Solutions to Practice Questions

Transcription

1 Solutions to Pactice Questions. The Hessian matix is g 00 g g (x; y) g g 6x 0 0 6y We see that g is stictly convex in its domain. The only stationay point is (; p 63). It is a global minimum point of g and g min 4 p 69:. The only stationay point is (0; 0; 0): The leading pincipal minos of the Hessian evaluated at (0; 0; 0) ae D ; D 3; and D 3 4, so (0; 0; 0) is a local minimum point of f: 3. The Lagangian function is concave as a sum of concave functions and (x ; y ; z ; ) (a6; a3; a6; a3) and f (a) f (x (a); y (a); z (a)) 00 a 6: Note that df da 4. Take the deivative of the Lagangian function evaluated at the solution with espect to nx i i p p i + When the solution is i p i 0 by the st-ode Similaly, take the deivative of the Lagangian function evaluated at the solution with espect to p i and simplify it using the st-ode @p i x i fo i ; : : : ; n:. Solve the poblem: max (x;y;z) x + y + z subject to x + y + 4z and x + 3y + z 0 Maximum is one at ( 3 p 00; p 00; 0) and (3 p 00; with ; 0: p 00; 0)

2 6. Please set up the Lagangian function and check the Kuhn-Tucke conditions case by case. 7. Lagangean function Kuhn-Tucke conditions L ax + y + [ (x + ) (y + ) ] L x a (x + ) 0; x 0; xl x 0 L y (y + ) 0; y 0; yl y 0 L (x + ) (y + ) 0, 0, L 0 Case ) Suppose that x 0 and y 0: Then, we know fom L x 0 and L y 0 that max[ a; ] : Since a > 0; this means > 0: Howeve, L 3 > 0: Theefoe, L 0 is violated Case ) Suppose that x 0 and y > 0: Then L y (y + ) 0 and L x a 0: Theefoe, > 0: This implies that L 4 (y + ) 0. Theefoe, y : If y ; then L y 4 0; so 4: Then L x a 0: Theefoe, (x ; y ; ) (0; ; 4 ) if 0 < a Case 3) Suppose that x > 0 and y 0: Then, L x a (x + ) 0 and L y 0: Theefoe, > 0: This implies that L 4 (x + ) 0: Theefoe, x : If x ; then L x a 4 0: Theefoe, a: Since fom L 4 y 4 0; it must be that a : Theefoe, (x ; y ; ) (; 0; a 4 ) if a Case 4) Suppose that x > 0 and y > 0: must be positive othewise L x a (x + ) 0 violates a > 0: We know that L x a (x + ) 0 and L y (y + ) 0: Thefoe, we have a x + y + > 0

3 This implies that (a + ) (y + ) : Theefoe, y a + x a a + > 0 > 0 Theefoe, (x ; y ; ) a a + ; a + ;! a + if < a < Theefoe, we have 8 >< (x ; y ) >: a q ; a + (0; ) if 0 < a (; 0) q if a () if < a < a + Now we need to show that () is the solution to the poblem. See the objective function is linea, so it is a quasiconcave function. Let s see the constaint. The Hessian of (x + ) + (y + ) is 0 H 0 Theefoe, the the function in the constaint is a convex function, so it is a quasiconvex function. Futhemoe, the st-ode deivatives of the objective function ae not all zeo. Theefoe, () is the solution to the poblem because Aow-Enthoven Su ciency Theoem holds. 8. If g is a concave function, then it satis es, fo all x; x 0 in a convex domain X and all [0; ]; g(x + ( ) x 0 ) g(x) + ( )g(x 0 ) If g is a quasiconcave if and only if fo all x; x 0 in a convex domain X and all [0; ] g(x + ( ) x 0 ) min[g(x); g(x 0 )] 3

4 Since g(x) + ( )g(x 0 ) min[g(x); g(x 0 )] fo all [0; ]; we have g(x + ( ) x 0 ) g(x) + ( )g(x 0 ) min[g(x); g(x 0 )] Theefoe, function g(x) is a concave function, then it is a quasiconcave function. 9. Note that x p < and if x k < ; then x k+ p x k + < p + ; so by induction x k < fo all k: Moeove because x k < ; one has x k+ p x k + > p q x k > x k x k so fx k g is (stictly) inceasing. By Theoem.0., the sequence is convegent. If x is its limit, then letting k! yields x p x +. This equation implies that x x + which has the two solutions and : Because is obviously not a solution of x p x + ; the only solution is x thus lim k! x k : 0. Detemine lim and lim. (a) Fo evey n; thee exists a numbe k n with ( ) k : Hence y n supf( ) k : k ng so lim n! y n : Thus lim k! x k : In the same way, we can see that lim k! x k (b) Aguments simila to those in (a) yields lim k! x k 3 and lim k! x k. Pove that the sequence of eal numbes fx k g with the geneal tems x k is a Cauchy sequence. 3 k Let n and m be natual numbes with m > n and de ne p m n. Then, jx m x n j jx n+p x n j (n + ) + (n + ) + + (n + p) < n(n + ) + (n + )(n + ) + + (n + p )(n + p) n n + n + n + n + p n n + p < n n + p 4

5 Thus, fo any > 0; if we choose n > ; then jx m m > n: This poves that fx k g is a Cauchy sequence. x n j < " fo all. Note that as k! ; x k is getting close to eithe o 0 depending on whethe k is odd o even. Theefoe, any convegent subsequence can have its limit equal to eithe o 0: 3. d(x; y) p P n i (x i y i ) P n i p (xi y i ) P n i jx i y i j 4. Look at [ ia i ; whee A i fig fo i ; ; : : :. The numbe zeo (0) belongs to the closue of [ ia i but not to [ ia i itself.. The convegence of the following sequence (a) x k ; + k k! (0; ) as k! (b) x k + ; + k! (; e) as k! k k k+ (c) x k ; ( )k 3! ; 0 3k k 6. Give examples of subsets S of R and continuous functions f : R! R such that (a) S is closed, but f(s) is not closed S R; f(x) e x ; f(s) (0; ) (b) S is open but f(s) is not open S R; f(x) e x ; f(s) [; ) (c) S is bounded, but f(s) is not bounded. S (0; ); f(x) x; f(s) (; ) 7. Conside the function f de ned fo all x in [0; ] by f(x) (x + ): Pove that f maps [0; ] into itself and nd a xed point. Since [0; ] is nonempty compact convex set in R and f(x) is a continuous function fom [0; ] into [0; ] itself. thee is a xed point by Bouwe s xed point theoem. The xed point is f(x) (x + ) x; so it is x Suppose that f is de ned fo all x in (0; ): Pove that f maps (0; ) into itself but f has no xed point. Why does not Bouwe s xed point theoem apply?

6 Note that fo any x (0; ); we have 0 < f(x) (x + ) < : Theefoe, f maps (0; ) into itself. If thee is a xed point, then it must satisfy (x + ) x; so it must be x : Howeve, x does not belong to the domain of f: This is an example that shows that when the domain is not closed, then the function may not have a xed point. 6