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

†hvMvkÖqx †cÖvMÖvg  Board CQ & MCQ Practice Sheet Solution 3 Rhombus Publications Y X Y X E(5, 0) O A(4,0) G( ) 4 3  4 3 F(0,5) D(0,4 ) C(2,0) B(0,2 ) †jLwPÎ n‡Z cvB mgvav‡bi Rb ̈ m¤¢ve ̈ AbyK‚j GjvKv AEFDG cÂf‚R| cÖvwšÍK we›`y ̧wj A(4, 0), E(5, 0), F(0, 5), D(0, 4), G     4 3  4 3 | A(4, 0) we›`y‡Z Z = 3  4 + 2  0 = 12 E(5, 0) we›`y‡Z Z = 3  5 + 2  0 = 15 F(0, 5) we›`y‡Z Z = 3  0 + 2  5 = 10 D(0, 4) we›`y‡Z Z = 3  0 + 2  4 = 8 G     4 3  4 3 we›`y‡Z Z = 3  4 3 + 4 3  2 = 20 3 = 6.67  G     4 3  4 3 we›`y‡Z Z Gi gvb me©wb¤œ A_©vr Zmin = 20 3 (Ans.) (M) †`Iqv Av‡Q, f(x)  0  ax + by + c  0  x – y + 2  0 [∵ a = 1, b = – 1 Ges c = 2] Avevi, g(x)  0  lx + my + n  0  x + y – 4  0 [∵ l = 1, m = 1 Ges n = – 4] DÏxcK-ii n‡Z Afxó dvskb: z = x + 2y (m‡e©v”PKiY) mxgve×Zv: x – y + 2  0, x + y – 4  0 Ges x, y  0 AmgZv ̧‡jvi Abyiƒc •iwLK mgxKiY: x – y + 2 = 0  x – y = – 2  x – 2 + y 2 = 1 ........ (i) Avevi, x + y – 4 = 0  x 4 + y 4 = 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) bs mgxKi‡Yi †jLwPÎ A1⁄4b Kwi| Y X Y X B(1,3) D O(0, 0) A(4, 0) (–2, 0) E(0,4) C(0,2) †jLwPÎ n‡Z cvB mgvav‡bi Rb ̈ m¤¢ve ̈ AbyK‚j GjvKv OABC PZzf‚©R| cÖvwšÍK we›`y ̧wj O(0, 0), A(4, 0), B(1, 3), C(0, 2)| GLb, O(0, 0) we›`y‡Z Z = 0 + 2  0 = 0 A(4, 0) we›`y‡Z Z = 4 + 2  0 = 4 B(1, 3) we›`y‡Z Z = 1 + 2  3 = 7 C(0, 2) we›`y‡Z Z = 0 + 2  2 = 4  B(1, 3) we›`y‡Z Z Gi gvb m‡e©v”P A_©vr Zmax = 7| (Ans.) 4| R‣bK f`a‡jvK m‡e©v”P 100 UvKv e ̈q K‡i wKQz msL ̈K Kjg I †cwÝj wKb‡Z Pvb| cÖwZwU Kjg I †cw݇ji g~j ̈ h_vμ‡g 12 UvKv I 8 UvKv| wZwb AšÍZ GKwU Kjg wKb‡eb wKš‘ 8 wUi AwaK †cwÝj wKb‡eb bv| (K) †hvMvkÖqx †cÖvMÖ‡gi 2 wU e ̈envi D‡jøL K‡iv| (L) DÏxc‡Ki kZ©vbyhvqx †hvMvkÖqx †cÖvMÖvg MVb K‡i mgvav‡bi AbyK‚j GjvKvi †K.wYK we›`y ̧‡jv wbY©q Ki| (M) H f`a‡jvK †Kvb cÖKv‡ii KZ ̧wj wRwbm wKb‡j GK‡Î me©vwaK msL ̈K wRwbm wKb‡Z cvi‡eb? mgvavb: (K) †hvMvkÖqx †cÖvMÖv‡gi `ywU e ̈envi wb‡P D‡jøL Kiv n‡jv: i) †hvMvkÖqx †cÖvMÖv‡gi gva ̈‡g e ̈emv-evwY‡R ̈ me©wb¤œ e ̈‡q m‡e©v”P gybvdv AR©b m¤¢e nq| ii) mxgve× ev‡R‡U A_©vr me©wb¤œ Li‡P cÖ‡qvRbxq cywó ev Lv` ̈gvb mg„× Lv` ̈ ZvwjKv cÖ ̄‘yZ I Zv mieiv‡ni Rb ̈ †hvMvkÖqx †cÖvMÖvg e ̈envi Kiv nq| (L) g‡b Kwi, f`a‡jvK x msL ̈K Kjg Ges y msL ̈K †cwÝj wKb‡eb| Afxó dvskb: Z = x + y kZ©: 12x + 8y  100, x  1, y  8, y  0 AmgZv ̧‡jvi Abyiƒc •iwLK mgxKiY: 12x + 8y = 100  3x + 2y = 25  x 25 3 + y 25 2 = 1 ...... (i) x = 1 ...... (ii) Ges y = 8 ...... (iii)
4  Higher Math 2nd Paper Chapter-2 Rhombus Publications QK KvM‡R g~jwe›`yO 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| †hLv‡b, x Aÿ y Aÿ eivei †QvU 1 e‡M©i evûi •`N© ̈ = 1 GKK| Y X Y X D C(3, 8) (1,8) E( ) 0 25 2 O (0,0) A (1,0) B( ) 25 3  0  †jLwPÎ n‡Z cvB mgvav‡bi m¤¢ve ̈ AbyK‚j GjvKv ABCD PZzf‚©R| cÖvwšÍK we›`y ̧‡jv A(1, 0), B     25 3  0 , C(3, 8), D(1, 8) (Ans.) (M) g‡b Kwi, f`a‡jvK x msL ̈K Kjg Ges y msL ̈K †cwÝj wKb‡j GK‡Î me©vwaK msL ̈K wRwbm wKb‡Z cvi‡eb| Afxó dvskb: z = x + y mgxe×Zv: 12x + 8y  100, x  1, y  8, y  0 ÔLÕ n‡Z cvB mgvav‡bi AbyK‚j GjvKvi †KŠwYK we›`y ̧‡jv h_vμ‡g A(1, 0), B     25 3  0 , C(3, 8) Ges D(1, 8)| A(1, 0) we›`y‡Z z = 1 + 0 = 1 B     25 3  0 we›`y‡Z z = 25 3 + 0 = 25 3 C(3, 8) we›`y‡Z z = 3 + 8 = 11 D(1, 8) we›`y‡Z z = 1 + 8 = 9  C(3, 8) we›`y‡Z z Gi gvb m‡e©v”P A_©vr, zmax = 11 myZivs, f`a‡jvK 3 wU Kjg Ges 8 wU †cwÝj μq K‡ib| (Ans.) 5| `„k ̈Kí- i: Afxó dvskb, z = 2x – y `„k ̈Kí-ii: f = 2x + 3y, g = 5x + 3y (K) †hvMvkÖqx †cÖvMÖvg KqwU Ask wb‡q MwVZ? (L) `„k ̈Kí-i Gi mgvav‡bi †K.wYK we›`y ̧‡jvi ̄’vbv1⁄4 (4t, 0), (0, 5 – 9t) Ges (2t – 1, 2 – 3t) n‡j Ges cÖ_g we›`y‡Z Afxó dvskb z Gi m‡e©v”P gvb _vK‡j t Gi Ak~b ̈ gv‡bi †mU wbY©q Ki| (M) `„k ̈Kí-ii Gi Av‡jv‡K f  12, g  15 Ges x, y  0 n‡j †jLwP‡Îi gva ̈‡g m¤¢ve ̈ †ÿÎwU wbe©vPb Ki| k‡Z© Kx cwieZ©b Ki‡j m¤¢ve ̈ †ÿÎwU PZzfz©R n‡e| [iv. †ev. 17] mgvavb: (K) †hvMvkÖqx †cÖvMÖvg wZbwU Ask wb‡q MwVZ: i. Afxó dvskb: z = ax + by ii. kZ© ev mxgve×Zv: a1x + b1y  c1 (•iwLK AmgZv) iii. AFYvZ¥K mgxe×Zv: x  0, y  0 (L) †`qv Av‡Q, Afxó dvskb: z = 2x – y GLb, (4t, 0) we›`y‡Z z = 2  4t – 0 = 8t (0, 5 – 9t) we›`y‡Z z = 2  0 – (5 – 9t) = 9t – 5 (2t – 1, 2 – 3t) we›`y‡Z z = 2  (2t – 1) – (2 – 3t) = 4t – 2 – 2 + 3t = 7t – 4 cÖkœg‡Z, (4t, 0) we›`y‡Z m‡e©v”P gvb we` ̈gvb,  8t > 9t – 5 Ges 8t > 7t – 4  – t > – 5  t > – 4  t < 5 myZivs, t Gi gvb: – 4 < t < 5, wKš‘ t  0  t Gi Ak~b ̈ gv‡bi †mU = {t: – 4 < t < 0}  {t: 0 < t < 5} (Ans.) (M) †`Iqv Av‡Q, f = 2x + 3y, g = 5x + 3y cÖ`Ë kZ©: f  12, g  15  2x + 3y  12  5x + 3y  15, x  0, y  0 AmgZv ̧‡jvi Abyiƒc •iwLK mgxKiY: 2x + 3y = 12  x 6 + y 4 = 1 ........ (i) Avevi, 5x + 3y = 15  x 3 + y 5 = 1 ....... (ii) x = 0 ........... (iii) y = 0 ........... (iv)