Getting on For all the preceding functions, discuss, whenever possible, whether local min and max are global.


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1 Problems Warming up Find the domain of the following functions, establish in which set they are differentiable, then compute critical points. Apply 2nd order sufficient conditions to qualify them as local ma, min, saddle or can t tell.. f, y = 3 + y y. 2. f, y = lny 2y 2 3. f, y = y 2 2y. f, y = y 3 y 2 + 3y + 2y f, y = 3 + y 5 3 3y f, y = 2 8y + e y 7. f, y = ln 2 + y 2 y 8. f, y = e 2 +y y 9. f, y = 2 y 2 0. f, y = lny 2 2 3y 2. f, y = y f, y = 2 y + y f, y = ln y 2 +y 2 Getting on For all the preceding functions, discuss, whenever possible, whether local min and ma are global.
2 Answer to problems Altogether, we answer the warming up and the getting on part. Solution to # The function f, y = 3 + y y, is defined by a polinomial, hence is differentiable infinte times, with continuous derivatives on R 2. Gradient is + y + 3 f, y = 2 + y 2 + 2y Critical points are solutions to + y y 2 = 0 + 2y = 0 that is the set Z = 0,, 0; 0, 23 }, 2 ; 2, 2 Hessian matri is D 2 6 2y + f, y = 2y + 2 det D 2 f0, 0 = 0 = 6 < 0, 0; 0 saddle point. 2 det D 2 f0, 0 = = 6 < 0, 0; saddle point. 3 det D 2 f2, 2 = = 32 > 0 and a, = 8 > 0, 2; 2 local min. 3 det D 2 f 2 3, 2 = = 32 3 > 0 and a, = 8 < 0, 2 3 ; 2 local ma. 3 Local are not global, ma and min. To prove it, it suffices to observe that and inf R f, = f, = while sup f, = + R Solution to #2. The function f, y = lny 2y 2, Hessian is defined on the open set D =, y R 2 : > 0, y > 0 }, y R 2 : < 0, y < 0 } where it is infinite times continuously differentiable. Gradient is f, y = y + y Critical points are solutions to = 0 y + y = 0 2
3 which solutions are [ [ =, y = 2], and =, y = 2] so that the set of critical points is singleton Z =, } 2 Hessian matri is D 2 f, y = y 2 so that D 2 f 0, 2 = which is negative definite,, local ma. Note that no global min eists: if it eisted, as D is an open set, it would be also a local min, but no local min eists. On the other hand, there is no global ma either. Indeed, for < 0 and y = we have lim f, = lim f t, = lim t + ln t + t 2 = +. t + Remark. Note that the function is *not* concave on D, since D is not a conve set; f is concave on the subset of the domain, y R 2 : > 0, y > 0 } where it attains maimum, but the same conclusion cannot be drawn on the whole domain. Solution to #3 f, y = y 2 2y,, 2 2 = 0 Gradient is f, y =, 2y 2 = = 0 Solution is: [ =, y = ] 2y 2 = Hessian is, f conve, global min 0 2 Ma does not eist f, y = 2 + y 2 2 so that Solution to # lim f, = = + sup f = + R 2. f, y = y 3 y 2 + 3y + 2y 3 2, 6 + 3y Gradient is f, y = 3 2y 3y y = 0 [ Critical points 3 2y 3y = 0, = 97+ 2, y = Hessian is D 2 f, y = 3 6y , determinant: 3 97 < 0 saddle, determinant: 3 97 a local ma 3 ] [, = 97 2, y = ] 97 2
4 Not a global ma, for sup f = lim f0, a = lim a 3 a 2 2a = + R 2 a + a + No global min otherwise it should be a critical point. Indeed we can say more, that is inf f = lim f0, a = lim a 3 a 2 + 2a = R 2 a + a + Solution to #5 f, y = 3 + y 5 3 3y + 2 2,, Gradient is f, y = 3 y = 0 Critical points 3 y 5 2, 3 = 0 Solution is: [ = 3, y = 6], [ =, y = 6], [ = 3, y = ], [ =, y = ] 6 0 Hessian is D 2 f, y =, 0 6y D 2 f 3, 6 =, determinant: 2, saddle point D 2 f, 6 =, determinant: 2, local min, not globlal for inf f = D 2 f 3, = D 2 f, = Solution to #6 f, y = 2 8y + e y y 2 8 Gradient is 8 + e y + 2 Critical points, determinant: 2, local ma, not a global ma for sup f = +, determinant: 2, saddle point y 2 8 = e y + 2 = 0, if y = = 0, Solution is: = 5, = 5 + if = e y 6 = 0, y = ln 6 2y 2 8 Hessian is D 2 f = 2 8 e y D 2 f 5, 0 = D 2 f + 5, 0 = 5, 0; + 5, 0,, ln , determinant: 60, saddle point, determinant: 60, saddle point
5 2 ln D 2 f, ln 6 = ln e Solution to #7. f, y = ln 2 + y 2 y lim f a, = lim a +, determinant: 32 ln 6 > 0, local min, but not global, as [ a 2 + 8a + e ] = inf f = a + The domain of f is D = R 2 \ 0, 0}, while the gradient is 2 f, y = 2 + y 2 y, 2y 2 + y 2 Note also that:  partial derivatives are defined and continuous at every point in D,  D is an open set then f is differentiable in D. Critical points are those P, y such that f, y = 0, that is, the set of solutions of the system 2 2 +y 2 y = 0 2y 2 +y 2 = 0 Case : Assume 0 and y 0 we have 2 2 +y y = 0 2 2y 2 +y = y y = y = 2 2y 2y = y 2 2 +y 2 = 2y 2 2 2y 2 = 0 2 +y = 2 2y + y y = 0 2 +y = 2 2y The first equation is satisfied either for a = y or b = y If a pluggin into second equation we get 2y 2 = 2 y2 = y =, y = so that the system is satisfied by, and,. If b, proceding similarly we get 2y 2 = 2 y2 = that has no real solution. Case 2: Let us wonder whether 0, y with y 0, or, 0 with 0 may be a solution. In the first case we get 0 y = 0 2y y 2 0 = 0 that has no solution. Similarly one derives that, 0 with 0 cannot be a solution. Critical points of f are, and,. We now compute Hessian matri. then 2 2 y 2 y D 2 f, y = 2 +y y 2 2 y 2 2 y 2 2 +y y 2 2 D 2 f, = D 2 f, =
6 whose determinant is , then the matri is indefinte and both critical points are saddle points. Solution to #8 f, y = e 2 +y y. a D = R 2 ; b f is differentiable in D, with gradient f, y = c Critical points are solutions to 2e 2 +y 2 2y = 0 2ye 2 +y 2 2 = 0 2e 2 +y 2 2y, 2ye 2 +y 2 2 The first equation is satisfied if one of the factors is null, that is: = 0, or 2 e 2 +y 2 2y = 0 In case, the 2nd equation gives 2ye y2 = 0 y = 0, that implies 0, 0 is critical. In case 2, we plug 2y = e 2 +y 2 into the 2nd equation, deriving 22y 2 2 = 0 2y 2 2 = 0 2y 2y + = 0, hence either 2a = 2y or 2b = 2y In both cases, plugging = ± 2y into the st equation and assuming, y 0, 0 without loss of generality, as 0,0 has been already computed as a solution, we derive e 3y2 = 2y. Now we prove that such equation has no real solution, by showing the function ϕy := e 3y2 2y, has no zeroes ϕ in R. In particular it suffices to show that ϕ is strictly positive everywhere. Surely ϕ has no zero in, 0] because in such interval e 3y2 > 0 while 2y 0, which implies ϕy > 0 for all y 0. There eist no zero of ϕ also in ]0, 2 ], that is for 0 < y /2, as y > 0 e 3y2 > and 2y so that y > 0 ϕy = e 3y2 2y > = 0, y ]0, 2 ]. What is left to check is the zeroes in ]/2, + [. Note that if we show that ϕ y > 0 for all y > /2 that is, ϕ increasing in ]/2, + [ we may derive that no zero eists in ]/2, + [, as ϕ/2 > 0 and ϕ increasing in ]/2, + [ implies ϕy ϕ/2 > 0, for all y ]/2, + [. Then we show now ϕ y > 0 for all y > /2. Note that ϕ y = 6ye 32 2, so that ϕ /2 = 3e 3/ 2 > 0. Moreover ϕ y = e 3y y 2 > 0 for all y. Then ϕ is itself increasing and positive in /2, the it is strictly positive for all y > /2. The only critical point is 0, 0. d Hessian is D 2 f, y = 2e 2 +y y ye 2 +y 2 ye 2 +y 2 2e 2 +y 2 + 2y 2 D 2 f0, 0 = Since f 0, 0 = 2 > 0, e det D 2 f0, 0 = > 0 we derive 0, 0 is a local min. Global minimization of f Weierstrass Theorem is needed, see lecture # We now discuss whether 0;0 is also a GLOBAL min. The idea is to prove that the originale problem is equivalent to maimizing f on a compact region. 6
7 As eponential function e 2 +y 2 grows, positively, at infinity faster than 2 y one may guess that indeed a global min does eist and is attained in a neighborhood of the origin. Nevertheless the assertion has to be precisely proven. We make use here of definition of global min and of Weierstrass Theorem in R n. If we show that f is coercive, that is in this case lim f = + + we may then infer that the f attains its minimum on R 2, and that the local minimizer is indeed a global minimizer. And consequently 0; 0 is shown to be the unique global minimum point for f on R 2. Why? If f coercive, there eists a radius R 0 > 0 such that inf R 2 f = inf B R0 f = min B R0 f where B R0 =, y R 2 : 2 + y 2 R 2 0}. Hence inf is indeed a min by means of Weierstrass Theorem, because f is continuous and B R0 is a compact subset of R n. Moreover, that is also a global min for f on the entire R 2. Ultimately, the minimizer has to be a critical point of f also, as it is a minimum attained on an open unbounded region. We then show that f is coercive. Let R > 0, and assume = R. Then 2 y R 3 so that for all, y B R 0; 0 one has f, y e R2 2R 3 =: ϕr Passing to limits as R +, we derive lim f, y lim + which implies that f is coercive. Solution to #9 ϕr = R + lim e R2 2R 3 = + R + f, y = 2 y 2 The domain is D =, y : y + y 0} =, y : y > 0 and + y > 0}, y : y 0 and + y 0} but f is differentiable only at D =, y : y + y > 0} =, y : y > 0 and + y > 0}, y : y < 0 and + y < 0} At any couple of type, or,, f attains the value 0 which is the minimum attainable value, hence global min points are, : R}, : R} Such points may not be met among critical points, as the function f is not differentiable at those points. A global ma does not eist, as lim f, 0 = lim =
8 Gradient is Critical points 2 y 2 y 2 y 2 Solution to #0. f, y = lny 2 2 3y 2, Domain D =, y : y > 2 + } y 2 Gradient is 2 6y + y 2 2 y 2 Critical points 2 = 0 6y + y 2 2 = 0, Solution is: P 0, P 2 0, / D, D Hessian is D 2 f 0, y 2 2 y y y = , whose determinant is: = , Then P 2 is a local ma. P 2 is the unique global maimizer. Alternative very epensive proof : f, y < ln y 3y 2 ma y> ln y 3y2 = f P 2 We now show that P 2 is a global ma. Indeed, let 0, y 0, y : y = }. Then lim f, y =,y ln = 0,y 0 so that there eists ε > 0 for instance, one may choose ε = 0.0 such that where the closed set D ε is given by for in the complement set sup D f = sup f D ε D ε =, y D : y ε}, y D D ε =, y D : < y < ε} f, y < ln ε 3y 2 < ln ε = ln 0.0 =. 60 < fp
9 Let s now show that If we do so then lim f, y =,y +,,y D sup D f = sup f = sup f D ε D ε B where B is a suitable closed ball centered at the origin and, being D ε B closed and bounded, hence compact, f would attain its ma in D ε B, hence in D, hence at P 2. Now note that y > 0, y + + y + + y + so that, if y +, then f, y < ln y 3y 2 ; if instead +, then y +, same conclusion, so that is proved. Solution to # Trivial: without any computation, 3, is the global minimizer and no local ma eists; sup f = +. Solution to #2 f, y = 2 y + y + 2,, The function is defined on D =, y : y 0} but differentiable in D o =, y : y > 0} 2 Gradient is y 2 y 2 y + 2 Critical points y 2 y = 0 y + = 0, Solution is: [ = 2, y = ] 8 Hessian is 2 = [ y2 y y 2 y 2 2 y 3 2 [ 3 ] ] = 2,y= 8 whose determinant is 2, so that 2, 8 is a local min. Nonetheless, it is not a global min. Indeed f 2, 8 = [ 2 ] y + y + 2 = = 2,y= 6 8 [ f, 00 = 2 ] y + y + 2 = 39 = Solution to #3 f, y = ln a Domain: y 2 +y 2. D =, y R 2 : =,y= 00 y 2 + y 2 > 0, 2 + y 2 0} =, y R 2 : y > 0,, y 0, 0} =, y R 2 : > 0, y > 0}, y R 2 : < 0, y < 0} 9
10 b f, y = y y 2, 2 y 2 2 +y 2 y differentiable in all D D open set and partial derivatives are continuous at every point of D. c The set of critical points Z is made of the solutions to that is y y 2 = 0 y 2 2 = 0 2 y 2 2 +y 2 y = 0 2 y 2 = 0 Z =, y D : y y + = 0} y y + = 0 =, y D : y = }, y D : y = } =, y D : y = } =, y D : y = } Let us try to apply 2nd order cpnditions. The Hessian matri is D 2 f, y = 2 y 2 y 2 +y y 2 +y 2 2 y 2 +y 2 2 y 2 2 +y y 2 y + and computed at points of Z becomes D 2 f, = so that f, < 0, but det Hf, = 0, so that 2nd order sufficient conditions do not apply. From 2nd order necessary conditions, we know that critical points cannot be local min, but could be ma or saddle. Note further that the Hessian is negative or null in the domain det D 2 f, y = y 2 + y 2 y y so that it is indefinite not conve nor concave in D Z, and negative semidefinite in Z. Indeed all points in Z are GLOBAL ma. Note that, since on D we have y > 0, then 0 < y 2 + y 2 2 as the second inequality is true y 2 0. Moreover 2 is attained at all points of Z. Hence f attains its GLOBAL ma with ma f = ln 2. The property is very easy to guess, nonetheless one may not get it at first sight. The one may start analyzing level sets Γ k of f, discovering that they are of type 2 + y 2 2αy = 0, with α =. Then, 2e k by a change of coordinates, one checks that the curve 2 + y 2 2αy = 0 describes two lines through the origin, so that all level curves are indeed of type, β for some positive β. Since f, β = ln β + β 2 and such function of β attains its ma at on the positive real ais, then the maimizing level curve was ;. 0
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