MA2112 Práctica 09 – Multiplicadores de Lagrange pdf
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(2) Ejercicio 1 Considere una placa circular plana de ecuación: x +y ≤1 2. 2. incluyendo la frontera. Esta placa se calienta tal que el calor en un punto (x,y) esta dado por T (x, y) = x + 2y − x . 2. 2. Halle la temperatura en los puntos mas calientes y mas fríos sobre la placa.. 2.
(3) Gráfica de nivel de la temperatura T(x,y). 3.
(4) Esquema aproximado de la placa. 4.
(5) Ejercicio 2. Sea f(x,y), tal que: ∂f (a, b) ∂f (a, b) = =0 ∂x ∂y. ¿debe tener f(x,y) un máximo o un mínimo en (a,b)? Razone su respuesta.. 5.
(6) Problema 1. Extremos sobre una elipse. Encuentre los puntos sobre la elipse de ecuación x + 2y = 1 2. 2. donde la función f (x, y) = xy. alcanza sus valores extremos. 6.
(7) ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, Cont 14.8 LAGRANGE MULTIPLIERS 2 ContourPlot[f[x,y], discriminant= fxx fyy " fxy {x, xmin, xmax}, {y, ymin, ymax} fx= D[f[x,y], x]; fx= y], D[f[x,y], x]; fxx} /.critical {{x, y}, f[x, discriminant, f (x, y) = xy fy= D[f[x,y], 1. ™y]; f œ yi ! xj and ™ g œ 2xi ! 4yj so that ™ f œ - ™ fy= D[f[x,y], y]; " fxy2 È2 # Ê x œ 8x- fy==0},{x, Ê - œfy==0},{x, „y}] œ 0. critical=Solve[{fx==0, 4 or xy}] 14.8 LAGRANGE MULTIPLIERS critical=Solve[{fx==0, minant, fxx} /.critical CASE 1: x]; If x œ 0, then y œ 0. But (0ß 0) is not on the fxx= D[fx, x];D[fx, fxx= È2 2 2 fxy= D[fx, y]; g(x, y) = x + 2y − 1 1. ™ f œ y i ! x j and ™ g œ 2x i ! 4y j so that ™„ gÈ Ê2yyiÊ !x CASE 2: y]; x Á 0 Ê - œ „ 4™ fÊœ x- œ fxy= D[fx, LIERS È2 # y]; fyy= D[fy, È2 " Ê x œ 8x-fyy= Ê D[fy, -œ „ y]; 4 or x œ 0. Therefore f takes on2 its extreme values at Š „ 2 ß # ‹ an discriminant= 2xi ! 4yj so that ™ f1:œ If -™ yifyy ! x" jfxx (2x(0 i ß!fxy œ 2x - andsoxxœÁ4y fyy " CASE xdiscriminant= œg 0,Êfxx then yœ 0.œfxy But 0)4yis2j)notÊonythe ellipse 0 È2 # È2 are „ . {{x, y}, f[x, y], discriminant, fxx} /.critical È 2 {{x, y}, f[x, # y], discriminant, fxx} È /.critical È or x œ 0.. CASE 2: x Á 0 Ê - œ „ 4 Ê x œ „ 2y Ê Š „ 2y‹ ! œ 0. Buty(0 ß 0)xjisœnot onithe ellipse ™ g Ê i! -(2x ! 4y j) ÊsoyxœÁ2x0.- and x œ 4y- È2 " È2 " Therefore f takes on its extreme values at „ ß and „ ß " 2. ™ f œ y i ! x j and ™ g œ 2x i ! 2y j so that ™ f œ ™ Š ‹ Š # MULTIPLIERS 14.8 LAGRANGE MULTIPLIERS 2 # 2 # 14.8 LAGRANGE È2 È2y x„j œ -(2x ix!œ4y„ j) È Ê2yy Ê œ 2x -„and x‹œ ! 4y2y -# # œ 1 Ê y œ „ " . 1 Ê Š Ê x œ 4x- Ê x œ 0 or - œ „ 4 # 2. È2 „ # . ellipse so x Á 0.are È È21: If"x œ 0, then y œ 0. But (0ß 0) is not on the 2 " CASE eme values1. " . iThe extreme values the ‹!and ™„f 1. œ jŠ and œ# ‹2x j so ™that fofœf on gellipse fxœ y„ i !™ and ™ g!œ4y 2x i !that 4y j so ™™ fœ -Ê™y # at Š 2 ßy#i™ 2xjß g 1 "‰ " 0.„ È2y‹ ! 2y# œ 1 Ê y œCASE ˆ 2: x Á 0 Ê œ „ Ê y œ 2x „ œ „ „ . Š È2 È2 2 # # # # 2. Ê ™ fxœœyi8x !-xxj and ™ gÊ œ„2x-i œ !or 2y j so that ™ f0.œ - ™ g Ê yi ! x Ê œ 8x „ or x œ Ê œ x œ 0. o 4 4 " # È 1 extreme values at Š „ È5ß È5‹ 2y œ 1 „ . 2Ê y # Therefore f takes on its " œ # Ê x œ 4x Ê x œ 0 or œ „ nd Š „ 2 ß " #CASE the But ‹ . The1:extreme CASE x of œyf0,on yellipse œ 0.ß 0) Butis(0 ß 0)onis the not ellipse on the elli 2 .(0 If x œ1:values 0,Ifthen œthen 0. not so 4. # # circle are 5 and " 5. CASE 1: If x œ 0, then y œ 0. But (0 ß 0) is not on the circle x ! y " È-2(2xi !È2y The of f™ong the ellipse 2 j) ÊÈy œ 2x ! 2yextreme j so that values ™fœ Ê y i ! x j œ and x œ 2y È2y È CASE 2: Ê x Á- 0œ Ê œ „ Ê x œ „ Ê „ Š CASE 2: x Á 0 „ Ê x œ „ 2y Ê „ 2y Š ‹ 1 " 4 # 4 1 ˆ ‰ r - œ „ 2 . CASE 2: x Á 0 Ê - œ „ 2 Ê y œ 2x „ # œ „ x Ê x ! a „ È2 " È32j so È 3. ™ f œ " 2x i " 2y j and ™ g œ i ! that ™ f œ " # # f takes on its extreme and-7 2 È ThereforeTherefore f takes on its extreme values atvalues „ atÈŠß „‹ andß Š‹„ ŠÈ. " ‹ .i # 2x.
(8) ,y], y];{x, [f[x,y], xmax}, {y,{y, ymin, ymax}, AxesLabel Ä {x, y, z}] t[f[x,y], {x,xmin, xmin, xmax}, ymin, ymax}, ContourShading Ä False, Contours Ä 40] fxx= D[fx, x]; rPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False, Contours Ä 40] olve[{fx==0, fy==0},{x, y}] ], x]; fxy= D[fx, y]; x]; x,[x,y], x]; ], y]; Caso 1: [x,y], y]; fyy= D[fy, y]; x,ve[{fx==0, y]; fy==0},{x, y}] =Solve[{fx==0, fy==0},{x, y}] Si x=0 entonces y=0, pero (0,0) no esta en la elipse discriminant= fx x]; y, y]; [fx, x]; 2 entonces x no puede ser igual a cero. y]; fxx fyy {{x, y}, f[x, y], ant= " fxy [fx, y]; y]; y], discriminant, fxx} /.critical [x, [fy, y]; nt= fxx fyy " fxy2 2 14.8 LAGRANGE MU inant= fxx fyy " fxy x != 0 , y], discriminant, fxx} /.critical , f[x, y], discriminant, fxx} /.critical E MULTIPLIERS. 1.. ™ f œ yi ! xj and ™. MULTIPLIERS GE j andMULTIPLIERS ™ gCaso œ 2xi2: ! 4yj so that ™ f œ - ™ g Ê yi ! xj œ -(2xi ! 4yj) Ê #y œ 2x Ê x œ 8x Ê È2 esor entonces: œœSi xsojœso 0.that™™ and ™-g™ 2x j4y that f de œf œ -cero, ™ g gÊ jj)) Ê 2x and0,xx tœ œ 4!i4y xÊ j and g„ œxi2x !diferente -™ Êyiy! i !xjxjœœ--(2x (2xi i!!4y 4y Ê y1: yœ œ If 2xx -œ and CASE. È2 È 2 x But œ 0, then y œ 0. (0 ß 0) is not on the ellipse so x Á 0. Ê - œ- „ or œ 0. -# Ê œ „ or x œ 0. 4. œx00,œÊ then œy0. 0, then œ -y œ „. 0ÁÊ - œ- „ 0 Ê œ. 4. # È 2 But (0 ß(0 0)x isœnot on the ellipse so È 0. But ß 0) is „ not on the ellipse soxÈ xÁ2y Á0.‹ 0. Ê 2y Ê „ Š 4 ##. È2 È2 x œx „ 4 4Ê Ê. CASE 2: x Á 0 Ê. ! 2y# œ 1 Ê y œ „ "# .. "" ## ÈÈ Therefore f takes on „È 2y2y Ê2ÊŠ" Š „„ 2y2y œ 1 Ê y œ „ . ‹‹!È œ „È !2y 2y œ 1 Ê y œ „ . È 2 ## ". kes on its extreme values at Š „. takes onextreme its extreme values Š s on its values at Šat„. ß # ‹ and Š „ 2 ß " # ‹ . The extreme values of f o 2 È2È2" " ÈÈ È2 2 2 "" „ ß and „ ß" . The extreme values of f on the elli ‹ 2 ß2 # ‹# and. Š Š„. . #‹ 2 2ß " # ‹. The extreme arevalues „ of.f on the elli #. El valor extremo de f 2. ™ f œ yi ! xj and ™ 8.
(9) Problema 2. Encuentre los puntos sobre la superficie de ecuación x y=2 2. mas cercanos al origen.. 9.
(10) ngly away yœgets to 2xy 0,2x noi j! points are 2x farthest away). œ a™ yclose b 2yÊ œ y and 2y œ 2 xy. i! yj sofar that ™#as f 6. -We g Ê j # # È optimize f(x ß y) œ x ! y , the squareaway). of the distance to the origin subjec x œ 3 and y œ 18 Ê x œ 3 and y œ „ 3 2.farthest easingly far away as y gets close to 0, no points are ## # not satisf CASE 1: If y œ 0, then x œ 0. But (0 ß 0) does œ x y " 2 œ 0. Thus ™ f œ 2x i ! 2y j and ™ g œ 2xy i ! x j sog(x thatß y)™ f # # satisfy the constraint xy œ 54 soofythe Á 0. mize f(x ß y) œ x ! y , the square distance to the origin subject to the constraint # # #curve xy œ 542 nearest nnt"the xy2 œ 54 has points " 2y the# origin (since " xy œ 54 so y Á 0. # # " -‰g(x ˆ -2‰œœ0.-yf(x ‰ œ , and since xof54 œthe 0Ê Ê ˆiy!‰œxˆ#0to (but g(0 ß™ 0) fÁ 0). Thus x Áconstraint 0œ and2xy 2x œan #Ê2x-i.#! y œ Then xy œ œ 54 ˆ " Thus ™ f œ 2y j ™ g œ 2xy j so that œ ™ g Ê 2x x optimize ßÊ y) œ x ! y , the square distance the origin subject to the CASE Ê 2 œ -2: If y Á 0, then 2 - œ- 2-x Ê x œ 2 " 2 # ˆÈ ‰ ˆ - ‰ œ 54 2y œ . Then xy œ 54 Ê 0, no points are farthest away). # # # # ## 2x # È ˆf2y ‰ 2 œ since x œ 0 Ê y œ 0 (but g(0 ß 0) Á 0). Thus x Á 0 and œ 2xy Ê x œ 2y Ê a 2y b y " 2 œ 0 Ê y œ 1 (since y 0) Ê x œ „ 2 . Therefore xy yxœ",g(x, 2œ 0. Thus ™ f œ 2x i ! 2y j and ™ g œ 2xy i ! x j so that ™ œ ™ g Ê 2x œŠ 18 Ê x œ 3 and y œ „ 3 2. x y) = x y − 2 " " $ # 2y 2y È œ Ê Ê x œ 3 and y #œ ˆ ‰ # Ê y œÊ # - xœ and y œ „ 3 2. #- œ È È , since x œ 0 0 (but g(0 ß 0) Á 0). Thus Á 0 and 2x œ 2xy Ê x # # 27 3 x origin y1 (since œ 2 nearest theÊ origin œ 2 has points farpoints awayœon as2 by " œnearest 0 Ê ythe œ y # 0) œ(since „ points 2x . yTherefore 2ß 1‹ are Š „ increasingly x the xy œx254 (since xy œx54 has re of# the distance to the# origin subject to the constraint g(xß y) # farthest away). È È2to #po aare 2yorigin b y " 2 œ 0 Ê y œ 1 (since y # 0) Ê x œ „ 2 . Therefore „ ß 1 are the È Š ‹ he (since xy œ 54 has points nearest the origin (since x y œ 2 has points increasingly far away as x gets close 0, no points Therefore Š$ß „ 3 2‹ are the points on the curve xy œ farthest away).. j and ™ g œ 2xyi ! x# j so #that ™ f œ - ™ g Ê 2x œ 2xy- and 2y œ x# away). œ (since x y œ 2 has points2yincreasingly far away as x gets close to 0, no . 2 nearest the origin # # 7.Thus (a) x ™ i ! 2x j and yi‰!Ê xj so that ™ f œ - ™ g Ê i ! j œ -(yi ˆœxy t g(0ß 0) Á 0). Á f0œand œ™ 2xygas xclose œ 2y increasingly far away gets to 0, no points are f. hest away). tance to the origin subject to" the constraint g(xß y) " " " x œ Ê œ 16 Ê œ „ . Use œ x 1#œ0 and y #10.œ T f œ i ! j and ™ g œ y i ! x j so that ™ g Ê i i ! j ) Ê y and - ™ f œ -È 4! j œ -(y 4 xsince # # È yœsubject #2xy 0)i ! Ê x œ „ 2 . Therefore „ 2 ß 1 are the points on the curve Š ‹ x j so that ™ f œ ™ g Ê 2x œ 2xy and 2y œ x g(x y)at the point " "to the constraint " isß 8 " value (4ß 4). Now, xy œ 0. 16,Then x # 0, y# 0 is ya œ branch oft œ™ Ê œ 16 Ê œ „ . Use œ since x # 0 and y # x œ 4 and 4 Ê 2y # # f œ -i x!Á j and ™2x g œ y2xy i! thatx4™œf 2y œ##- ™ g# Ê i ! j œ -(yi ! xj) Ê 1 œ -y an - Thus 4 ˆxj so ‰ Ê 0). 0 and x at ™8 fat œthe-point ™ g (4optimize Ê 2x œfar 2xy - and 2y œ xasymptotes. -0 is 2 has points increasingly away as x gets close to 0, no points are with the x-and y-axes as The equations x !the y œ distance c give a fam 6. We f(x ß y) œ x ! y , the square of ue is ß 4). Now, xy œ 16, x # 0, y # a branch of a hyperbola in the first yquad " " " " x œ Ê œ 16 Ê œ „ . Use œ since x # 0 and y # 0. Then x œ 4 and œ 2y È È # # 4 4 As these lines move away from the origin, the number c increases. Thus œ 2 y-axes . ˆTherefore 2ß The 1‹ are the points on curvea family of parallel lines with m Š„ ‰ dxh 2x œ 2xy Ê x œ 2y # the „ x-and as asymptotes. equations x ! y œ c give xœ x y " 2 œ 0. Thus ™ f œ 2xi ! 2yj and ™ g œ 2xy value is 8 at the point (4where ß 4). Now, x # 0, # hyperbola's 0 is a branch of a hyperbola in the fir x ! yxy œœ c is16, tangent toythe branch. lines move away from the origin, the number c increases. Thus the minimum value of c occ sthese increasingly far away as xpoints gets close to 0, no points are È fore „ 2 ß 1 are the on the curve Š ‹ 2y with the x-and y-axes as™ asymptotes. The equations x j!soythat œ c™ give afamily of parallel lines (b) f œ y i ! x j and ™ g œ i ! f œ ™ g Ê y i ! x j œ( " Ê œ , since x œ 0 Ê y œ 0 (but g(0 ß 0) Á 0). T ere x ! y œ c is tangent to the hyperbola's branch. that ™ f œ - ™ g Ê xi ! j œ -(yi ! xj) Ê 1 œ -y and 1 œ -x Ê y œ - and As these lines move away origin, c increases. Thus the Êfrom x points œthe 8 Ê f()ß )the ) œnumber 64 is the maximum value. Theminimum equationsvalue xy œ way as x gets close to 0, no are fUse œ yi-!œ xj "and ™ gxœ#i ! j so that ™ f Then œ - ™xgœÊ4 and yi !yxjœœ4-(Ê i ! jthe ) Êminimum yœ- œ xy!yœ since 0 and y # 0. # get where x ! y4 œ c isatangent toa" the hyperbola's branch. to maximum value) give a family of hyperbolas in theÊ first x andœthird Ê 2y b y 2 œ 0 Ê y œ 1 (since y # 0) „ " x œ 8 Ê f( )ß ) ) œ 64 is the maximum value. The equations xy œ c (x # 0 and y # 0 or x $ 0 œ - ™ g Ê i ! j œ -(yi ! xj) Ê 1 œ -y and 1 œ -x Ê y œ - and ™œ f œ16, yi x!# xj 0, andy # ™ 0gaxes œ aias! j so that f œ maximum - ™ ginÊthe yifirst !ofxjcquadrant œ -(i !where j) Ê the y œhyperbo -œx xy is branch of a™hyperbola asymptotes. The value occurs et a maximum value) give a family of hyperbolas in the first and third quadrants with the xand since x # 0 and y # 0. Then x œ 4 and y œ 4 Ê the minimum # f()ß )) œx 64 # ! yis œgive 16.maximum xœ 8 Ê the value. The equations xym œœ c (x # 0 and y incr #0o x y œ 2 nearest the origin (since x y œ 2 has points s.s Ê The equations x ! y œ c a family of parallel lines with " 1. " as asymptotes. The maximum value of c occurs whereand the hyperbola xy œ c is tangent to the li œ (y i ! x j ) Ê 1 œ y and 1 œ x Ê y œ - in the first and third quadrants with the # 0, y a#maximum 0 is a branch of give a hyperbola inofthe first quadrant to get value) a family hyperbolas origin, c increases. Thus the minimum value of c occurs y œ 16.the number farthest away). # the # 0.as 4 Let and ymaximum minimum tions xThen ! y œx cœ a family ofxÊ parallel lines with mof œthe "1. 8.give f(x ßœ y) 4 œ !value y# be thec square distance from the origin. Then ™t axes asymptotes. The of occurs where the hyperbola xy œ c is tangent 10 yperbola's branch..
(11) Caso 1: artial Derivatives. Si square x=0 entonces cero, entonces: œ x# ! y# , the of the distancey=0, to thepero origin, g(0,0) subject tono thees constraint œ 0. Thus ™ f œ 2xi ! 2yj and ™ g œ y# i ! 2xyj so that ™ f œ - ™ g Ê 2xi ! 2yj Ê 2x œ -y# and 2y œ 2-xy. x != 0 # hen x œ 0. But (0ß 0) does not satisfy the constraint xy œ 54 so y Á 0. hen 2 œ 2-x Ê x œ -" Ê 2 ˆ -" ‰ œ -y# Ê y# œ -2 . Then xy# œ 54 Ê ˆ -" ‰ ˆ -2 ‰ œ 54 Caso 2: " " Ê - œ Ê x œ 3 and y# œ 18 Ê x œ 3 and y œ „ 3È2. 27. 3. È2‹ are the points on the curve xy# œ 54 nearest the origin (since xy# œ 54 has points Si x es distinto de cero, entonces:. ! are"farthest away). y as y gets close to 0, no points 2x = 2xy. 2y x2. #. ⇒ x = 2y ⇒ 2y 2. 2. 2. $. y−2=0⇒y =1. œ x# ! y# , the square of the distance to the origin subject to the constraint g(xß y) us ™ f œ 2xi ! 2yj and ™ g œ 2xyi ! x# j so that ™ f œ - ™ g Ê 2x œ 2xy- and 2y œ x# como y>0 # # ‰ œ 0 Ê y œ 0 (but g(0ß 0) Á 0). Thus x Á 0 and 2x œ 2xy ˆ 2y Ê x œ 2y x. √È Ê y œ 1 (since y # 0) Êx x=œ±„ 2 2 . Therefore Š „ È2ß 1‹ are the points on the curve. origin (since x# y œ 2 has points increasingly far away as x gets close to 0, no points are. 11.
(12) Ejercicio 3. Utilice el método de los multiplicadores de Lagrange para encontrar el valor mínimo de la función f (x, y) = x + y sujeta ala restricción xy = 16, x > 0, y > 0 .. 12.
(13)
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