solutions sundaram
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Chapter 3 (pp. 9099): 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 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 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 converging) 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 onetoone 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) = ex 
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 9293), 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. 100111): 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 =
saddle, D2f (1, 74
) = 5
4 7
7 0 =
saddle,
D 2f (1, 74 ) =
54 7
7 0 = saddle.
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.500501).
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. 112144): 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 positivedenite, negativedenite 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. 145171): 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 squareroot 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 KuhnTucker 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 KuhnTucker because min is continuous 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 KuhnTucker 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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