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Inaeqvationes in plano et programmatio linearis

Svmmarivm

Haec lectio stvdivm inaeqvationvm in planvm transfert. Demonstrat rectam planvm in dvo semiplana dividere, in qvorvm vtroqve expressio linearis qvae eam definit signvm svvm servat, et inde methodvm pvncti experimenti dedvcit. Regionem solvtionvm systematis inaeqvationvm linearivm vt intersectionem semiplanorvm describit, cvm verticibvs svis, et eam convexam esse ostendit; deinde regiones conicis terminatas considerat. Programmatione lineari conclvditvr: inqvisitione maximi vel minimi fvnctionis linearis in regione polygonali, cvivs theorema fvndamentale affirmat in polygono convexo circvmscripto vtrvmqve in verticibvs attingi et, in regione non circvmscripta qvae aliqvem verticem habet, omne optimvm qvod exstet etiam in vno eorvm attingi; et problematis de prodvctione, de victv et de vectvra.

Proposita discendi

  1. Copiam solvtionvm inaeqvationis linearis dvarvm variabilivm vt semiplanvm, apertvm vel clavsvm, ope pvncti experimenti repraesentare.
  2. Regionem solvtionvm systematis inaeqvationvm linearivm, vertices eivs et vtrvm circvmscripta sit necne determinare, et convexitatem eivs ivstificare.
  3. Regiones rectis et conicis terminatas per inaeqvalitates describere, et regiones sic descriptas repraesentare.
  4. Problemata programmationis linearis dvarvm variabilivm constitvere et solvere theoremate verticvm et rectis libramenti, distingvendo optimvm vnicvm, optima mvltiplicia et optimi absentiam.

Semiplana

In lectione de aeqvatione rectae visvm est omnem rectam esse copiam pvnctorvm qvae aeqvationi Ax+By+C=0Ax + By + C = 0 satisfacivnt, vbi AA et BB non ambo nvlla svnt: si B≠0B \neq 0, explicatvr y=−ABx−CBy = -\frac{A}{B}x - \frac{C}{B}, recta declivitatis −AB-\frac{A}{B}; si B=0B = 0, aeqvatio ad x=−CAx = -\frac{C}{A} redigitvr, recta verticalis. Si pro aeqvalitate inaeqvalitas postvlatvr, copia solvtionvm iam non est cvrva sed regio fit. In lectione de aeqvationibvs et inaeqvationibvs primi gradvs, inaeqvatio linearis vnivs incognitae intervallvm rectae realis pro solvtione habebat; cvm dvabvs incognitis, solvtio est semiplanvm.

Sit rr recta non verticalis et sit P=(x0,y0)P = (x_0, y_0) pvnctvm. Dicemvs PP esse svpra rr si y0y_0 maior est qvam ordinata pvncti rectae rr cvivs abscissa est x0x_0, et infra si minor est. Si rr verticalis est, eodem modo de pvnctis ad dextram eivs et ad sinistram sitis dicitvr. Vtraqve harvm dvarvm copiarvm est semiplanvm apertvm cvivs ora est rr; si recta additvr, est semiplanvm clavsvm.

TheoremaSemiplana

Sit rr recta Ax+By+C=0Ax + By + C = 0, vbi AA et BB non ambo nvlla svnt, et sit E(x,y)=Ax+By+CE(x, y) = Ax + By + C. Expressio EE signvm constans habet vtrimqve ab rr: in omnibvs pvnctis alterivs e dvobvs semiplanis apertis positiva est et in omnibvs alterivs negativa. Accvrativs, si B>0B > 0, svpra rr positiva est et infra negativa; si B<0B < 0, contra; et si B=0B = 0, ad dextram rectae rr positiva est cvm A>0A > 0, et ad sinistram cvm A<0A < 0.

Demonstratio

  1. Ax0+By1+C=0\dato{q}{A x_0 + B y_1 + C = 0}

    Sit P=(x0,y0)P = (x_0, y_0) pvnctvm qvod in rr non est, et ponamvs primvm B≠0B \neq 0. Recta verticalis per PP transiens secat rr in vnico pvncto Q=(x0,y1)Q = (x_0, y_1), cvivs ordinata obtinetvr explicando yy ex aeqvatione rectae rr; pvnctvm QQ eidem aeqvationi satisfacit.

  2. E(x0,y0)=(Ax0+By0+C)−(Ax0+By1+C)=B⋅(y0−y1)E(x_0, y_0) = \resaltar{(A x_0 + B y_0 + C) - (A x_0 + B y_1 + C)} = \dato{dif}{\resaltar{B \cdot (y_0 - y_1)}}

    Hoc nihilvm a valore expressionis EE in PP svbtrahimvs: termini in x0x_0 et constantes se mvtvo tollvnt, et prodvctvm remanet.

  3. B>0→(E(x0,y0)>0↔y0>y1)B > 0 \rightarrow \left( E(x_0, y_0) > 0 \leftrightarrow y_0 > y_1 \right)

    Si B>0B > 0, per regvlam signorvm in lectione de tabvlis signorvm et inaeqvationibvs explicatam, prodvctvm signvm habet expressionis y0−y1y_0 - y_1: positivvm est cvm PP svpra rr est, et negativvm cvm infra.

  4. B<0→(E(x0,y0)>0↔y0<y1)B < 0 \rightarrow \left( E(x_0, y_0) > 0 \leftrightarrow y_0 < y_1 \right)

    Si B<0B < 0, prodvctvm signvm contrarivm habet signo expressionis y0−y1y_0 - y_1: expressio svpra rr negativa est et infra positiva.

  5. E(x0,y0)=(Ax0+C)−(Ax1+C)=A⋅(x0−x1)E(x_0, y_0) = \resaltar{(A x_0 + C) - (A x_1 + C)} = \resaltar{A \cdot (x_0 - x_1)}

    Si B=0B = 0, tvm A≠0A \neq 0 et rr est recta verticalis x=x1x = x_1, vbi x1=−CAx_1 = -\frac{C}{A}. Calcvlvs iteratvr cvm pvncto rectae rr qvod ordinatam pvncti PP habet.

  6. A>0→(E(x0,y0)>0↔x0>x1)A > 0 \rightarrow \left( E(x_0, y_0) > 0 \leftrightarrow x_0 > x_1 \right)

    Si A>0A > 0, expressio positiva est prorsvs cvm x0>x1x_0 > x_1, id est ad dextram rectae rr; si A<0A < 0, ad sinistram. In omnibvs casibvs signvm expressionis EE in pvncto pendet tantvm ex latere rectae rr in qvo pvnctvm est.

Theorema plvs dicit qvam qvod in vsv adhibetvr: non opvs est meminisse qvod signvm cviqve lateri respondeat, qvoniam satis est signvm expressionis EE in vno pvncto alicvivs lateris novisse vt in omnibvs noscatvr. Haec est methodvs pvncti experimenti.

Ad inaeqvationem Ax+By+C>0Ax + By + C > 0, vel cvm <<, ≥\ge avt ≤\le, repraesentandam, dvcitvr recta Ax+By+C=0Ax + By + C = 0 et valor expressionis compvtatvr in pvncto experimenti qvod in ea non est, plervmqve in origine. Si pvnctvm inaeqvationi satisfacit, copia solvtionvm est semiplanvm qvod id continet; si non satisfacit, est altervm. Si inaeqvalitas stricta est, recta ad copiam solvtionvm non pertinet et linea interrvpta dvcitvr; si non est, ad eam pertinet et linea continva dvcitvr.
ExemplvmSemiplanvm per pvnctvm experimenti

Copiam solvtionvm inaeqvationis 2x−3y+6>02x - 3y + 6 > 0 repraesentare.

Demonstratio

  1. 2x−3y+6=0,(0,2),(−3,0)2x - 3y + 6 = 0, \quad (0, 2), \quad (-3, 0)

    Rectam finientem per intersectiones cvm axibvs dvcimvs: cvm x=0x = 0 obtinetvr y=2y = 2, et cvm y=0y = 0, x=−3x = -3.

  2. 2⋅0−3⋅0+6=6>02 \cdot 0 - 3 \cdot 0 + 6 = \resaltar{6} > 0

    Origo in recta non est. Valorem expressionis in ea compvtamvs: positivvs est, ita vt origo inaeqvationi satisfaciat.

  3. 2x−3y+6>0↔y<23x+22x - 3y + 6 > 0 \leftrightarrow \resaltar{y < \frac{2}{3}x + 2}

    Copia solvtionvm est semiplanvm qvod originem continet. Theorema id confirmat: cvm B=−3<0B = -3 < 0, expressio infra rectam positiva est.

  4. 2⋅(−3)−3⋅0+6=02 \cdot (-3) - 3 \cdot 0 + 6 = \resaltar{0}

    Inaeqvalitas stricta est: pvncta rectae, vt (−3,0)(-3, 0), expressionem ad nihilvm redigvnt neqve ei satisfacivnt. Ora exclvditvr et linea interrvpta dvcitvr.