Bai Tap Toan Kinh Te Them 7682

8/2/2019 Bai Tap Toan Kinh Te Them 7682
1/20
BI TP TON KINHT
CHNG I: QUY HOCH TUY N TNH
I.
Lp m hnh ton hc cho cc bi ton d i
y1.Mt x nghip c 4 my A, B,C, D dng sn xut ra 6 loi sn phm nh mcthi gian cho mi sn phm i vi tng my, qu thi gian ca tng my c cho trong
bng sau:SP
MYnh mc thi gian cho sn phm ( gi)
SP1 SP2 SP3 SP4 SP5 SP6A: 132 1 2 1 3 1 0B: 120 2 1 1 2 3 2C: 100 1 2 0 2 1 3
D: 144 2 2 2 1 4 2Gi 1 SP(1000 )
40 30 35 75 65 60
Hy lp phng n sn xut sao cho tng thu nhp l ln nht (Max) m vn m bo anton cho my.
2. Mt phn xng c nhim v sn xut ra 3 loi sn phm cn s dng 2 loinguyn liu A v B. Hao ph nguyn liu, d tr nguyn liu cho trong bng di y.Hy lp k hoch sao cho tng chi ph nh nht vi iu kin khng c s dng qu snguyn liu, tng s sn phm c 3 loi khng t hn 350 s sn phm 1, khng t hn25% tng s sn phm.Nguyn liu Sn phm D tnh nguyn liu
(kg)I II IIIA 2 1 3 1500B 1 2 2 650
Chi ph/ SP(1000)
8 5 10
3. Hai a phng Ninh Bnh v Hng Yn cung cp Khoai vi khi lng 200 tnv 300 tn cho 3 a phng tiu th Khoai l Hi Phng, Ngh An v Nam nh vi yucu tng ng l 170 tn, 200 tn v 130tn cc ph (nghn/ tn) cho trong bng diy. Hy lp k hoch sao cho tng chi ph nh nht.
Ni tiu thHi Phng Ng An Nam nh
Ninh Bnh 20 12 25Hng Yn 12 24 14
B mn Ton - Bi tp Ton Kinh T - Dng cho SV cc lp tn ch! 1

2/20
1 2
4. Ba khu t ngi ta d tnh trng 3 lo cy lng thc: Ng, khoai v sn. Nngsut v chi ph trn 1ha vi mi loi cy tng khu t v nng sut (t) chi ph1000/ha.Din tch(ha) Ng Khoai SnKI:30 150 80 80KII:50 120 75 90
KIII:27 140 70 70Yu cu 1000 1200 1500Hy phn b cy trng cho ht t ai sao cho m bo yu cu v s lng v tng chi
ph l nh nht.5. C 2 x nghip sn xut 2 loi sn phm. Bit rng u t 1000 vo x nghip I
trong 1 n v thi gian lm c 1,2 sn phm A v 1,8 sn phm B cn u t 1000vo x nghip II lm c 1,8 SP A v 0,8 Sp B. Nguyn liu v lao ng cung cp chovic sn xut, nh mc nguyn liu lao ng v d tr nguyn liu cho trong bng diy.Hy tm phng n u t sao cho tng s vn b ra t nht.
Nguyn liu d tr: 400 tn v gio lao ng l 200000 giXN nh mc hao ph LN(kg/SP), lao ng(gi/SP)
A BNL L NL L
I 4 2 10 4II 4,2 3 9 4,5
Yu cu slng SP
21000 20000
II .Giicc bi ton sau bng phng php hnh hc 2x1 +x2
8
2x +x 10
1. f(X) = x1+x2Min
tha mn x1
x2 4
2
x1+
x2
4
x1 ,x2 0
x1 +x2 2
3x1 2x2 6
2. f(X) =6x1
+ 5x2Max
tha mn x1+
2x 1
x2 2
x2 0
x1 ,x2 0

3/20

4/20
1 2
1 2
1
1
3
3. Max ( x1 - x2 ) tha mn
2x1 3x2 62x +x 6
x1 2x2 1
x1 + x2 5
x1 ,x2 0
x1 2x2 0 x +2x 6
4. Min ( 4x1 + x2 ) tha mn 3x1 +x2 62x +x 4 1 2
x1 ,x2 0
III. Gii cc bi ton sau bng phng php n hnh
Bi1.
f(X) = x1 x2 +x3 +x4 +x5 x6Min
x1
3x1 +x2 4x
3
+x4 + 6x6 = 9
+ 2x6 = 2
x + 2x +x5 + 2x6 = 2
x j 0 ; j = 1, 6
Bi2.
f(X) = 5x1 + 2x2 + 4x3 +x4 + 5x5 + 3x6Min
2x1 +x2 +4x3
+ 3x5 = 152
4x1
3x
+ 2x3 +x4 + 3x5+x5
= 60
+x6 = 36
x j
0,
j = 1, 6

5/20
Bi3.
f(X) = 4x2 2x3 13x4 +7x5 + 10x6Min
x1 3x2 + 2x3 + 13x4 7x5 12x6
x2+x
3+ 4x
4x
54x
6
= 9
= 3
2x2 +x3 2x4 + 2x5 4x6 = 9
x j 0,
j = 1, 6
Bi
4.
f(X) = x1 +x2 x3Max

6/20
2x1 +x2 + 3x3 +x4 = 1
x1
+ 2x2 2x3
x j 0
+x5 = 1
j = 1, 5
Bi5.
f(X) = 3x1 + 2x2 + 14x3 + 13x4 + 4x5Max
x1
+ 2x3 +x4 +x5 = 1
x2 +x3 x j
0
+ 2x4 +x5 = 1
j = 1, 5
Bi6.
f(X) = 2x1 +x2 + 5x3 + 3x4Min
x1 + 2x2 +3x3
= 15
2x1 +x2 + 5x3 = 20
x1 + 2x2 +x3 +x4 = 10
x j 0
j = 1, 4
Bi7.
f(X) = 2x1 + 8x2 + 2x3Max
4x1 x2 +
3x3 7x1 + 5x2 +
12x3
= 7
= 19
x j 0 j = 1, 3

7/20
1
Bi8.
f(X) = 3x1 + 7x2 + 6x3 + 5x4Max
x1 + 2x2 + 5x3 +
3x4 4x1 + 3x2 + 2x3 x4
= 12
= 7
x j 0 j = 1, 4
Bi9.
f(X) = x1 +x2 + 2x3 2x4 4x5Min
x1 +x2 3x3 + 2x4 2x5 = 8 2x1 x
3 +x4 x5 21
3x1 + 5x3 3x4 + 2x5 = 25
2x x j
0
+x4 + 4x5
j = 1, 5
20

8/20
1
2
Bi10.
f(X) = 2x1 + 3x2 x3 4x4 + 6x5Max
2x1 +x2 x3
= 40
5x1 x1
2x3+ 2x3
+ 2x44x4
x5 0
+x5 12
3x
+ 2x3 + x4 + 2x5 14
x j 0 j = 1, 5
Bi11.
f(X) = 2x1 +x2 3x3 +x4Min
2x1 +x2 +
4x3
+x4 = 6
5x1 + 2x2 + 2x3 +
x4
= 10
x j 0 j = 1, 4
Bi12.
f(X) = 2x1 +x2 + 3x3 x4Min
x1 2x2 +x4 = 16
x2 +x3 4x4 8
x x j 0
+ 3x3 2x4
j = 1, 4
20
Bi13.
f(X) = 2x1 + 4x2 +x3 +x4Max
x1 + 3x2

9/20
+x4 1 2x1 +
x2 3
x2 + 4x3 +x4 3
x j 0 j = 1, 4
Bi14.
f(X) = 2x1 + 4x2 +x3 +x4Max
x1 + 3x2 +x4 1
2x1 +
x2 3
x2 + 4x3 +x4 3
x j 0 j = 1, 4
Bi15.
f(X) = 2x1 + 5x2 +x3 2x4Max

10/20
1
1
2
2
3x1+ 2x3 5x4 23
4x1
+x3 4x4 36
x2 2x3 + 2x4 = 14
2x +x3 3x4 12
x j 0 j = 1, 4
Bi16.
f(X) = x1 + 5x2 x3Min
x1 x2 + 4x3 + 3x4 8
3x1 5x2 + 2
x3
2x4 2
6x 4x3 5x4 2
x j 0 j = 1, 4
Bi17.
f(X) = 2x1 2x2 + 3x3 +x4 3x5Min
4x1
2x3+x
44x
5= 8
x1
+ 2x4 x56x3 +x4 + 3x5
20
= 15
x+x + x3 +x4 2x5 = 14
x j 0 j = 1, 5
a) Gii bi ton trn bng phng php n hnh.b) C th kt lun g trong trng hp f(X)Max
c) Tm phng n X sao cho

11/20
1 4
f(X)= 27
Bi 18. Cho bi ton vi tham s : f(X) = x1 +x3 x4Min
x1 +x2 +12x
3
2x4 + 4x5 = 9
2x1 + 8x3 + (1 )x4 +2x5
= 14
x + (1)x = 4
x j
0,
j = 1, 5
a) Bit x l mt phng n cc bin ng vi c s A1, A2, A5. Hy lp bng nhnh ng vi x .
b) T bng n hnh lp, hy tm tt c cc gi tr ca sao cho x l phng nti u.
c) Gii bi ton cho vi =1 v =3.
Bi 19. Cho bi ton vi tham s : f(X) = x2 3x4 +x5 + 6x6Min

12/20
2 4
x1 x2 +4x3
2x4 + 12x5 +11x6 = 9
2x2 + 2x3 + (1 )x4 + 8x5 +14
x6
= 14
x + (1)x + 3x6 = 4
x j
0,
j = 1, 6
1) Bit x l mt phng n cc bin ng vi c s A1, A2, A3. Hy lp bng nhnh ng vi x .
2) Tm tt c cc gi tr ca sao chox l phng n ti u.3) Gii bi ton cho vi =1 v =-1.
CHNG II: BI TON I NGU
Bi 1. Vit bi ton i ngu? Ch ra cc cp i ngu ca cc bi ton sau:
a) g(Y) = 7y1
+6y2
+4y3
Maxtho mn
y1 + 2y2 + 5y3 2 3y1
y2 +y3 3
2y13y2 + 2y3 4
y1 + 3y2 + 2y3 2
b) f(X) = 2x1 +x2 3x3 +x4Max
tho mn
2x1 +x2 +
3x3
+ 2x4 = 2
4x1 x2 + 2x3
x4 = 5
x j 0 j = 1,4
3x1 +x2 2x3 4

13/20
4x1 x2 + 3x3 2 c) f(X) = 6x1 + 3x2 x3Min
tho mn 2x1 +x2
x3 4
x1 2x2 + 3x3 5
x j 0 j = 1, 3
4x1 x2 + 3x3 x4 22x1 + 3x2 x3 +2x4 5
x1 +x2 + 3x3 +x4 3
d) f(X) = 2x1 x2 + 3x3 4x4Max
tho mn 2x
x+ 2x x = 6
1 2 3 4
3x1 +x2 +4x3 +2x4 10x j 0 j = 1, 4

14/20
1
)
Bi 2. Cho biton
f(X) = x1 3x2 +x3 2x4
Min
tho mn
4x1 + 12
x2
+ 3x4 = 24
x1 + 3x2 x3 3
4x 18x2 + 2x3 + 3x4 33
x j 0
;
j = 1, 4
Chng minh rng x = (0;1;0; 4) l phng n ti u ca bi ton.
Bi 3. Cho biton
f(X) = 15x1 + 10x2 + 6x3
Min
tho mn
3x1x1
+2x3 2
+ 2x2 + 2x3 3
2x1 +x2 4x1 +2x2 x1
+x3 2
2x3 1
1
x2 0 , x3 0
Bit rng x = (1,5,
11l phng n ti u ca bi ton cho. Hy tm phong n ti u
4 4ca bi ton i ngu.
x1 +2x2 +x3 7
4x1 +3x2 6x3 9
2x1
x2
8x3
6
Bi 4. Cho biton
f(X) = x1 +x2 x3Max
tho mn
2x2 +x3 2

15/20
2x1 x2 + 5x3 1
a) Chng t rng cc phng n
x1
x = (4,6,1) ,y = (4,0,
3,0,
0,1)5 5
+ 3x3 1
x3 0
theo th t l
phng n ti u ca bi ton cho v bi ton i ngu ca n.b) Tm cc tp phng n ti u ca bi ton cho v bi ton i ngu ca n.

16/20
1
2 2
Bi 5. Cho biton
f(X) = x1 +x2 2x3 3x4 + 4x5
Min
tho mn
9x 2x +
3x 7
2 1 4 2 5 x + 4x +x
3x 42
1 2 4 5 2 1
x +x 1x = 5
2 3 5
x j 0 , j = 1, 5
a) Gii bi ton bng phng php n hnh.b) Tm li gii cho bi ton i ngu.Bi 6. Cho biton
f(X) = 4x1 4x2 + 2x3 3x4 2x5 2x6Max
tho mn
3x1 2x2 +x3 + 2x5 = 8
x1
+ 5x3 + 3x4 x5 = 3
5x1 4x3 +x5 +x6 = 5
x1,x3 ,x5
0
a) Vit bi ton i ngu, ch ra cp rng buc i ngu.b) Xc nh phng n ti u ca bi ton gc.
Bi7. Cho bi
tonf(X) = 5x
19x
2+ 5x
3+ 7x
4+ 3x
5Min
2x1 + 6x2 2x3
2x4 +x5 4
8x1 + 2x3 + 4x4 x5 = 20
x
x2
x j 0
+x3
j = 2,5
x5 1

17/20
(
a) Chng t rng phng n x = (0,1,0,5, 0) l phng n ti u ca bi ton cho.
Tm tp phng n ti u ca bi ton i ngu.b) Hy tm tt c cc phng n ti u ca bi ton cho c thnh phn
Bi 8. Cho bi ton vi tham s f(X) = 4x1 + 10x2 +2x3 8x4 + 9x5 +x6 4x7Min
x3= 4 .
2x1 +x3 2x4 + 2x5 6x7 = 7
2x1 2x2 + 2x3 + 4x4
3x5
+x6 = 7
4x1 + 3x2 2x3
2x4
+x5 +x6 22x7 = 20
x j 0
j = 1, 7
a) Tm gi tr ca x = 11
,0,0,9,0, 0,
3) l phng n ti u ca bi ton
cho.
2 2 2b) Trong trng hpx khng l phng n ti u, c th ni g v bi ton cho vbi ton i ngu ca n.

18/20
CHNG III: BI TON VN TI
Bi1.
Gii cc bi ton vn t i sau:
TP 20 50 50 2030 6 2 4 640 4 6 3 870 1 5 8 7
Bi 2.T
P50 90 90 70
110 7 8 5 3
110 2 4 5 980 6 4 1 2Bi 3.
TP
220 320 540 650
250 4 5 3 5790 7 5 4 9670 9 8 3 4340 7 6 6 2
Bi 4.T
P150 80 180 60 80
50 7 4 17 5 8200 7 15 10 11 30100 13 6 14 16 20200 14 22 17 16 35
Bi 5.T
P35 75 40 70 80
60 7 10 1 9 1285 9 4 10 11 548 8 7 5 6 8
107 10 7 3 13 9
Bi 6.

19/20
TP
40 80 75 50 55
60 8 6 4 5 775 7 3 9 6 1045 2 10 7 8 480 11 5 8 12 9
Bi 7.T
P35 15 65 25 40 20
70 4 7 7 10 11 630 10 5 2 3 8 960 12 6 4 11 9 740 13 5 11 6 7 5
Bi8.
TP
54 37 49 120 70
80 4 5 3 2 7120 7 5 3 4 M80 9 7 M 2 550 2 6 4 3 6
Bi 9.T
P
14 15 25 16 17 20 23
27 6 3 4 3 5 7 932 12 11 16 14 16 13 1743 5 6 4 7 4 5 628 7 9 10 11 9 6 9
Bi1.
CHNG IV: BI TON SN XUTNG B G ii cc bi ton ngbsau y:
Chi titMy
CT11
CT21
CT31
A:1 63 240 160B:1 103 61 200C:1 720 360 161D:1 173 120 240

20/20
Bi 2.Chi tit
MyCT1
1CT2
1CT3
1A:1 120 80 61B:1 36 100 91C:1 180 82 36
D:1 60 120 102Bi 3.
Chi titMy
CT12
CT21
CT33
CT42
A:1 76 50 105 80B:1 160 34 72 48C:1 180 36 108 60D:1 200 32 96 160
Bi 4.Chi tit
MyCT1
1CT2
1CT3
1CT4
1A:5 3 5 7 8B:4 8 6 6 6C:6 6 3 6 5D:8 5 2 4 10
Bi5.
Chi titMy
CT12
CT21
CT31
CT43
A:1 54 0 108 54B:5 96 32 14 15C:2 62 28 0 51D:1 242 128 96 192

Bai Tap Toan Kinh Te Them 7682

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