Tiêu đề Linear Programming Board CQ & MCQ Practice Sheet Solution.pdf ✅

2  Higher Math 2nd Paper Chapter-2 Rhombus Publications 2| A I B `yB ai‡bi Lvevi Av‡Q| hvi g‡a ̈ †cÖvwUb I †k¦Zmvi wb¤œiƒc: [w`. †ev. 17] Lv` ̈ †cÖvwUb †k¦Zmvi cÖwZ GK‡Ki `vg A 4 5 40 UvKv B 6 3 50 UvKv •`wbK b~ ̈bZg cÖ‡qvRb 16 11 (K) †hvMvkÖqx †cÖvMÖvg ej‡Z wK eyS? (L) mgm ̈vwUi †hvMvkÖqx †cÖvMÖvg MVb Ki| (M) †jLwP‡Îi mvnv‡h ̈ †hvMvkÖqx †cÖvMÖvgwUi mgvavb Ki| mgvavb: (K) me©wb¤œ wewb‡qv‡Mi wewbg‡q m¤¢ve ̈ m‡e©v”P gybvdv AR©‡bi j‡ÿ ̈ †Kvb cwiKíbv‡K D‡Ïk ̈ dvskb, wm×všÍ PjK I kZ© ev mxgve×Zv GB wZbwU Z_ ̈‡K GKNvZ PjK wewkó mgxKiY ev AmgZvq iƒc`vb Ki‡j †h MvwYwZK mgm ̈v cvIqv hvq Zv‡K †hvMvkÖqx †cÖvMÖvg e‡j| (L) g‡b Kwi, x †KwR A Lv` ̈ Ges y †KwR B Lv` ̈ cÖ‡qvRb| Afxó dvskb: †gvU LiP, Z = 40x + 50y (me©wb¤œKiY) mxgve×Zv: 4x + 6y  16 (†cÖvwUb) 5x + 3y  11 (†k¦Zmvi) Ges x  0, y  0 (Ans.) (M) ÔLÕ Gi AmgZv ̧‡jvi Abyiƒc •iwLK mgxKiY: x 4 + y 8 3 = 4 ......... (i) x 11 5 + y 11 3 = 1 ......... (ii) x = 0 ......... (iii) y = 0 ......... (iv) QK KvM‡R g~jwe›`y O Ges x Aÿ I y Aÿ‡K h_vμ‡g XOX Ges YOY Øviv wPwýZ K‡i (i), (ii), (iii) I (iv) mgxKi‡Yi †jL A1⁄4b Kwi| (†hLv‡b, x Aÿ eivei †QvU 5 e‡M©i evûi •`N© ̈ = 1 GKK) I y Aÿ eivei †QvU 3 e‡M©i evûi •`N© ̈ = 1 GKK) X Y X E(1,2) O A(4,0) D( ) 0 11 3 ( ) 0 8 3 B C( ) 11 5  0 Y AE Ges ED †iLvsk؇qi Dci ̄’ I Zv‡`i Wvbcvk¦© ̄’ we›`ymg~‡ni †mU cÖ`Ë mKj kZ©‡K wm× K‡i weavq H AÂjwUB mgvav‡bi AbyK‚j GjvKv|  cÖvwšÍK we›`y ̧wj A(4, 0), E(1 2) Ges D     0 11 3 A(4, 0) we›`y‡Z Z = 40  4 + 50  0 = 160 E(1, 2) we›`y‡Z Z = 40  1 + 50  2 = 140 D     0 11 3 we›`y‡Z Z = 40  0 + 50  11 3 = 183.33  E(1, 2) we›`y‡Z Z Gi gvb me©wb¤œ A_©vr me©wb¤œ LiP 140 UvKv|  A Lvevi n‡Z 1 †KwR Ges B Lvevi n‡Z 2 †KwR Lv` ̈ cÖ‡qvRb| (Ans.) 3| DÏxcK-i: Z = 3x + 2y, x + 2y  4, 2x + y  4, x + y  5 I x, y  0| DÏxcK-ii: f(x) = ax + by + c, g(x) = lx + my + n, a = 1, b = – 1, c = 2, f(x)  0, l = 1, m = 1, n = – 4, g(x)  0 Ges x, y  0| (K) †hvMvkÖqx †cÖvMÖv‡gi 2 wU kZ©vewj D‡jøL Ki| (L) DÏxcK-i Gi kZ© mv‡c‡ÿ †jLwP‡Îi mvnv‡h ̈ Afxó dvskb z Gi me©wb¤œ gvb wbY©q Ki| [Kz. †ev. 17] (M) DÏxcK-ii n‡Z †jLwP‡Îi mvnv‡h ̈ z = x + 2y Gi m‡e©v”P gvb wbY©q Ki| [Xv. †ev. 19] mgvavb: (K) †hvMvkÖqx †cÖvMÖv‡gi 2 wU kZ© wb‡¤œ D‡jøL Kiv n‡jv: i) mgm ̈vi Afxó dvskb Aek ̈B _vK‡Z n‡e Ges Zv wm×všÍ Pj‡Ki •iwLK (GKNvwZK) dvskb iƒ‡c cÖKvk‡hvM ̈ n‡Z n‡e| ii) wm×všÍ PjK ̧wj G‡K Ac‡ii mv‡_ m¤úK©hy3 I AFYvZ¥K n‡e hv‡Z G‡`i g‡a ̈ MvwYwZK m¤úK© •Zwi Kiv hvq| PjK ̧wj FYvZ¥K n‡e bv| (L) Afxó dvskb: Z = 3x + 2y (me©wb¤œKiY) kZ©: x + 2y  4, 2x + y, x + y  5, x, y  0 AmgZv ̧‡jvi Abyiƒc •iwLK mgxKiY: x 4 + y 2 = 1 ......... (i) x 2 + y 4 = 1 ......... (ii) Ges x 5 + y 5 = 1 ......... (iii) QK KvM‡R g~jwe›`y 0 Ges x-Aÿ I y-Aÿ‡K h_vμ‡g XOX Ges YOY Øviv wPwýZ K‡i (i), (ii) I (iii) bs mgxKi‡Yi †jLwPÎ A1⁄4b Kwi|