Solutions_to_PracticeFinal.pdf

Solutions_to_PracticeFinal.pdf - 3 For each of the...

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Unformatted text preview: 3. For each of the following vector fiekls find the divergence and curl: (a) (b) This vector field represents an attractive force generated at 61W = _i the origin and decaying according to an inverse; square law. [513 Instead of the input being written as three coordinate var!" ables, it is written as a, generic displacement vector, 7". H. 693""? E(%.z),%\ ."w .a (fia‘b‘m’ff’a 0433,?» 1 ma. vuw ”I'M. w .5 nut-Luci}v MN Ol'V any, fine.“ V“??? :1 V?” F +. pv‘fi Pr 0"" dwwfigcdnr (WW 5 o “flak W12 5“? "£1. I». 1, {3%. ~92 V5 2 3(K‘Jro“a%) «39.43.?» J» 4M3 m Elihh \7 [(4 “Ham“ 1.) a. 1‘92. 4.. I. L ‘ 1' L f) L :0 'Rsofi m {Krb¢3\“3ck*'fl* -,-. ”~37; V F” :9 E +~ W75? %cx‘+om (WM 1: \’-“— “a x - . ,3 97m Mm Mohk WA "'1‘ V“ ‘3 wan) ‘ "“ ‘ ' 37 f 7 k ' .3 a" 2.1 E; “H: M L L‘ V J” 3 '- \ ":7 A 3E 210: rm) "“5 1b 1’ H CD ‘35:": [ Hz- 4 H<¢gfil:z) = ~213+ (cf—F213 V‘fi 7‘ (shah 3:“X «3'3ng "1 D+o+ “‘1 "a 1:. ., fi 3% 3;; c (own -—(o.o\7§ + (WHEN: V E M3 :2“: I?(w, y, z) = «axp(~w2 H y2 w 2:2)? ”f ‘5‘ c. v' r'L 1' o3 VXR : 9. fig. 3; ._ 0"? w—(Dv-szflflb Hf) 2w by M: v“ . nthy‘nl" a, (5%}? o o 4— (0 ,6 (4.3 a )) k :16w(—><°vgha.")<olm a, :3) Which of these vector fields is h'rotationa]? Which could represent the flow of an incom- pressible fluid? Wk: ck a A an. aha/Ave? .9 “FE MJ‘ G are Nrfiwkww‘ and M cmsarvL-WQ, “\H a» (0A4 19¢ ’IM rflw “1L M Vuw‘qnmmk ’flwf'Al (bx/(,u/I yaks; 7w '}"°L ‘Ng, glaJm'F-Mw. a; q .614“, $1M P‘) “4%? “’4" ha. C’WIQNI‘L'VM \M wkb‘. WV‘N: I’w’)’ Wake/‘7 vaafirs) cm % Find the average value of wyz on each of the. following curves. (Hint: you will need to divide the line intergral of the function $313 on tlle'cm‘ve by the length of the curve. Finding the length of the curve may or may not require doing an arc length integral, Lo. 21 line integrai of the constant function 1, over the same mn‘ve). (to) The line segment from (0,0,0) to (1,2,6) WW 4 curve» W = «is wamu WH- 1:va FL“: (4,11%. 54;.) é, 63:0, Y.) Me: 2. <5. 2.5) mm; m | 1 w ’ 25 [am 5‘ Sivan; at: ¢ Swan: {2391b m lofiaséi 0H: .. 1L 0 C}. 55:30 7‘ w J: we Mm raj“. cf 5“ m fifi 2“ (h) The portion of the. circular helix given by {(cos mm m) It 6 [0, 4w] ] a“ “a, We. we. a‘w W k at; a“ an» Leuo'hx bake; M (1‘. hmfl. NHL°(H\ Uf‘ 1L) +0 ’Ghak M ‘5“‘34’k. W; Mut- e, [\wam‘MWM “FHA: <¢°Hh 3N3!“ ‘5) +61%) tmfi] 9 f“ H51 1 «macaw D an» W’mw {enammceumqw 1“ fi‘fi -::- ('2: 39 1033+“ v vam “A 5410‘?!) '4 éfl’h’ . 5W a 3mm almsb am? ”Rh” 8 fififids '3“ 3 665(5) SM?) "3'?- 1f: 0”; .'. (Casi: “A; a ‘ ° s. 336 3;.be 7. 4‘3“ '% 5mm» 15 c/Ié m {w}: o ld‘%¢£v dv:SMHiw% ‘ W M w V5wfi-MM) '1 % (wfiécoumh +£icesxzmdfl 4* R ‘m 10 ‘I’h r" L: [’CLJT ¢°b{%)1o &”12:«'%«$MCM)\O ) '59 5mg. value. 1} $wfifl~0+o~o 2-4"qu #mgfiq‘, (c) The circle of radius 1 in thq plane z 2 1 with center at (0,0,1) HUG- mu ”$0119? m curut LS 411:. da‘uwmf‘umce‘ °F W awk Gar-1. 2 MY . 3 (Pmrqwh’w M Cum/l 'm' 17‘”) 5: (cast, Ss‘k'f?‘ 1> reelioflv’fi' mm) MAM) (N we. v’hW‘ W J“ M“ 7%., 7 W76! 1 =- (~sm’c314Lcwk1‘ato‘“ 5,. "57%;!“ 43mg cam 0) ,fi ply—25H: (aooL warmed ’fka MAW“ , $0 '2.“- @vaMPB’vJ‘mV‘, yaw. As 2. Scosft\$\\~{+').l .315 a a lei" um SW) (Au: ml’flcll: 6 5‘» F Yacht {5‘0 :2; uwndo) b 0 €92.11 m u:sM/W)='o t #9 H L Gwea vv\vL r‘ 1’“, fig C we wk, "be ska/M Lq okuwvy who# «H 01A 7%) N Work? 11 g. Let C! be the curve givanparamem-icafly by #(t) 2 253+ g} _. 25$ for t 6 [0,4] Find each of the following line integrates (notice line integrals of various types occur here). Which of them would be useful to compute work done by a. force vector field on a particle moving- along the curve? Which of them wouhl be useful as a, step in finding the average value of a ftmution'on the curve? ( l .3 hi a) fmds “F"(éi'r 23 +473 ~21: at? =- l‘?’(6lldfi ‘3 “L‘aatz +923“ at“ \‘E’H‘q’ ‘H" X/B)’¢-'Lb . a" ‘ ’r (a; u: em? 15w 5» u A 330“ ’1- 314: J 94"L'2" #14; ALL?» We‘d": {‘"fv-‘g flirts-t G- o . ' 2‘? I ' ,5 14' (A '2‘ a: -u"’a!u lira“ 8) c 3.2«m — WM? 8 :fié‘f LE 7 ' ‘ "W USQ'fiJ (m-filmy 3 ”he (In/“V3,? VJVQ o§ '\ {VAC/+30% (b) where 1%», y, z) a 3127+ 1:; + 11" “(i/(£3) :— (1‘ (a, "1) WWW 4; 2+. 1:) ”fit Ftr)~rA£“ L "L ’37.. Offim . 7‘" dt 1+ 7.6: {r +1MJ 0,5 4» Decide which of the following vector fields are conservahivo and find a, potential funcflan for each. (If the descrip§ion of the geator field has only two variables and only mentions the bask: vectors 3 and 3", but not A! it gives a. vector field on the guy—plane, otherwise it gives a vector field on 3—space.) For each vector field which is conservative, find apotential function. 9" ,, T 94* ‘- L H, t: \ (1») 1303,21) = (H1071 + (w + 2w)? 9‘3 3) ‘ ’3“ 2-. W (m ”'3‘ 5° «3:; waawc {S- o‘ pakgk'uk fiwhm (b) @(w.y)m—%vi’~l'$3 T a~. r; _, “" *5; ("W ”’6“ .3.”— X r; } ax Maj“ tow Eflrva‘LW 14 (c) H(w,y,z)= Zuzana-(3:2 2 2F 1:2? 1- 2ycos(m3+y2 +333 "I chos(m3+ 312 +25% +64. . mic-50‘" M31" 9% 23. “1'33 V” H m VQ’L “5 WW8 “- WK {Viva 2) + Zanflflgfif) W} WWHWW l3)\ 7- {Wrmiu‘wn +2?) WWW‘WWI 7% Shfk‘w‘» ‘3)” “I 25 $9 “‘ a: 3L“ .me'cufi‘h.‘ 5121.)}(03 a 1 a ‘0:- . ‘ a fins»; “ ‘3’ 2’ '5“. 0 ‘flwoiw‘fi K a» '25. “”7““ .33 ’g‘HA m Po‘kfi’k‘A \'6;M8\\5V‘ k k y 5- HI! [7 ”PM Mt‘kg 1M- WW)’ " :5 k n g ‘Lx as {xi—Lamvé') olx <3 Sakfafigahfi) + NEW] :. sxkcx’dwgfifi) Sirfihl'gl 5., 53th Wyn?) + COW) (70w NgL-f la-uuefivbmfi vap‘ 5}; 9M} (1MP g7 waved; MA Dixie/1c,“ \V‘fi‘fiavp 9L Pervk‘fll lXJtJMI-CM )I. 15 g For ouch of the vector fields in the previous question assume the coordinafsos of displaco~ ' mont vectors (the inputs) are IIIGRSUI'Gd in meters and tho vectors represent a force in Newtons at each point. Find the work done by the vector fields 13 and (3" in moving a. particle from (1,1) to (~1,-1) along the posh fit) :4 (cos t,81nt}, and the work done by H' in moving a particle from (1,0,0) to (1,0,471') aiong the port Ht) : (oost,sint,t). (Hint: the probiem will be much easier for the conservative vector fields than for the homeonservative vector fields.) 6) WHIL. pint)... ‘07 E Wfi‘n [Ivlv‘fi‘h‘wd ‘fihd’h’h sgx’i'fi‘fl'f‘flm f3; §(“"*h "- 'F’th : (”4 +1 '1'” "" (”l4-¥\ 1:. ”*1 N-w C”- Nfinfi. Wfia flan“? hi. mwtmv" ) - G) HMS M” CMdeM 5° w“ W 3W OLD t-l-wo .ufiwfll Emb’w " ’49! tr 'x'j‘ W473: (cash m~£> 6:91}!le ”f? w: 83:4?" 3“ g (“awhcob ' (How-MD (:49 a 0 ‘ fr: 1 m ‘P S 93,9135 +~ caf’k 0“? 7“ [ 4,:+ 2,60: (70% r: (W 0 alt- " '1 . ME f; z; |+uca521r fl Sn L 3~ L ,1 Sb I“ 16 - 242+»2Esm2fl 2» 1r+Ow£0~°i 0 3‘ ’11 NW E. continuéd H: 75 can 6W‘3VVQ' 4.. 17 6 Compute each of the following in two ways: as given and by using Green’s Thgorcm to - rewrite it. f1? 0 df" 0 (a) ' where 0' is the chefs an“ +y2~ =36 (oriented positively) and f(<v,y)== (~y,:¢). (Hint or waning: you’ll need to give a pammetnization of the circle to do this dimctly, but once it’s l‘ewfltten using Green’ 3 Theorem the double integral should be easy.) C “max; ova if. gummewma ‘27 ML): gwsimw HEW] In dff= < 656* 6”» 6AA fi(81&‘\ '2, Zwéfi'i‘n} 605£> 7m Q's! “~— ' @fid? :3 S3Gsm7‘t +'§.(,c.fi£~alt .5 £36563 .. 712,374 0 O D 7‘ ~ WA“ :1» ”21114)- / f </"“” Gr“ okfibl’ (b) ' ff fligdli R {labels R is the disk of mdius 2 centeled at the migkn. (Hint m wmning: how the _ dhect wayw— do the double integzal -— is fahly easy, and a, review of stufi‘ fiom Exam 2. Using Gwen’s theoram involves picking (13,62) so that swam—0131392 :32 Any P and Q wifih that property will do, and again, chasing & pammetiization of the muse, in this case $118 circle bounding the disk with positive mientation. ) l'. (A A {om-AH \m’f'UH—‘i your cp'énlrnJL, BK'NerH gig"? Rafe Mam rvlrcm M Nraam .3 (#:qu v: 0 er 9.1 o 99 9%” 11-“ 7.7 ”“10” 3 S Ware Ar 019 & 111 ($1 L '3» :8“ 0;?sz d9“; Avgm (:7 A9 0 fi (70 - s . m : ‘ I '34:“) :2, 1"” M124 4% M" ' Am a ,, <ng o> To ow MM (“Wm ‘91“! M” 'Q‘F—O , ?¢”){8 S4 F " , “F (u v; @w. am» e econ“? Y” . fim’mnfi Z” (16%le \e 2: em 4a] m“ ' . " '” 1L, 5-9 @Ed? '" S \bcofi’csmzfdi %‘ ’37 bad ”OR. 0 2 r :2 w»? w \M 54‘;be M‘ \(Ww Q0” 3 \,, 0“ Fa». 1.}. Use Green’s Theorem to computofihe area, inside the cardioid given in polar coax-(Imam. ‘ by a“ = 1 + 31110 (Hint: this can be given parametricany by using 7: = 6', thus givlng the parametrization in rectangular coordinates fit) :: (1 + sint) costg+ (I + aint)slnt§ for t E {0, 271’]. And, don’t. be afraid to use an intagrgt table.) 34 0% o{ N WW‘JKD‘M we» "FNWVM w.) A? g) x43 M M clav— éiJcLH'mmmjfli-b 'afl ' '- 53? (wt 4 M) 4+: .' . T, LCD-S": ,4. 15Mf655+14 (3 5: WW . .. P‘ 7’ S U '3' gktw (RYE (gvyt rylswwk (95+) ah, ‘ o I it 1 “I: . 1g“ “I: +« Coswc mi; .5. 23m as”; + 23ak+m +91 ‘ '7‘" CO) ‘ ‘ 0 ...- 28“ an”: "1‘ '3) WEE SW1: Jr, '9» Sh-Q'lc of”: 0H: 0 [PE 271 W ' sh . flank; if +§E§f a 05%;; _+ 11% ,, T b \k’T uuffis‘v‘k dig 2 I 5 M? $ (£15 (WWW‘V‘LO “(€ch 538%)“ t w.” 9% Jr ‘r ' \"‘ Mrs) 41;“ [w 9/ /$\‘: 1.”. (W I} (j; r“ Ef: A.» f/ wk f, (57$ ‘3. The ellipsoid 6‘ usually given as the locus of 9:53 + ”13 +252 =2 1 can be given parametrlcally by t 11052)) I: (3 slnueosu, 2alnushw, cosy} (um) e [0, 1r] X [0, 2w] (Aside: this was obtained from the parametrization of the unit sphere about the ortgln using the angles from spherical coordinates as parmnetem by stretching by at factor of 3 in the se-dlrectioa and a factor of 2 in the y—directien. All that stuff about stretching, re~ fleeting amt translntlng curves from high school algebra goes on worklng ln 3 dimensions.) Use the parametrization do do the following: (it) Find 9"“ X 7"}. (this will be helpful in later parts). If": 1- (3 co; ucysv, 2,60)” sew; ~ Sl‘hu) l; .1! "” <“3-Tl‘tlt3an‘, 7-3?“ka v! o> ' ' a Smu§ NJ '* "L c v 33.6%”.le 6cm mi, — (mm a. t J | i (v 1. WW”! 3““! 0’) V l. 3‘ (la) Find the cquntlon of the plane tangent to the ellipsoid at (fi, Egg, %§), Then check your answer by findlng the tangent plane using the fact that the ellipsoid 13 the level set for a. function and 113ng the gradient of that functlon at the point of tnngenoy as in Chapter 14. t Ls: c. 1:: “+l§% %l umm‘%‘ “fi”%% fifiwlfim “ ‘ i (jig: 15"? 9% In To V9 as"? 'MRT’: o O ((3) Write an ktesated integia-l (with the palame’mm as vaxlables of integxfiatlon in some convenient ordea) that gives the sunface men of the ellipsoid. Do not evaluate the itelfited Integral (1th very unpleasant: ) , EMS ‘== WNW» E. 1'4 z“. T! . ' ‘ fl” 3 S fl'S‘uA‘LLcos‘v 4 ‘lsfiu smzv Hens‘mmwa'aalv o o . . ‘ ((1) Write an iieralted bitegt'al equal fa the surface integral ff Eats 15 5 . Perform any obvious simplificaitons of the integrand, but do not evaluate the iterated integral. (Again it is very unpleasant). w w w? M "H 4% M‘M * ’2. I . . ' ., 6 5m kw (\‘Cug‘h ouq‘v *i‘ (T Smd‘h Suqu’I' 3“” ”mm" cam dw‘v (e) Let; 5 be fake vector field glven by 15‘(m,y, z) 5-4554. 2; + :35. Write an iterated integral giving the flux of 5‘ through 8, that. is‘equal to the sm‘face integral, ff if: 9 45” a Do not evaluate the integral. (Emu though It’s not; nearly as unpleasmt asjhe '38 others.) E {W’HNH ”v (”15mm $3th 0524; 38§nupos v} 2"" w ‘ N 14.99 V> {(1. Suki.“ (.oJV 3 S’IVI‘TL'SMV 6C!Jh5;hh> Wild-3 7. S S (v’LS-mevy (‘6le w} .{ , J a ' ° " am 5L, “PR ‘1' , , d ' d I“ S S '“ 4‘353k. S'W‘VOJV &S$\'“1u 6.6311.st + {Esmmnmu my I; v i) U . (E) Now, use she Diverggnce g‘heorem to evaluate the surface integral giving the flux of A - 13mm) = mm: + 23’ + wk thxough a. SSfi'GLE 1- SSS VHF: 01V ' 9‘1E(fv.3t$)‘ ”2% f‘sgg}'-LLH]3 e R \7-("5 ’= fimw MMfiX 'h OHM F. .. lee pm‘mnetrlmtions of each of the following curves or surfaces, including what. interval of region the parameter or pair of parameters lie in: {a} The 031310 of radiua 3 about the: origin in the plane. 5M0» )‘y. iJ'M #5. «'2 r“: ’3 wt. °“," we (L, WW {,Hr‘cJ-mk ‘3) t ginozdr .' m «a we, 29% 9> 9 WW (1)) The sphere of radius 2 about; the origin (i113napaca). 5km. ‘k‘b‘c— lava—"Vb"\a,. 56’Nr-‘Ctl £2.13wa W32. 50 “Q “ml 9 auvt ‘3 f‘r’nmfirfilfimf W9); '<1sh\c{;cc$9l1.svxxcf$m6,_ 2c&$(‘?> . cg €— Qonfl‘ 9 GENT]; (o) The fine segmept from (1,~2 1) to {..2 3 2) . (IQ. khk. pf +‘VJN3 Cmvfix cmhx'kw‘} wawlus Lire. . aU—aOH-a 13+ £492 3 Z) ' “'9“? :{1'942 2.5}..1mrbdw3b I 154%) :{1 at wag wi> (d) The part; of the pm'abofic hyperboioid z :2 ya w 2 lying Inside the cylinder of radius 1 about the zvmds. ”’1’ k ‘Ufig‘w. UK N fitcikfenlr‘bf'k VWLLI/ a) mkkfv‘lfln ‘/ WED 0330 (e) The ta‘latigie with verbices (1;, ~2, 1), (—2, 3, 2) and (o, o, 0). ' ' ' Could ufir‘NvJ w”‘7" 7}“ A} Mr ‘ 7 w M we“ . ‘ .Q'M W 04% ' ‘FH’ [Na 3*“. Irwim'ué ”IL,‘ “”43; kak W M ~> w Mm +m 4m awe, . hwy, M“ (”HUM \- WUK’L‘ \> + ’LLévti'arz) .+ V<0tola> w V}? (ya/M,“ Hui 03v, away} [11M G Em") \ 094’ . . h!) (”IL s ’9‘"; °3£~m\35“é M mahwcl {V IN. RUM” 75 OWE ' ' ' 10 [0. Evaluate the foilowing sm‘face integrals: (a) fa'S )3 where B is the part of the plane 3m+2y+ 52: a 301y111g lusida the cylinder of radius 4 about: the z~axis. (Le. {ind bhg surface area of 2) Nb. Mask ‘(v @Qqfi-‘aMm 1k S‘W'Fh-LQ, N )x‘MA. Ni- Cu».- Wbat 'FW :3. ‘. we, C-A w» 9&3th 11,0.) dy‘erM- Mb) '5 '2. T; 3";er . % 6 ‘5 53' . XLHjLQ‘i‘Q‘ 4‘ 5’ “fir" {It 0: ~’$/§> F9": (‘3; I, ”IL/37> fixfl: <34517f1 W m" C 3 W? A “as — lgfiwix 2 may)“ M \/ 4; d =v @dfir j . 3 F ’2— (194k 5" 830B ‘1‘ Sggdfl 2 %‘W4 (cw Nkmé FL (D4? 11 '"" 'erfi‘fié ‘:v::'wr'23,w M 5" ,a,‘ \cmwh (b) - . f/wg—zdS' x: .where )3 Is the same surface as far the previous part. , FY’M“ (A) wu NM ‘51,? *:. @dfs g, Magma Wam‘l’nfizqflfim mt WM" ~ ‘ Naval» by “Awake“ NV ’10» Why'd ‘VkQaL . I Mint M $st 2: SSW—m 6 *%x*%a)€_§ 51A 010:2 «:2 I ' 'D ' ' ' €7NM'HZ' . - 'L or I ctr—«w aka/J” W» ' U 4km.7(a~{ ":2 Hoe: m Ema/A W. :3.sz .7“va :Mkrg‘m‘a 9+ \R “:in m “M W, 1 {Aik‘a‘n 9+ .--' 38 Q. r“ C ”w... ‘TF 4‘ ex)»,- vvrfybq ’1‘ 83 5* JP? . 5" W”, D W O in: 3 WM WW 9 6 o). r ”g 5‘ F, ggé Y c I) 7'“ cm [“3 6-1? Sam 49 ., 5— 0 m ' .3. M65, 9 53:39 ,. fiéwfi? 7’ t“ 13" 3 5" (Afisr 'quwzg- : Mam"??? a? "F” ‘ ””37" 5‘ (c) _ fffioaflg' U where U is the part of this graph of z = af"+y2 iying above the Square [~1, 1] x {—‘I, 1} in the ivy-plate. (is. find the flux of a flow with constant velocity 1 upward through the surface). ‘ Pflmyfl hksflg¢x. _ PM v3 ~— <14. v. 243w“ (11,va E~\‘1'3x£-:',n KM) '4 <h 0. “0. a (14“!) hr- (0‘. \’LV> fixfi mm: (rm, "2N . J>_ (70 .9— ;ww an my M71 SQ) \ ) .... fi » 1.; _§ S @1033» <~m,~2v,\3 MW 35%» M N .4 ‘ u . 4 , : S§_\AuAv. «I n ‘ g. 4 (W <+ .w E’WWl-y 13 (Q. Consider the'xrcctor field Waging) : (wyfiumg), and £113 parametrizations of tha unit circle 6‘ about hhopxigln in the w—plane given by . fit) m (cost,siut,0) t G [0,21r} . and of the part of the unit sphere about the origin lying above the (By-plane H given by Rum) : (siliucosv, aln 2935212), cosy) (um) 6 {9, g} x {55%} Notice that 0‘ is film boundmy of H, and thus Stokes’ Theorem 21336163 that the welkntype ‘line integral of 15' along 0 equals a certain smfzwo integm! mm H Write an or‘dinmy defimte integlal equal to the: welkntypa line integral along 0 and an oldinaly iterated integral equal to the amfaca integml oval H which Stokas’ Thcoxem assures us is equal ' to It You do not need to evaluate eithm lntcgmi. VN1"%\’1L LN mJ-Obm‘.‘ N C firm-46} < 334:»: (”fin o3 Phat)? <15“? 0'3" °> [53mg $4.9M” 9 (—wé m ow K mg C» . MM WW :, m w} if v); E-ab‘ 3.) $fi~ ‘wl. M124 WK? 7“ 0"? Jag)? + 1E h> N 0V5? ("1.1M V)?~ (cosucosv (3051A ”w, ~smu§ FWVW) ".4 4-» swx saw} _5~"*"'~ cwv) 0 § "(a x?" 1* 43mm cesv} sin a saw; ovsuswwx) . 'I\ v ., “\Wz .ngxi“. MAE "j, g S (agglu) ' <9?»th 0:5“... pM’Lk-ymvl CJJM§§A¢L> 5114.61" 13: 9 ° ' 14 “1&7“ VF: J§S 7.59msmha’1wlv "7:: 5? min 61qu MW PW “a Find the absolute (81km. global) maximum and minimum of the function 031 the region ROW?) = $2 —- 233 —I- '43; {@wW+wsm (and at what points they ocem‘). There is only one way to look for candidates points in the interior, but for the boundary, try this two ways: use Lagrange multipliers, and (as another way) turn it into a, one-variable problem by pm‘antebriamg the boundary “£3,301.43” + 2:3 a 2} as {(x/icosfl, fish) (1)19 e {Dari} and doing the waiting onto 1 problem with 3 as the variabie. Igfin}%lx R%(K0)¢'“‘31*% $6 [by {a} g, hm kwmawjz 5’57 Ufofl My" Vim/YR PVQ (5 1 @5493“ X143“ ¢ ’2. I; 4w W$~¢ <fl~£»65‘+4> Q?) 1' 41$: 2'3) (Orig) “mt. C(l}‘a.v'\ ovum") *- lya'bx I f MJM‘H-l S» ‘1" 9-“W*. 0" )D‘. [rev/“Jul! u NW!%‘¢MLSM& )9 a? crfix‘ad po’w‘r) *th-Wi'fiu 4¢Cb* U“ 4 1K=9I X310 n-\ 91H 1:70:va M “W. ihfg on. bu mam. ...
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    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

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    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

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    Dana University of Pennsylvania ‘17, Course Hero Intern

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    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

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    Jill Tulane University ‘16, Course Hero Intern