Title Liner Programming Engg Practice Sheet Solution [HSC 26].pdf ✅

†hvMvkÖqx †cÖvMÖvg  Engineering Practice Sheet Solution (HSC 26) 1 02 †hvMvkÖqx †cÖvMÖvg Linear Programming WRITTEN weMZ mv‡j KUET-G Avmv cÖkœvejx 1| wg. ingvb dj †Kbvi Rb ̈ m‡e©v”P 130 UvKv e ̈q K‡i Kgc‡ÿ GK †KwR K‡i Av‡cj I Av1⁄2yi wKb‡Z Pvb| wZwb †`vKv‡b wM‡q †`L‡jb Zvi UvKvi g‡a ̈ wZwb m‡e©v”P wZb †KwR Av1⁄2yi I Pvi †KwR Av‡cj ev `yB †KwR Av1⁄2y&i I mvZ †KwR Av‡cj wKb‡Z cv‡ib| Zvi H UvKvi g‡a ̈ wZwb m‡e©v”P KZ IR‡bi dj wKb‡Z cvi‡eb? [KUET 19-20] mgvavb: awi, cÖwZ †KwR Av1⁄2y‡ii g~j ̈ m, Av‡c‡ji g~j ̈ n UvKv| Zvn‡j, 3m + 4n = 130 ....... (i) 2m + 7n = 130 ....... (ii) (i) I (ii) n‡Z, m = 30, n = 10 Y x = 1 (1, 10) (4, 1) (1,1) 3x + y = 13 y = 1 X O x †KwR Av1⁄2yi I y †KwR Av‡cj wKb‡j Afxó dvskb z = x + y AmgZvmg~n: 30x + 10y ≤ 130  3x + y ≤ 13; x, y ≥ 1 AmgZv eR©b K‡i cÖvß mgxKiY: 3x + y = 13, x = 1, y = 1 mgvavb †ÿ‡Îi †KŠwYK we›`ymg~n (1, 1), (4, 1) I (1, 10) (1, 1) Gi Rb ̈ Z = 1 + 1 = 2 (4, 1) Gi Rb ̈ Z = 4 + 1 = 5 (1, 10) Gi Rb ̈ Z = 1 + 10 = 11  Zmax = 11 m‡e©v”P 11 †KwR IR‡bi dj wKb‡Z cvi‡eb| (Ans.) 2| GKRb e ̈emvwq Zvi †`vKv‡bi Rb ̈ †iwWI Ges †Uwjwfk‡b wg‡j 100 †mU wKb‡Z cv‡ib| †iwWI †mU Ges †Uwjwfkb †mU cÖ‡Z ̈KwUi μqg~j ̈ h_vμ‡g 40 I 120 Wjvi| cÖwZ †iwWI Ges †Uwjwfkb †m‡U jvf h_vμ‡g 15 I 20 Wjvi| m‡e©v”P 10800 Wjvi wewb‡qvM K‡i wZwb m‡e©v”P KZ jvf Ki‡Z cv‡ib? [KUET 08-09; CUET 07-08] mgvavb: awi, x wU †iwWI I y wU TV wKb‡e, (100, 0) X O (270, 0) (15, 85) (0, 100) (0, 90) X Y Y x + y ≤ 100; 40x + 120y ≤ 10800  x + 3y ≤ 270 ; x ≥ 0; y ≥ 0 jvf, Z = 15x + 30y †KŠwYK we›`y ̧‡jv, (0, 0), (100, 0), (15, 85) I (0, 90) (0, 0) Gi Rb ̈ Z = 15  0 + 30  0 = 0 (100, 0) Gi Rb ̈ Z = 15  100 + 30  0 = 1500 (15, 85) Gi Rb ̈ Z = 15  15 + 30  85 = 2775 (0, 90) Gi Rb ̈ Z = 15  0 + 30  90 = 2700  m‡e©v”P jvf = 2775 Wjvi (Ans.) weMZ mv‡j CUET-G Avmv cÖkœvejx 3| wb¤œwjwLZ k‡Z© F = 3x + 4y Gi m‡e©v”P gvb wbY©q Ki: x + y ≥ 7, 2x + 5y ≥ 20, x ≥ 20, x ≥ 0 Ges y ≥ 0| [CUET 04-05] mgvavb: Y (0, 7) 2x + 5y = 20 (0, 4) O(0, 0) (5, 2) (10, 0) (7, 0) x + y = 7 X †KŠwYK we›`y ̧‡jv (0, 7); (5, 2) Ges (10, 0) (0, 7) Gi Rb ̈ F = 3  0 + 4  7 = 28 (5, 2) Gi Rb ̈ F = 3  5 + 4  2 = 23 (10, 0) Gi Rb ̈ F = 3  10 + 4  0 = 30  F Gi m‡e©v”P gvb = 30 (Ans.)
2  Higher Math 2nd Paper Chapter-2 MCQ weMZ mv‡j KUET-G Avmv cÖkœvejx 1. GKRb e ̈emvwq 40 UvKv †KwR `‡i †cqviv Ges 120 UvKv †KwR `‡i Av‡cj wKb‡Z cv‡ib| Dfq cÖKvi wg‡j wZwb Zvi †`vKv‡b †gvU 120 †KwR dj ivL‡Z cv‡ib| D3 e ̈emvwq †cqviv wewμ K‡i cÖwZ †KwR‡Z 16 UvKv Ges Av‡cj wewμ K‡i cÖwZ †KwR‡Z 32 UvKv jvf Ki‡Z cv‡ib| hw` wZwb m‡e©v”P 12000 UvKv wewb‡qvM Ki‡Z cv‡ib, Zvn‡j †Kvb cÖKv‡ii dj KZ †KwR wb‡j wZwb m‡e©v”P jvf Ki‡Z cvi‡eb? [KUET 16-17] 120 †KwR †cqviv 120 †KwR Av‡cj 30 †KwR Av‡cj I 90 †KwR †cqviv 90 †KwR Av‡cj I 30 †KwR †cqviv 60 †KwR Av‡cj I 60 †KwR †cqviv DËi: 90 †KwR Av‡cj I 30 †KwR †cqviv e ̈vL ̈v: x, y ≥ 0 x + y = 120 ...... (i) 40x + 120y ≤ 12000 ...... (ii) Z = 16x + 32y Y X X Y O(0, 0) D(120, 0)(300, 0) C(30, 90) (0, 120) (0, 100) †KŠwYK we›`ymg~n (30, 90), (120, 0), (0, 100) m‡e©v”P jvf, Zmax = 16  30 + 32  90 = 3360 2. A I B cÖKvi †Ljbv •Zwi‡Z h_vμ‡g 5 I 3 GKK kÖg Ges 3 I 4 GKK KuvPvgvj jv‡M| A cÖKv‡ii cÖwZwU †_‡K 10 UvKv I B cÖKv‡ii cÖwZwU †_‡K 12 UvKv jvf Kiv m¤¢e nq Ges †Kv¤úvwbwU 165 GKK kÖg I 132 GKK KuvPvgvj †hvMvb w`‡Z cv‡i, Z‡e m‡e©v”P †h jvf n‡e Zv n‡jvÑ [KUET 14-15] 330 taka 360 taka 420 taka 448 taka 650 taka DËi: 420 taka e ̈vL ̈v: awi, h_vμ‡g x I y GKK A I B †Ljbv •Zwi Ki‡Z n‡e| 5x + 3y ≤ 165 ; 3x + 4y ≤ 132 ; x, y  0 (24, 15) (33, 0) (44, 0) (0, 0) (0, 33) (0, 55) Y X †KŠwYK we›`ymg~n (0, 0), (33, 0), (24, 15), (0, 33)  m‡e©v”P jvf, Zmax = 10x + 12y = 10  24 + 12  15 = 420 taka 3. A I B cÖKvi hš¿ •Zwi‡Z h_vμ‡g 15 I 5 GKK mgq Ges 5 I 10 GKK KuvPvgvj jv‡M| 105 GKK mgq I 60 GKK KvuPvgvj w`‡q m‡e©v”P †h jvf n‡e (hLb A Gi cÖwZ GK‡K jvf 50 UvKv Ges Zv B Gi Rb ̈ 30 UvKv), Zv n‡jvÑ [KUET 13-14, 12-13] 390 UvKv 420 UvKv 380 UvKv 400 UvKv 350 UvKv DËi: 390 UvKv e ̈vL ̈v: awi, x GKK A I y GKK B •Zwi Ki‡ev|  15x + 5y ≤ 105; 5x + 10y ≤ 60; Z = 50x + 30y (6, 3) (7, 0) (12, 0) (0, 0) (0, 6) (0, 21) Y X †KŠwYK we›`ymg~n, (0, 0), (0, 6), (7, 0), (6, 3)  m‡e©v”P jvf, Zmax = 50  6 + 30  3 = 390 weMZ mv‡j IUT-G Avmv cÖkœvejx 4. Alal and Dulal shopped at the same store. Alal bought 5 kg of apples and 2 kg of bananas and paid altogether BDT 22. Dulal bought 4 kg of apples and 6 kg of bananas and paid together BDT 33. Find the cost 1 kg bananas. [IUT 18-19] BDT 3.5 BDT 3 BDT 2.5 BDT 4.5 DËi: BDT 3.5 e ̈vL ̈v: cost of 1 apple = x; cost of 1 banana = y As per Overstion 5x + 2y = 22; 4x + 6y = 33 Solving, x = 3; y = 3.5 5. A company produces 2 types of product. It uses 3 plants for the production. The 1st product requires one hour in plant 1 and 3 hours in plant 3 for producing 1 item. The 2nd product requires 2 hours each in plant 2 and for one production. Total available hours in plants 1, 2 in a week are 4, 12 and 18 respectively. Profit for each item of product 1 is 3 thousand, and for product 2 is 5 thousand. The possible maximum weekly profit for the company is- [IUT 16-17] 18 thousand 33 thousand 36 thousand 3 thousand DËi: 36 thousand
†hvMvkÖqx †cÖvMÖvg  Engineering Practice Sheet Solution (HSC 26) 3 e ̈vL ̈v: x = 4 (4, 3) (6, 0) (0, 0) (0, 9) (2, 6) Y X (4, 0) (0, 6) Intersection Points (0, 0), (4, 0), (2, 6), (0, 6) Maximum weekly profit, Zmax = 3  2 + 5  6 = 36 weMZ mv‡j BUTex-G Avmv cÖkœvejx 6. x ≥ 0, y ≥ 0, 3y – x ≤ 10, x + y ≤ 6, x – y ≤ 2 kZ©vbymv‡i Z = 2y – x Gi m‡e©v”P gvb KZ? [BUTex 15-16] – 5 – 4 – 3 – 2 DËi: mwVK DËi †bB e ̈vL ̈v:    0  10 3 (2, 4) x = y = 6 (2, 0) (– 2, 0) (4, 2) X Y X Y (0, 0) †KŠwYK we›`ymg~n (0, 0), (2, 0), (2, 4), (4, 2),    0  10 3 m‡e©v”P gvb, Zmax = 2  10 3 – 0 = 20 3 7. 100 iæB gv‡Qi †cvbvi `vg 60 UvKv Ges 100 KvZjv gv‡Qi †cvbvi `vg 30 UvKv n‡j, GK e ̈w3 1200 UvKvq KZ ̧‡jv iæB gv‡Qi †cvbv wKb‡Z cvi‡eb? hLb †gvU μqK...Z gv‡Qi msL ̈v me©vwaK 3000 n‡e| [BUTex 13-14] 500 1000 1500 2000 DËi: 1000 e ̈vL ̈v: awi, x GKK iæB Ges y GKK KvZjv|  60 100 x + 30 100 y  1200  2x + y = 4000; x + y  3000; x, y  0 Ges Awfó dvskbwU, Z = x + y (1000, 2000) (0, 0) (2000,0) (3000,0) (0, 3000) (0, 4000) Y X (0, 3000) we›`y‡Z, Z = 0 + 3000 = 3000 (1000, 2000) we›`y‡Z, Z = 1000 + 2000 = 3000 (2000, 0) we›`y‡Z, Z = 2000 + 0 = 2000 †h‡nZz wZwb iæB Ges KvZjv DfqB †K‡bb|  iæB, x = 1000 ; KvZjv, y = 2000 8. x ≥ 0, y ≥ 0, x + y ≤ 7, 2x + 5y ≤ 20 k‡Z©i mv‡c‡ÿ z = 3x + 4y Gi m‡e©v”P gvb KZ? [BUTex 13-14] 21 22 23 24 DËi: 23 e ̈vL ̈v: (5, 2) (0, 0) (7,0) (10,0) (0, 4) (0, 7) Y X †KŠwYK we›`ymg~n, (0, 0), (0, 4), (5, 2), (7, 0)  m‡e©v”P gvb, Zmax = 3  5 + 4  2 = 23 weMZ mv‡j SUST-G Avmv cÖkœvejx 9. x + 2y  10; x + y  6; x  4; x, y  0 ; kZ©vaxb Z = 2x + 3y Gi m‡e©v”P gvb KZ? [SUST 18-19] 8 14 16 17 DËi: 16 e ̈vL ̈v: (4, 2) (0, 0) (4,0) (0, 5) Y X (2, 4) †KŠwYK we›`ymg~n, (0, 0), (4, 0), (4, 2), (2, 4), (0, 5)  m‡e©v”P gvb, Zmax = 2  2 + 3  4 = 16 -----