theinequality is not strict,then the feasible set w H is closed and bounded because intersection of closedand bounded sets. 13.

max (x) = maxx

R +xp(x) c(x)

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) = p2 x, now (x) = p2 x, so that the maximization problem isnot bounded above.

14. a)

max v(c(1), c(2)) subject to c(1) 0, c(2) 0, p(1) c(1) + p(2)1 + r c(2) W 0

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

c(1) 0, c(2) 0, p(1) c(1) + p(2)1 + r c(2) W 0

become

c 0, p c W 0we are in the conditions of example 3.6 (pages 92-93), and p 0 if and only if p(1) 0 and p(2) 0. 15.

max (x1) + (x2) + (x3) subject to0 x1 y10 x2 f (y1 x1)0 x3 f (f (y1 x1) x2)

lets call A the subset of R 3+ that satises the conditions, A is compact if it is closedand 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])])]R

3+

and maxima exist because of continuity.Consider (x1, x2, x3) Ac in R3+ , then (if we reasonably suppose that f (0) = 0 ),becasue of continuity, at least one of the following holds:x1 > y 1, x2 > f (y1 x1) or x3 > f (f (y1 x1) x2).If we take r = max {x1 y1, x2 f (y1 x1), x3 f (f (y1 x1) + x2)}, r > 0 andB (( x1 , x2 , x3), r )Ac .Ac is open = A is closed.

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

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

f (x) = 1 2x 3x2 = 0 for x = 1 1 + 3

3 = 11

3

f (x) = 2 6x , f (1) = 4f ( 1

3) =

4

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

4. a)

f x = 6x2 + y2 + 10 xf y = 2xy + 2 y

= y = 0x = 0

D 2f (x, y) = 12x + 10 2y

2y 2x + 2(0, 0) is a minimum;

b)

f x = e2x + 2 e2x (x + y2 + 2 y) = e2x (1 + 2( x + y2 + 2 y))f y = e2x (2y + 2)

= y = 1

1 + 2( x + 1 2) = 0 = x = 12

= f (12

, 1) = 12

e

since limxf (x, 1) = 0 and limx+ f (x, 1) = + , the only critical point( 12 , 1) is a global minimum;c) the function is symmetric, limits for x or y to are + , aR :

f x = ay 2xy y2f y = ax 2xy x2= (0, 0)(0, a)(a, 0)(

25

a, 15

a)(15

a, 25

a)

D 2f (x, y) = 2y a 2x 2ya 2x 2y 2x12

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(

25 a,

15 a) and (

15 a,

25 a) are local minima

a

R ;

d) when sin y = 0 , limits for x are ,f x = sin yf y = x cos y

= (0,ky ) kZ = {. . . , 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 + , aR ,

f x = 4x3 + 2 xy2

f y = 2x2y 1= y =

12x2

= 4x3 =

1x

= impossible

there are no critical points;

f) limits for x or y to are + ,f x = 4x3 3x2f y = 4y3 = (0, 0)(

34

, 0)

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

g) limits for x or y to are 0,f x = 1x

2 + y2(1+ x 2 + y2 )2

f y = 2xy(1+ x 2 + y2 )2= (1, 0)(1, 0)

since f (1, 0) = 12 and f (1, 0) = 12 , the previous is a maximum, the latter aminimum;h) limits for x to are +,limits for y to depends on the sign of (x2 1),

f x = x3

8 + 2 xy2 1f y = 2y(x2 1)

=

y = 0 = x = 2

x = 1 = y =

74

x = 1 = impossibleD 2f (x, y ) =

38 x

2 + 2 y2 4xy4xy 2y(x2 1)

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D 2f (2, 0) = 32 0

0 0 =

) = 5

4 7

7 0 =

D 2f (1, 74 ) =

54 7

5. The unconstrained function is given by the substitution y = 9 x:f (x) = 2 + 2 x + 2(9 x) x2 (9 x)2 = 2x2 + 18 x 61

which is a parabola with a global maximum in x = 92 = y = 92 .

6. a) x is a local maximum of f if R + such that yB (x, ), f (y) f (x),considering then that:lim inf

yx f (x) f (y) = liminf yB (x , )x

f (x) f (y) 0

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

1 lim inf = limsupwe have the result;

b) a limit may not exist, while liminf and limsup always exist if the function

is dened on all R

;c) if x is a strict local maximum the inequality for liminf 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 isnot constant in x = 0 .

8. f is not constant null, otherwise also f would be constant null,since limyf (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 dx i

= (f ) df dx i

, df

dx i= 0 =

d f dx i

= 0

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

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b)

d2 f dx i dx j

= ddx j

(f ) df dx i

= (f ) df dx i

df dx j

+ (f ) d2f dx i dx j

df dx i

= 0 = d2 f

dx i dx j= (f )

d2f dx i dx j

Df ( x) = 0 = D2 f ( x) = (f ( x)) D 2f ( x)

a negative denite 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).

D 2f ( x) =f x 1 x 1 . . . f x 1 x n

... . . . ...

f x 1 x n . . . f x n x nis positive denite if:

1) f x 1 x 1 > 0,2) f x 1 x 1 f x 2 x 2 f 2x 1 x 2 > 0,. . .n) |D

2

f ( x)| > 0;

D 2g( x) =gx 1 x 1 . . . gx 1 x n

... . . . ...

gx 1 x n . . . gx n x n

= D 2 f ( x) =f x 1 x 1 . . . f x 1 x n... . . . ...f x 1 x n . . . f x n x n

is negative denite if:1) f x 1 x 1 < 0,2) (f x 1 x 1 ) (f x 2 x 2 ) (f x 1 x 2 )2 > 0,. . .n)

|D 2

f ( x)

|< 0 if n is odd,

|D 2

f ( x)

|> 0 if n is even;

the i th 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 2f ( x) positive denite = D

2f ( x) negative denite.

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

1. a)

L(x,y, ) = x2 y2 + (x2 + y2 1)Lx = 2x + 2 x

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

=0= = 1 y = 0L = x2 + y2 1

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

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

f (x) = x2 (1 x2) = 2 x2 1 = min for x = 0max for x =

the solution is different from (a) because the right substitution is y 1 x2,admissible only for x[1, 1]. 2. a) Substituting y 1 x:

f (x) = x3 (1 x)3 = 3 x2 3x + 1 = max for x = b)

L(x,y, ) = x3 + y3 + (x + y 1)Lx = 3x2 +

=0= = 3x2Ly = 3 y2 +

=0= = 3y2 = x = y

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

12

x = y = 12 is the unique local minimum, as can be checked in (a).

3. a)

L(x,y, ) = xy + (x2 + y2 2a2)Lx = y + 2 x

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

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

L = x2 + y2 2a2 =0=

x = 0 y = 2ay = 0 x = 2ay = 2x x = 2y

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(0,

2a) and (

2a, 0) are saddle points, because f (x, y) can be positive or neg-

ative 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 aminimum.

b) substitute x 1x , y 1y and a 1a

L(x ,y, ) = x + y + (x2 + y2 a2)

L x = 1 + 2 x =0

= x = 1

2L y = 1 + 2 y

=0= y = 12

= x

L = x2 + y2 a2 =0= x = y =

22

a

for x and y negative we have a minimum, otherwise a maximum.

c)

L(x,y,x, ) = x + y + z + (x1 + y1 + z1

1)

Lx = 1 x 2 =0= x =

. . . =0= y =

. . . =0= z =

we have a minimum when they are all negative ( x = y = z = 3) and a maximumwhen all positive ( x = y = z = 3).d) substitute xy = 8 (x + y)z = 8 (5 z)z:

f (z) = z(8 (5 z)z) = z3 5z2 + 8 z

f z = 3z2 10z + 8 =0= z =

5 25 243

= 243

as z , f (z) ,f zz = 6 z 10 is negative for z = 43 (local maximum)and positive for z = 2 (local minimum),

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when z = 43 , x + y = 5

z = 113 and xy = 8

(x + y)z = 289

= one is also 43 and the other 73 ,

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 16x :f (x) = x + 16x = f x = 1 16x 2

=0= x = 4when x = 4 and y = 14 we have a minimum, maxima are unbounded for limx0and limx;

f) substituting z

6

x and y

2x:

f (x) = x2 + 4 x (6 x)2 = 16 x 36 ,f (x) is a linear function with no maxima nor minima.

4. Actually a lemniscate is (x2 + y2)2 = x2y2 , while (x2y2)2 = x2 + y2 identiesonly 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 + y2)2 = x2 y2x

4+ 2 x

2y

2+ y

4

x2

+ y2

= 0y4 + (2 x2 + 1) y2 + x4 x2 = 0

y2 = 2x2 1 + 4x4 + 4 x2 + 1 4x4 + 4 x22

y = f (x) = 2x2 + 1 + 8x2 + 12computing f (x) and nding where it is 1 we nd the arg max = ( x, y),(x, y) will be the arg min .

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5. a) (x

1)3 = y2 implies x

1,

f (x) = x3 2x2 + 3 x 1 = f x = 3x2 4x + 3 which is always positive forx 1,since f (x) is increasing, min f (x) = f (1) = 1 (y = 0);b) the derivative of the constraint D((x 1)3 y2) 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)

L( x,

) = c x + 12 x Dx +

(Ax

b)

=n

i=1

ci x i + 12

n

i=1

n

j =1

x j D ji x i +m

i=1

in

j =1

Aij x j bi

Lx i = ci + D ii x i + 12

n

j =1 ,j = i

x j D ji + 12

n

j =1 ,j = i

x j D ij +m

j =1

j A ji

= ci +n

j =1

x j D ij +m

j =1

j A ji

L i =n

j =1

Aij x j

bi

b).. .

7. The constraint is the normalixation |x| = 1 , the system is:

f ( x) = x Ax

=n

i=1

n

j =1

x j A ji x i

f x i

= 2n

j =1x

jA

ji

xi = n j =1 ,j = i x j A ji

Aii

x =

0 A21A 11 . . . An 1

A 11A 12A 22 0 . . .

An 2A 22

... ... . . .

...A 1nA nn

A 2nA nn . . . 0

x B x

such that | x| = 1

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

since D 2f ( x) = 2A11 . . . An 1

... . . . ...

A1n . . . Ann

, x is constant,all critical points are maxima, minima or saddles, according wether A is positive-denite, negative-denite or neither.

8. a)

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

in this period the average stock isR T 0 s(t )

T = x

T

2T = x2 ,

in the long run this will be the total average;

b)

L(x,n, ) = C hx2

+ C 0n + (nx A)Lx = C h2 + nLn = C 0 + x

=0= n = C h2 x =

C 0

L = nx

A =0=

C h C 0

22 = A =

=

C h C 0

2Ax and n negative have no meaning so must be negative.

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 limx 10 x

1 + x

2 = + .

10.

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min w1x1 + w2x2 s.t. (x1, x2)X {(x1 , x2)R 2+ | x21 + x22 1}a)

if w1 < w 2 it is clear from the graph that (1, 0) is the cheapest point in X ,similarly, if w2 < w 1, the cheapest point is 0, 1,if w1 = w2 they both cost w1 = w2;

b) if nonnegativity constraints are ignored:

minx 21 + x

22

1

w1x1 + w2x2 = lim(x 1 ,x 2 )(,)

w1x1 + w2x2 = similarly for (x1 , x2) (+ , + ) if w1 and w2 are positive. 11. [x

12 is not dened if x < 0. . . ]

max x12 + y

12 s.t. px + y = 1

L(x,y, ) = x12 + y

12 + ( px + y 1)

Lx = 12 x12 + p

Ln = 12 y12 +

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

L = px + y 1 =0= p(2p)2 + ( 2)2 = 1 p

42 p2 +

142

= p2 + p42 p2

= 1 = = p2 + p2 pthe sign of does not matter to identify x = 1 p2 + p and y =

p2 p2 + p ,

the unique critical point, with both positive components,

x12 + y

12 = 1+ p p2 + p = 1+ p p is greater e.g. than 0

12 + 1

12 = 1 ,

being the unique critical point it is a maximum.

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

1. Since the function is increasing in both variables, we can search maxima on theboundary,substituting y = 1 x2 (which means x[0, 1] =y [0, 1]):

f (x) = ln x +ln 1 x2 = ln x+12

ln(1x2) = f x = 1x

12

2x1 x2

= 1 2x2(1 x2)x

=0=

x = 22

=

y = 22

which is the argument of the maximum.

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 = x( py cos pz sin )=0=

sin cos

= py pz

= y

z =

py pz

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

sin = py

p2y + p2z cos =

pz p2y + p2z

y = x py

p2y + p2z z = x

pz p2y + p2z

( x py p2y + p2z , x pz p2y + p2z ) is a maximum and ( x py

p2y + p2z, x pz p2y + p2z ) is a minimum.

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

max x1x2x3 s.t. x 1 [0, 1], x2 [2, 4], x3 = 4 x1 x2

max f (x1, x2) = 4 x1x2 x21x2 x1x22 s.t. x 1 [0, 1], x2 [2, 4]consider only x1,f (x1) = 4 x1x2 x21x2 x1x22 has a maximum for x1 =

4x 2x 222x 2 = 2 12 x2,by simmetry a maximum is for x1 = x2 = 43 , 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)

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max x1x2x3 s.t. x 1 [0, 1], x2 [2, 4], x3 = 6 x1 2x23

max f (x1, x2) = 2 x1x2 13

x21x2 23

x1x22 s.t. x 1 [0, 1], x2 [2, 4]

f (x1) has a maximum when x1 = 2x 223 x 222

3 x 2= 3 x2,

f (x2) when x2 = 2x 113 x 224

3 x 2= 32 14 x1 ,

the two together give x1 = 3 32 + 14 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 allvariables, so we can use Lagrangean method:

L( x, ) =T

t=12t x t +

T

t=1x t 1

Lx i = 2i1x12

i +

if = 0 all x i are 0 (minimum);

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

2i1x12

i = 2 j 1x12

j =x ix j

12 =

2 j +1

2i+1 = 2 j i =

xix j

= 2 2( j i )

the system becomes x1 = x1, x2 = 14 x1, . . . xT = 1

22( T 1) x1,since they must sum to 1:

x1 = 1 /T 1

i=0

4i , x2 = 1

4/

T 1

i=0

4i , . . . x T = 1

4(T

1) /

T 1

i=0

4i

as T increases the sum quickly converges to 1114= 43 .

5. a)

min w1x1 + w2x2 s.t. x 1x2 = y2, x1 1, x2 > 0the feasible set is closed but unbounded as x1 +

b) substituting x2 = y2

x 1 we get:

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f (x1) = w1x1 + w2 y2

x 1

f (x1) = w1 w2 y2

x 21

=0= x1 = + w2w1 y since x1 > 1f (x1) = 2 w2 y

2

x 31> 0 since x1 > 1

if w2w1 y 1, this is the solution, and x2 = w1w2 y,otherwise 1 is, and x2 = y2,in both cases satisfying Kuhn-Tucker conditions:

L(x1, x2 , 1 , 2) = w1x1 + w2x2 + 1(y2 x1x2) + 2(x1 1)

for w2w1 y, w1w2 yLx 2 = w2 1x1

=0= 1 = w2

y w1w2Lx 1 = w1 1x2 + 2

=0= 2 = 0

for (1, y2)

Lx 2 = w2 1x1 =0= 1 = w2Lx 1 = w1 1x2 + 2

=0= 2 = w2y2 w1 .

6.

max(x 1 ,x 2 )

R 2+

f (x1, x2) = max(x 1 ,x 2 )

R 2+

p1x121 + p2x

121 x

132 w1x1 w2x2

substitute x x121 and y x

132 , the problem becomes

max(x,y )R2+ f (x, y) = max(x,y )R

2+ x( p1 + p2y) w1x

2

w2y3

f x = p1 + p2y 2w1x =0= x =

p1 + p2y2w1

f y = xp2 3w2y2 = p2 p12w1

+ p22y2w1 3w2y

2 =0= y =

p222w1 p222w1 2 + 6 w2 p2 p1w1

3w2only the positive value will be admissible;for p1 = p2 = 1 and w1 = w2 = 2 we have:

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y = 14 + 116 + 66 = 97 124 = x = 9796it is a maximum, because D2f (x, y) = 4 11 12y

is negative semidenite for

every y 0,it is a global one for positive values because it is the only critical point. 7. a) Substituting x x

121 and y x2, considering that utility is increasing in bothvariables, the problem is:

max(x,y )R2+ f (x, y ) = x + x

2y such that 4x

2+ 5 y = 100

substituting y = 20 45 x2 the problem becomes:maxx

R +f (x) = x + 20 x2

45

x4

f x = 1 + 40 x 16

5 x3

with numerical methods it is possible to calculate that the only positive x for whichf x = 0 is x 3.5480,where f (x) 128.5418,since f xx = 40 485 x2 < 40 485 5 < 0, it is a maximum,we obtain x1 12.5883 and x2 9.9294;

b) buying the coupon the problem is:

max(x,y )

R 2+

f a (x, y) = x + x2y such that 3x2 + 5 y = 100 awhich becomes:

maxx

R +f a (x) = x +

100 a5

x2 35

x4

f a = 1 + (200 2a)x 12

5 x3

the value of a that makes the choice indifferent is when f a (x) 128.5418 (aswithout coupon) and f a = 0 ,for lower values of a the coupon is clearly desirable.

8. a)

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

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if is the amount of income spent in food, f = wl p and e = (1 )wlq :

max uwl p

, (1 )wl

q , l s.t. l [0, H ], [0, 1], uf > 0, u e > 0, ul < 0 ;

b)

L(l,, ) = uwl p

, (1 )wl

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

L l = du

dl + 1 2

L = du

d + 3 4and the constraints;

c) the problem is

maxl[0,16], [0,1]

f (l, ) = ( 3l)13 ((1 )3l)

13 l2

= ( )13 (1 )

13 (3l)

23 l2

we can decompose the two variables function:f (l, ) = g()(3l) 23 l2, where g() ( )

13 (1 )

13

since (3l)23 is always positive, for l[0, 16],and g() is always positive, for [0, 1],g() maximizes alone for = 12 , where f (

12 ) = 64 ;

now we have:

f (l) 64(3l)23 l2 = f l = 128 (3l)

13 2l

=0= l = 64 313 l

13

=

l43 64

0.6934 =

l

17.1938

f ll = 128 (3l)43 2 always negativel 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 forleisure in the utility) and splitting the resources on the two commodities.

9.

max x131 + min {x2, x3} s.t. p 1x1 + p2x2 + p3x3 I

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

max x131 + x2 s.t. p 1x1 + ( p2 + p3)x2 I

now also Kuhn-Tucker applies.

10. a)

max p

f (L + l)

w1L

w2l s.t. l

0, f

C 1 is concave =

f l < 0

b)

L(l, ) = p f (L + l) w1Lw2l + lL l = p f l(L + l) w2 + L = l

l = 0 and = w2 p f l(L + l);c) pf (L+ l)w1Lw2l maximizes once (by concavity) for pf l(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)

max pyx141 x

142 p1(x1 K 1) p2(x2 K 2) s.t. x 1 K 1, x2 K 2

L(x1, x2, 1, 2) = pyx141 x

142 p1(x1 K 1) p2(x2 K 2) + 1(K 1 + x1) + 2(K 2 + x2)

Lx 1 = 1

4 pyx

34

1 x142 p1x1 + 1x1

Lx 2 = 1

4 pyx

141 x

34

2 p2x2 + 2x2L 1 = K 1 + x1L 2 = K 2 + x2

b) the unbounded maximum of f (x1, x2) = x141 x

142 x1 x2 + 4 is for:

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f x 1 = 14 x34

1 x142 1 = 0 = x

341 =

14 x

142

f x 2 = 14 x141 x

34

2 1 = 0 = x342 =

14 x

141

= x1 = x2 =14

2=

116

bounds are respected,the rm sells most of x1 and buys only 116 units of x2 to produce

14 units of y;

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

12. a)

max py (x1(x2 + x3)) w x

L( x, ) = py (x1(x2 + x3)) w x + xLx 1 = py (x2 + x3) w1 + 1

Lx i i= {2,3} = py x1 wi + iL i = xi

x1 = w2 2 py = w3 3 py = 3 2 = w3 w2x2 + x3 =

w1 1 py

b) [ w4 ??? ] xR3+ and R

3 are not dened by the equations;

c) the problem can be solved considering the cheapest between x2 and x3,suppose it is x2, the problem becomes:

max pyx1x2

w1x1

w2x2

which has critical point ( w2 py , w1 py ) but its Hessian

0 py py 0

is not negative semidef-

inite.The problem maximizes at (+ , + ) for any choice of py R + and wR 3+ .

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