Programacion Lineal

November 23, 2022 | Author: Anonymous | Category: N/A
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2. Zl bvj gj rbpdòb ljojsdtb ljojsdtb pbrb sumsdstdr bn gàb ?5 uldgbgjs gj prctjàl prctjàlbs, bs, =5 gj

hrbsbs y 1 gj vdtb`dlbs. _us prjsbs scl gcs tdpcs gj bld`bnjs; rbtcljs quj nj prcpcrodclbl ? uldgbgjs gj prctjàlbs, 7 gj hrbsb y 2 gj vdtb`dlbs9 y pbnc`bs, quj nj prcpcrodclbl 6 uldgbgjs gj prctjàlbs, = gj hrbsbs y 2 gj vdtb`dlbs. _d obzbr y oc`jr ul rbtÿl nj oujstb 8 uldgbgjs gj jljrhàb y ulb pbnc`b 2= uldgbgjs gj jljr jljrhàb hàb,, ³ouâ ³ouâltb ltbs s prjs prjsbs bs gj obgb obgb onbsj onbsj gjmj gjmj obzbr obzbr pbrb pbrb sbtds sbtdsfb fbojr ojr sus ljojsdgbgjs, ocl jn `jlcr hbstc gj jljrhàb3 _cnuodÿl ]jstrdoodcljs; tjlj`cs : rjstrdoodcljs nbs oubnjs scl ? x 2 + 6 x = ≪ ?5 7  x 2 + = x = ≪ =5

 x 2 + x = ≪ 1

 x 2 ≪ 5  x = ≪ 5

Nb fulodÿl cmkjtdvc f  ( ( x 2 , x = )0 8 x 2 + 2= x =

 

]/tb; Ubrb sbtdsfbojr sus ljojsdgbgjs jn bvj gjmj obzbr 6 rbtcljs y = pbnc`bs =.  ocl 15 Ih. gj bojrc y 2=5 Ih. gj bnu`dldc sj qudjrj abojr mdodonjtbs gj

`cltbòb y gj pbsjc quj sj vjlgjrâl b =55 jurcs y 2:5 jurcs rjspjotdvb`jltj. Ubrb nb gj `cltbòb scl ljojsbrdcs 2 Ih. Gj bojrc y ? Ih. gj bnu`dldc, y pbrb nb pbsjc gj = Ih. gj obgb ulc gj ncs `jtbnjs. ³Ouâltbs mdodonjtbs gj pbsjc y gj `cltbòb sj gjmjl fbmrdobr pbrb cmtjljr jn `âxd`c mjljfdodc3

 

_cnuodÿl  x 2 0lu`j lu`jrcgj rcgj mdodonj mdodonjtbs tbsgj gj pbsjc

 x = 0lu`j lu`jrcgj rcgj mdodonjtb mdodonjtbss gj gj`clt `cltbòb bòb

]jstrdoodcljs = x 2 + x = ≧ 15 = x 2 + ? x = ≧ 2=5

 x 2 ≪ 5  x = ≪ 5

Fulodÿl cmkjtdvc f  ( ( x 2 , x = )0 2:5 x 2 + =55 x =

 

]/tb; Ubrb cmtjljr jn `bycr mjljfdodc quj js gj 1:55 jurcs sj gjmjl fbmrdobr =5

mdodonjtbs gj `cltbòb y ?5 mdodonjtbs gj pbsjc

?. ?.   Ubrb bmclbr ulb pbrojnb gj aujrtb sj ljojsdtbl, pcr nc `jlcs, 1 ih gj ldtrÿhjlc y 2= ih gj fÿsfcrc. _j gdspclj gj ul prcguotc @ ouyc prjodc js gj ? jurcs pcr idnchrb`c y quj ocltdjlj ul 25% gj ldtrÿhjlc y ?5% gj fÿsfcrc y ctrc prcguotc L quj ocltdjlj ul =5% gj ldtrÿhjlc y ul =5% gj fÿsfcrc y ouyc prjodc js gj 7 jurcs pcr idnchrb`c. ³Ruä obltdgbgjs sj gjmj tc`br gj @ y gj L pbrb bmclbr nb pbrojnb jl jn `jlcr  hbstc pcsdmnj3  x 2 0 prcg  prcguotc uotc @ 

 x = 0 pr  prcguotc cguotc L 

]jstrdoodcljs 5.2 x 2 + 5.= x = ≪ 1 5.? x 2 + 5.= x = ≪ 2=

 x 2 ≪ 5  x = ≪ 5

Fulodÿl cmkjtdvc 2

f  ( ( x , x

=

)

0?

 x

2+7

=

 x

 

]/tb; Ubrb bmclbr nb pbrojnb bn `jlcr ocstc pcr vbncr gj 215jurcs sj rjqudjrj =5

gjn prcguotc @ y ?5 obltdgbgjs gjn prcguotc L 7. Zl oc`jrodbltj gjsjb oc`prbr gcs tdpcs gj frdhcràfdocs, F2 y F=. Ncs gjn tdpc F2

oujstbl ?55 jurcs y ncs gjn tdpc F=, :55 jurcs. _cnc gdspclj gj sdtdc pbrb =5 frdhcràfdocs y gj 8555 jurcs pbrb abojr nbs oc`prbs. ³Ouâltcs frdhcràfdocs ab gj oc`prbr gj obgb tdpc pbrb cmtjljr mjljfdodcs `âxd`cs jl nb vjltb pcstjrdcr, sbmdjlgc quj jl obgb frdhcràfdoc hblb jn ?5 % gjn prjodc gj oc`prb3 ]jstrdoodcljs  F 2 + F = ≧ =5 ?55

2 + :55

 F 

=

 F  ≧

8555

 

Fulodÿl cmkjtdvc (5.? rjprjsjltb jn ?5%) f  ( ( F 2 , F = )0 (?55 Ø 5.? ) F   2 2 + ( :55 Ø 5.? ) F   = = ↝ f  (  ( F 2 , F  =  = )045 F 2 + 2:5 F =

 

:. Zlb dlgustrdb vdlàocnb prcguoj vdlc y vdlbhrj. Jn gcmnj gj nb prcguoodÿl prcguoodÿl gj vdlc

js sdj`prj `jlcr c dhubn quj nb prcguoodÿl gj vdlbhrj `âs oubtrc uldgbgjs.  Bgj`âs, jn trdpnj gj nb prcguoodÿl prcguoodÿl gj vdlbhrj `âs oubtrc vjojs nb prcguoodÿl prcguoodÿl gj vdlc js sdj`prj `jlcr c dhubn quj 21 uldgbgjs. Abnnbr jn lû`jrc gj uldgbgjs gj obgb obgb prcgu prcguot otc c quj quj sj gjmjl gjmjl prcg prcguo uodr dr pbrb pbrb bnoblz bnoblzbr br ul mjljf mjljfdod dodc c `âxd` `âxd`c, c, sbmdjlgc quj obgb uldgbg gj vdlc gjkb ul mjljfdodc gj 1 jurcs y obgb uldgbg gj vdlbhrj = jurcs. ]jstrdoodcljs = x 2∝ x = ≧ 7 7  x 2 + ? x = ≧ 21

 x 2 ≪ 5

 

 x = ≪ 5

Fulodÿl cmkjtdvc f  ( ( x 2 , x = )0 1 x 2 + = x =

 

]/tb;  Ubrb Ubrb cmtjl cmtjljr jr ul mjlj mjljfdo fdodc dc `âx `âxd`c d`c gj =1 jurc jurcs s sj ljoj ljojsdt sdtb b prcg prcguo uodr dr ? uldgbgjs gj vdlc y = gj vdlbhrj

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