Rangarajan K. Sundaram - A First Course in Optimization Theory - Solution Manual PDF

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Paolo Pin [email protected] http://venus.unive.it/pin

August, 2005

Rangaraian K. Sundaran, 1996, “A First Course in Optimization Theory”, Cambridge University Press:

Solutions of (some) exercises Appendix C (pp. 330-347): Structures on Vector Spaces. 1.   1 1 kx + yk2 − kx − yk2 = < x + y, x + y > − < x − y, x − y > 4 4  1 = < x, x > +2 < x, y > + < y, y > −(< x, x > −2 < x, y > + < y, y >) 4  1 = 4 < x, y > = < x, y > ✷ 4 2. kx + yk2 + kx − yk2 = kx + yk2 − kx − yk2 + 2kx − yk2

Polarization Identity =⇒ = 4 < x, y > +2kx − yk2

= 2 < x, x > +2 < y, y >   = 2 kxk2 + kyk2 ✷

3. Let’s take e.g. R2 , ~x = (1, 0) and y~ = (0, 2):   kx + yk12 + kx − yk12 = 32 + 32 = 18 6= 10 = 2 · (12 + 22 ) = 2 kxk21 + kyk21 ✷

1

4. a) {xn } → x in (V, d 1 ) if and only if: lim d 1 (xn , x) = 0 =⇒ lim d 2 (xn , x) ≤ lim c1 · d 1 (xn , x) = c1 · lim d 1 (xn , x) = 0

n→∞

n→∞

n→∞

n→∞

since d 2 (xn , x) ≥ 0, ∀ n: limn→∞ d 2 (xn , x) = 0 ; b) A is open in (V, d 1 ) if and only if: ∀ x ∈ A, ∃ r > 0 s.t. B1 (x, r) = {y ∈ V | d 1 (x, y) < r} ⊂ A =⇒ ∀ x ∈ A, ∃

r r > 0 s.t. B2 (x, ) = {y ∈ V | c2 · d 2 (x, y) ≤ r} ⊆ B1 (x, r) ⊂ A c2 c2

the reverse of (a) and (b) is equivalent.

✷

5. Consider R with the usual metric d(x, y) ≡ |x − y| and with the other one defined as δ(x, y) ≡ max{d(x, y), 1}; δ is a metric, to prove triangle inequality consider that: δ(x, z) = max{d (x, z), 1} ≤ max{d (x, y) + d (x, z), 1} ≤ max{d (x, y) + d (x, z), 1 + d (x, y), 1 + d (y, z), 2}

= max{d(x, y ), 1} + max{d(x, z ), 1} = δ(x, y) + δ (y, z)

∀ Bδ (x, r) ⊂ R, Bd (x, r) ⊆ Bδ (x, r), similarly ∀ Bd (x, r) ⊂ R, Bδ (x, min{r, 12 }) ⊆ Bd (x, r ), so they generate the same open sets; nevertheless 6 ∃ c ∈ R+ such that d(x, y ) < c · δ(x, y ), ∀ x, y ∈ R, suppose by absurd it exists, then d(0, c) = c < c · δ(x, y) is impossible since δ(x, y) ≤ 1, ∀ x, y ∈ R.

✷

6. d ∞ (~ x, ~y ) = max{|x1 − y1 |, |x2 − y2 |} is equivalent to any  1 p d p (~ x, ~y ) = |x1 − y1 |p + |x2 − y2 |p , with p ∈ N, p ≤ 1, because:

1 d∞ ≤ dp ∧ dp ≤ d∞ 2 all the metrics are equivalent by symmetry and transitivity (see exercise 7).

7. If d 1 and d 2 are equivalent in V , ∃ b1 , b2 ∈ R such that ∀ x, y ∈ V :

2

✷

;

d 2 (x, y) ≤ d 1 (x, y) ≤ b2 · d 2 (x, y) b1 similarly if d 2 and d 3 are equivalent in V , ∃ c1 , c2 ∈ R such that ∀ x, y ∈ V : d 3 (x, y) ≤ d 2 (x, y) ≤ c2 · d 3 (x, y) c1 hence ∃ (b1 · c1 ), (b2 · c2 ) ∈ R such that ∀ x, y ∈ V : d 3 (x, y) ≤ d 1 (x, y) ≤ b2 · c2 · d 3 (x, y) b 1 · c1

=⇒ d 1 and d 3 are equivalent. ✷ 8.

ρ(x, y) =



1 x 6= y 0 x=y

Positivity and Simmetry come from the definition; for Triangle inequality consider the exhaustive cases: x = y = z =⇒ d (x, z) = 0 = 0 + 0 = d (x, y) + d (y, z) x = y 6= z =⇒ d (x, z) = 1 = 0 + 1 = d (x, y) + d (y, z) x 6= y = z =⇒ symmetric

x = z 6= y =⇒ d(x, z) = 0 ≤ 1 + 1 = d(x, y) + d (y, z ) all different =⇒ d(x, z) = 1 ≤ 1 + 1 = d(x, y) + d (y, z) ✷ 9. We generalize from R to any V ; every subset X in (V, ρ) is closed because X C is open (every subset is open in (V, ρ)); in (V, ρ) the only convergent sequences are those constant from a certain point on (not only constant sequences as stated at page 337), consider a finite subset A of V , every sequence in A have a constant (hence converging) subsequence, consider now an infinite subset B of V , (by the Axiom of choice) we can construct a sequence with all different elements, where all subsequence will have all different elements, =⇒ compact subsets in (V, ρ) are all and only the finite ones. ✷ 10. Since that every subset X in (V, ρ) is closed (exercise 9), then cl(X) = X, ∀ X ⊆ V , and ∀ X ⊆ V, 6 ∃A ⊂ X s.t. A 6= X and cl (A) = X

3

neither for V itself, so that if V is uncountable, there exists no countable subset of V whose closure is V . ✷ 11. The correct statement of Corollary C.23 (page 340) is: A function f : V1 → V2 is continuous if and only if, ∀ open set U2 ⊆ V2 , f −1 (U2 ) = {x ∈ V1 |f (x) ∈ U2 } is an open set in V1 . (The inverse function of every open set is an open set.) Since every subset of (R, ρ) is open, every function (R, ρ) → (R, d) and (R, ρ) → (R, ρ) is continuous, that is F = G = {f | f : R → R}; clearly not every function (R, d) → (R, d) is continuous (here continuity has its standard meaning), then H ⊂ F = G and H 6= F . ✷ 12. See exercise 11 for the correct statement of Corollary C.23 (page 340); the open sets in (Rn , d ∞ ) and (Rn , d 2 ) coincide, because: d∞ ∀ Bx,r d2 ∃ Bx,r

= {y ∈ Rn | max{|yi − xi |}i∈{1,...n} ≤ r} n X 1 d∞ = {y ∈ Rn | ( (yi − xi )2 ) 2 ≤ r} ⊆ Bx,r i=1

and n X 1 d2 ∀ B x,r = {y ∈ Rn | ( (yi − xi )2 ) 2 ≤ r} i=1

∃q≡

r

r2 d∞ s.t. Bx,q n

d2 = {y ∈ Rn | max{|yi − xi |}i∈{1,...n} ≤ q} ⊆ Bx,r

;

hence also the continuous functions in (Rn , d ∞ ) → (R, d) and the continuous functions in (Rn , d 2 ) → (R, d) coincide. ✷ 13. [The p-adic valuation here is not well defined: if r = 0, a = 0 and any n would do;   if r = 1 there are no a, b, n such that 1 = p−n ab , a is not a multiple of p. For a survey: http://en.wikipedia.org/wiki/P-adic_number ] 14. Since V with the discrete topology is metrizable from (V, ρ), topological compactness is equivalent to compactness in the sequential sense (page 343), hence exercise 9 proves that the compact subsets in (V, ρ) are all and only the finite ones. ✷

4

15. As seen in exercise 11, every such function is continuous. ✷ 16.   12 d [(x, y), (x′ , y′ )] = d 1 (x, y)2 + d 2 (x′ , y ′ )2

positivity and symmetry are straightforward from positivity of d 1 and d 2 ; triangle inequality :  1 2 d 1 (x, z)2 + d 2 (x′ , z ′ )2  1 2 ≤ (d 1 (x, y) + d 1 (y, z ))2 + d 2 (x′ , z ′ )2  1 2 ≤ (d 1 (x, y) + d 1 (y, z ))2 + (d 2 (x′ , y ′ ) + d 2 (y′ , z ′ ))2

d [(x, z), (x′ , z ′ )] =

= = ≤

=

=

 1 2 d 1 (x, y )2 + 2d1 (x, y )d 1 (y, z) + d 1 (y, z )2 + d 2 (x′ , y ′ )2 + 2d 2 (x′ , y ′ )d 2 (y′ , z ′ ) + d 2 (y′ , z ′ )2 h  i 12 d 1 (x, y )2 + d 2 (x′ , y′ )2 + d 1 (y, z )2 + d 2 (y′ , z ′ )2 + 2 d 1 (x, y)d 1 (y, z) + d 2 (x′ , y′ )d 2 (y′ , z ′ ) h d 1 (x, y)2 + d 2 (x′ , y′ )2 + d 1 (y, z)2 + d 2 (y′ , z ′ )2 + . . .  i 1 2 . . . + 2 d 1 (x, y)d 1 (y, z) + d 1 (x, y )d 2 (y′ , z ′ ) + d 2 (x′ , y ′ )d 1 (y, z ) + d 2 (x′ , y ′ )d 2 (y′ , z ′ ) h   i 1 2 d 1 (x, y )2 + d 2 (x′ , y′ )2 + d 1 (y, z )2 + d 2 (y′ , z ′ )2 + 2 d 1 (x, y) + d 2 (x′ , y ′ ) d 1 (y, z) + d 2 (y′ , z ′ ) q √ √ √ since a + b + 2 ab = a + b 1   12  2 + d 1 (y, z)2 + d 2 (y′ , z′ )2 d 1 (x, y )2 + d 2 (x′ , y ′ )2

= d[(x, y ), (x′ , y′ )] + d [(y, z ), (y′ , z ′ )]

. ✷

17. The base of a metric space (X, d 1 ) is the one of its balls B(x, r ), 1 every ball in (X, d 1 ) × (Y, d 2 ) = (X × Y, (d 21 + d 22 )2 ) is of the form: 1

B((x∗ , y ∗ ), r) = {(x, y ) ∈ X × Y | (d 1 (x, x∗ )2 + d 2 (y, y ∗ )2 ) 2 ≤ r}

⊆ {(x, y) ∈ X × Y | max{d 1 (x, x∗ ), d 2 (y, y ∗ )} ≤ r} = B(x∗ , r) × B (y∗ , r)

for every open set in (X, d 1 ) × (Y, d 2 ) there is a couple of open sets, one in (X, d 1 ) and one in (Y, d 2 ), whose product contains them, moreover

5

B (x∗ , r1 ) × B (y∗ , r2 ) = {(x, y ) ∈ X × Y | d 1 (x, x∗ ) ≤ r1 ∧ d2 (y, y ∗ ) ≤ r2 } 1

⊆ {(x, y) ∈ X × Y | (d 1 (x, x∗ )2 + d 2 (y, y ∗ )2 ) 2 ≤ r1 + r2 } = B((x∗ , y ∗ ), r1 + r2 )

then for every couple of open sets in (X, d 1 ) and (Y, d 2 ), there is an open set in the product space containing their product, then the two topologies coincide. ✷ 18. [(a) is not hard only for f compact =⇒ f sequentially compact, the other way round and (b) are really non trivial; see e.g.: http://www-history.mcs.st-and.ac.uk/˜john/MT4522/Lectures/L22.html ] 19. f topological continuous =⇒ f sequentially continuous: consider xn → x ∈ V , and an open Of(x) containing f (x) ∈ V ′ , f −1 Of(x) contains x and all of the xn after a certain n0 , then all of the f (xn ), after the same n0 , are in Of(x) , f is sequelntially continuous; f sequentially continuous =⇒ f topological continuous: non trivial. . . ✷ 20. a) 1) Clearly ∅, [0, 1] ∈ τ ; 2) if O1 , O2 ∈ τ , O1C and OC 2 are finite, C C but then also their union OC 1 ∪ O2 = (O1 ∩ O2 ) is, =⇒ O1 ∩ O2 ∈ τ ; 3) if {Oα }α∈A ⊆ τ , OCα is finite T ∀ α ∈ A, S S but then also their intersection α∈A OCα = ( α∈A Oα)C is, =⇒ α∈A Oα ∈ τ ; b) suppose ∃ x ∈ [0, 1] S with a countable base {Bi }i∈N , C is finite ∀ i ∈ N, since BC i i∈N B i is countable, S since [0, 1] is uncountable, ∃ y 6= x, y ∈ [0, 1] such that y 6∈ i∈N BCi , =⇒ y ∈ Bi ∀ i ∈ N, {y}C contains x but is not contained in any element of its base. ✷

6

Chapter 3 (pp. 90-99): Existence of Solutions. 1. As a counterexample consider f : (0, 1) → R such that f (x) = sup f (x) = limx→0 (1,0)

1 : x

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 6 ∃x ∈ D such that x ≥ x ¯, =⇒ 6 ∃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:  1 if x 6= 0 x f (x, y) = 0 if x = 0 max f on D is limx→0 f (x, y) = limx→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 P elements (because otherwise a∈A ak is different from every a ∈ A but should be in A). Suppose A is convex and ∃ a, b ∈ A such that f (a) 6= f (b), then f[a,b](λ) = f (λa + (1 − λ)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]

7

(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 ∈ sup(f )−1 , moreover, since limx→∞ f (x) = 0, ∃x ¯ s.t. ∀ x > R+ | f (xn ) = sup(f ) − n x ¯, f(x) < 1. {xn } is then limited in the compact set [0, x], ¯ hence it must converge to x ˆ. f (ˆ x) is however sup(f ) and then f (ˆ x) = max(f ). f (x) = e−x , 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 g −1 (f −1 (A)) ⊂ V1 is. g −1 ◦ 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 x≥1

,

f(x) = e−|x|

arg max(f ) = 0, max(f ) = e0 = 1 and also sup(f ◦ g) = 1, which is however never attained. ✷ 8. n | u(y ~) ≥ u ¯} min p~ · x ~ subject to ~x ∈ D ≡ {~ y ∈ R+

s.t. u : Rn+ → 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 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;

8

(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. n max p · g(~ x) − w ~ ′ · ~x s.t. x ~ ∈ R+

with p > 0, g continuous, w ~ ≫ ~0

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

✷

11. ~ subject to φ ~ ∈ {ψ ~ ∈ RN | p~ · ψ~ ≤ 0, y~(φ) ~ ≥ 0} max U (~ y (φ)) PN S ~ = ωs + where ys (φ) i=1 φi zis , U : R+ → R is continuous and strictly increasing ′ S if y ≥ y ′ ∀i ∈ {1, . . . , S} and at least one inequality is strict). (~ y ≫ ~y in R+ i i ~ ∈ RN such that p~ · φ ~ ≤ 0 and Z ′ φ ≫ ~0. Arbitrage: ∃ φ A solution exists ⇐⇒ no arbitrage (=⇒) (by absurd: (A =⇒ B) ⇔ (¬B =⇒ ¬A) ) ~ ∈ RN such that p~ · ψ~ ≤ 0 and Z t ψ ~ ≫ 0~, then any multiple k · ψ ~ of ψ~ suppose ∃ ψ has the same property,

9

~ increases linearly with k and then also U (~y (k · ~ψ)) increases with k, y~(k · ψ)  ~ max U (~ y (k · ψ)) is not bounded above; N ~ ~ (⇐=) PN∀ ψ ∈ R such that p~ · ψ ≤ 0 there are two possibilities: either i=1 ψi zis = 0 ∀ s ∈ {1, . . . S}, PN ψi zis < 0; or ∃ s ∈ {1, . . . S} such that i=1 in the first case the function of k U(~y (k · ~φ)) : R → R is constant, then a maximum exists, in the second case ~y ( ~φ) ≥ 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)) with h(x) ≥ 0, (~x, x) ∈ [0, w ]n+1 , w − x =

n X

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; P 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 π(x) = max xp(x) − c(x) x∈R+

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∗ , π(x) ≤ π(0) = 0, the solution can then be found in the compact set [0, x∗ ] ∈ R+ ; b) now ∀ x > x∗ , π(x) = π(0) = 0, as before; c) consider c(x) = p2¯ · x, now π(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) ≫ ~0, p~(1) · ~c(1) +

10

p~(2) · ~c(2) ≤ W0 1+r

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

p ~(2) ), 1+r

the

p~(2) ~c(1) ≫ ~0, ~c(2) ≫ ~0, p~(1) · ~c(1) + · ~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 ≫ ~0 if and only if ~ ✷ p~(1) ≫ ~0 and p~(2) ≫ 0. 15.   0 ≤ x1 ≤ y1   0 ≤ x2 ≤ f (y1 − x1 ) max π(x1 ) + π (x2 ) + π (x3 ) subject to  0 ≤ x ≤ f (f (y − x ) − x ) 3 1 1 2

let’s 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 3 A ⊆ [0, y1 ] × [0, max f ([0, y1 ])] × [0, max f ([0, max f ([0, y1 ])])] ⊂ R+

and maxima exist because of continuity. Consider (ˆ x1 , x ˆ2 , x ˆ3 ) ∈ Ac in R3+ , then (if we reasonably suppose that f (0) = 0), becasue of continuity, at least one of the following holds: x ˆ1 > y1 , x ˆ2 > f(y1 − x ˆ 1 ) or x ˆ3 > f(f (y1 − x ˆ1 ) − x ˆ2 ). If we take r = max{ˆ x1 − y1 , x ˆ2 − f (y1 − x ˆ 1 ), x ˆ3 − f (f (y1 − x ˆ1 ) + x ˆ 2 )}, r > 0 and B((ˆ x1 , x ˆ2 , x ˆ3 ), r) ⊂ Ac . 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 maxx∈D f (x) = 1, but Df (1) = 1 6= 0. ✷ 2. ′

2

f (x) = 1 − 2x − 3x = 0 for x = ′′

f (x) = −2 − 6x ,



1±

√  1+3 −1 = 1 −3 3

f ′′ (−1) = 4 = −4 f ′′ ( 13 )

−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 + y2 + 10x =⇒ y = 0 ∧ x = 0 fy = 2xy + 2y   12x + 10 2y 2 D f (x, y) = 2y 2x + 2

(0, 0) is a minimum; b)

fx = e2x + 2e2x (x + y2 + 2y) = e2x (1 + 2(x + y2 + 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 ( 12 , −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 − y2 fy = ax − 2xy − x2 2



1 1 2 2 =⇒ (0, 0) ∨ (0, a) ∨ (a, 0) ∨ (− a, − a) ∨ (− a, − a) 5 5 5 5

D f (x, y) =



2y a − 2x − 2y a − 2x − 2y 2x

12



(− 25 a, −51a) and (− 15 a, − 25 a) are local minima ∀ a ∈ R; d) when sin y 6= 0, limits for x ±∞ are ±∞,  fx = sin y =⇒ (0, kπy) ∀ k ∈ Z = {. . . , −2, −1, 0, 1, . . .} fy = x cos y f is null for critical points, adding any (ǫ, ǫ) to them (ǫ < adding (ǫ, −ǫ) to them, f is negative, critical points are saddles;

π ), 2

f is positive,

e) limits for x or y to ±∞ are +∞, ∀ a ∈ R, fx = 4x3 + 2xy 2 fy = 2x2 y − 1



=⇒ y =

1 1 =⇒ 4x3 = − =⇒ impossible 2x2 x

there are no critical points; f) limits for x or y to ±∞ are +∞, fx = 4x3 − 3x2 fy = 4y3



3 =⇒ (0, 0) ∨ ( , 0) 4

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, fx = fy = since f (1, 0) = minimum;

1 2

1−x2 +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),

fx fy

y = 0 =⇒ x = 2 ∨ 3 √ = x8 + 2xy 2 − 1 =⇒ x = 1 =⇒ y = ± 47 2 = 2y(x − 1)    ∨   x = −1 =⇒ impossible  3 2  x + 2y2 4xy 2 D f (x, y) = 8 4xy 2y(x2 − 1) 

     

13

  5 √  √ 0 7 7 2 √4 =⇒ saddle, =⇒ saddle, D f (1, 4 ) = 0 0 7 0 √   5 √ − 7 4 √ =⇒ saddle. ✷ D2 f (1, − 47 ) = − 7 0 D2 f (2, 0) =



3 2

5. The unconstrained function is given by the substitut...