Bab 2. Program LinearKelas 11, Agustus – September 2019

→ mempelajari cara mendapat nilai optimum (maks - min) dari

bbrp model pertidaksamaan (= sistem pertidaksamaan).

Yang harus dikuasai: Ø Pers garis lurus

Ø Pertidaksamaan linear 2 var

𝑨. 𝑷𝒆𝒓𝒔. 𝒈𝒂𝒓𝒊𝒔 𝒍𝒆𝒘𝒂𝒕 𝟐 𝒕𝒊𝒕𝒊𝒌 𝒃𝒆𝒃𝒂𝒔

𝑥1, 𝑦1

𝑥4, 𝑦4

∆𝒙

∆𝒚

𝑔𝑟𝑎𝑑𝑖𝑒𝑛:

𝑝𝑒𝑟𝑠 𝑔𝑎𝑟𝑖𝑠:

𝑎𝑡𝑎𝑢:

∆𝑦∆𝑥

𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1

𝑦 = 𝑚𝑥 + 𝒄

𝒄

1) 𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑝𝑒𝑟𝑠 𝑔𝑎𝑟𝑖𝑠 𝑦𝑔 𝑚𝑒𝑙𝑎𝑙𝑢𝑖:

𝑎) 4, 2 & (1, 8)

S(4, 2)(1, 8) 𝑚 =

2 − 84 − 1 = −2

𝑦 − 8 = −2(𝑥 − 1)

𝑦 = −2𝑥 + 10

𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1 𝑦 = 𝑚𝑥 + 𝒄

2 = −2 . 4 + 𝒄10 = 𝒄

𝑦 = −2𝑥 + 10

𝑏) 1, 13 & (7, 5)

S(1, 13)(7, 5) 𝑚 =

13 − 51 − 7 = −

43

𝑦 − 5 = −43 (𝑥 − 7)

𝑘𝑎𝑙𝑖 3 → 3𝑦 − 15 = −4(𝑥 − 7)

3𝑦 + 4𝑥 = 43

𝑩. 𝑷𝒆𝒓𝒔. 𝒈𝒂𝒓𝒊𝒔 𝒍𝒆𝒘𝒂𝒕 𝒕𝒊𝒕𝒊𝒌 𝒅𝒊 𝒔𝒃. 𝒙 & 𝒔𝒃. 𝒚

𝒂

𝒃

𝒂𝑥 + 𝒃𝑦 = 𝒂𝒃

“ 𝑎𝑛𝑔𝑘𝑎 𝑑𝑖 𝑠𝑏. 𝑦 𝑘𝑎𝑙𝑖 𝑥𝑡𝑎𝑚𝑏𝑎ℎ

𝑎𝑛𝑔𝑘𝑎 𝑑𝑖 𝑠𝑏. 𝑥 𝑘𝑎𝑙𝑖 𝑦= ℎ𝑎𝑠𝑖𝑙 𝑘𝑎𝑙𝑖 𝑘𝑒𝑑𝑢𝑎𝑛𝑦𝑎 “

𝑟𝑢𝑚𝑢𝑠 𝑏𝑢𝑙𝑎𝑛:𝐽𝑜𝑛𝑜

𝑝𝑒𝑚𝑏𝑢𝑘𝑡𝑖𝑎𝑛 ?

(0, 𝒂)

(𝒃, 0)

𝒂𝑥 + 𝒃𝑦 = 𝒂𝒃

𝑚 =𝒂 − 00 − 𝒃 = −

𝒂𝒃

𝑦 − 𝒂 = −𝒂𝒃 𝑥 − 0

𝒃𝑦 − 𝒂𝒃 = −𝒂𝑥 →

𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑝𝑒𝑟𝑠 𝑔𝑎𝑟𝑖𝑠 𝐿1, 𝐿2, 𝐿3

𝐿1 ≡

𝐿2 ≡

5𝑥 + 7𝑦 = 35

4𝑥 − 2𝑦 = −8 → 2𝑥 − 𝑦 = −4

S(−1, 1)(4, 4) 𝑚 =

1 − 4−1 − 4

=35

𝑦 − 4 =35(𝑥 − 4) → 5𝑦 = 3𝑥 + 8

c 3

5

𝑪. 𝑴𝒆𝒏𝒈𝒈𝒂𝒎𝒃𝒂𝒓 𝒈𝒂𝒓𝒊𝒔

1) 𝐺𝑎𝑚𝑏𝑎𝑟𝑘𝑎𝑛 𝑦 = 2𝑥 + 4

𝒙 𝒚0

04

−2→ (0, 4)→ (−2, 0)

𝑦=2𝑥+4

2) 3𝑥 + 2𝑦 = 10

𝒙 𝒚0

05

10/3

𝑫. 𝑴𝒆𝒏𝒈𝒈𝒂𝒎𝒃𝒂𝒓 𝒅𝒂𝒆𝒓𝒂𝒉 𝒑𝒆𝒓𝒕𝒊𝒅𝒂𝒌𝒔𝒂𝒎𝒂𝒂𝒏

1) 𝐺𝑎𝑚𝑏𝑎𝑟𝑘𝑎𝑛 𝑦 ≤ 2𝑥 + 4

𝒙 𝒚0

04

−2

𝑐𝑒𝑘 𝑡𝑖𝑡𝑖𝑘 (0, 0)

0 ≤ 2 . 0 + 4?

𝐵𝐸𝑁𝐴𝑅𝐻𝑃

2) 𝐺𝑎𝑚𝑏𝑎𝑟𝑘𝑎𝑛 2𝑦 + 𝑥 < 0

𝒙 𝒚0 04 −2

𝑐𝑒𝑘 (1, 0)

2 . 0 + 1 < 0?

𝑆𝐴𝐿𝐴𝐻

×

×

×

×

×𝐻𝑃

3) 𝑥 + 2𝑦 ≥ 6 ; 3𝑥 + 2𝑦 < 12 ; & 𝑥 ≥ 0

𝒙 𝒚0 36 0

𝒙 𝒚0 64 0

×

××

××

×

×

𝑐𝑒𝑘 (0, 0)×

×

×

×

×

×

×

×

𝐻𝑃

×

×

×

×

×

×

4) 2𝑥 + 3𝑦 ≤ 12 ; 2𝑥 + 𝑦 ≥ 8 ; 𝑦 ≥ 0

xx

x

x

x

xx

x

x

x

x

x

x

x

xx x x x

HPx

𝐾𝑒𝑟𝑗𝑎𝑘𝑎𝑛 𝑈𝐾 2.2 ℎ𝑎𝑙𝑚 46

𝟑𝒂) 𝑥 + 𝑦 ≤ 6

2𝑥 + 𝑦 ≤ 10

5𝑥 + 9𝑦 ≤ 45

𝑥 ≥ 0 ; 𝑦 ≥ 0

𝐸.𝑀𝑒𝑛𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑝𝑒𝑟𝑡𝑖𝑑𝑎𝑘𝑠𝑎𝑚𝑎𝑎𝑛 𝑑𝑎𝑟𝑖 𝑔𝑎𝑚𝑏𝑎𝑟

1) 𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑠𝑖𝑠𝑡𝑒𝑚:

6𝑥 + 2𝑦 = 12

3𝑥 + 𝑦 ≤ 6

𝑥 ≥ 0

𝑦 ≥ 0

2) 𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑠𝑖𝑠𝑡𝑒𝑚:

𝑨

5𝑥 + 3𝑦 15≥

𝑦 ≥ 1

𝑨 →

5𝑥 + 6𝑦 30≤𝑩 →

𝑩

3.

–x + y ≥ –6

x + y ≥ –2

x ≥ 0y ≤ 0

−6𝑥 + 6𝑦 − 36≥𝑨 →

𝑨

−2𝑥 − 2𝑦 4≤𝑩 →

𝑩

4.3x + 2y ≥ 6

2x + 3y ≤ 6

-x + 2y ≥ -4

= fungsi dari tujuan yg dicari

Nilai optimum terletak di salah satu

titik pojok dari daerah yg diarsir

Optimum / Obyektif Fungsi F.

1) Tentukan nilai maks & min f(x, y) = 5x - 4y

𝟓𝒙 − 𝟒𝒚(2, 0)(3,2)(2, 3)(0, 2)(0, 0)

𝑚𝑎𝑘𝑠

𝑚𝑖𝑛

𝟓𝒙 − 𝟒𝒚(2, 0) 10(3,2) 7(2, 3) −2(0, 2) −8(0, 0) 0

2) maks & min: z = 2x + 3y

𝟐𝒙 + 𝟑𝐲(4, 0) 8(8,0) 16(2, 3) 13

𝑚𝑎𝑘𝑠

𝑚𝑖𝑛

ïïþ

ïïý

ü

³³£+£+

00462

yxyxyx3. Dari sistem:

tentukan nilai maks z = 3x + 2y

yxyxf

yxyxyx

23),(

00462

+=

ïïþ

ïïý

ü

³³£+£+

3x + 2y

(3, 0)

(2, 2)

(0, 4)

(0, 0)

9

10

8

0

ümaks

𝐾𝑒𝑟𝑗𝑎𝑘𝑎𝑛 𝑈𝐾 2.4 ℎ𝑎𝑙𝑚 51 − 52

1𝑎) 3)

𝑥 + 3𝑦 ≥ 4𝑦 − 𝑥 ≤ 2

𝑥 + 2𝑦 ≤ 10𝑥 ≥ 0 ; 𝑦 ≥ 0

𝑛𝑖𝑙𝑎𝑖 𝑜𝑝𝑡𝑖𝑚𝑢𝑚𝑧 = 3𝑥 + 4𝑦

𝑓(𝑥, 𝑦) = 𝑥 + 3𝑦

x + 3y4, 0 4

10, 0 102, 4 140, 2 6

0, 4/3 4

1𝑎) 𝑥 + 3𝑦 ≥ 4𝑦 − 𝑥 ≤ 2

𝑥 + 2𝑦 ≤ 10𝑥 ≥ 0 ; 𝑦 ≥ 0

𝑓(𝑥, 𝑦) = 𝑥 + 3𝑦

4/3

4

××

××

××

××

×

2

−2

×

××

××

××

×

5

10

××

×

××

×

××

××

××

××

××

𝐴

𝑥 + 2𝑦 = 10−𝑥 + 𝑦 = 2 � 𝐴(2, 4)

𝑚𝑎𝑥 = 14 ; 𝑚𝑖𝑛 = 4

3) 𝑛𝑖𝑙𝑎𝑖 𝑜𝑝𝑡𝑖𝑚𝑢𝑚 𝑧 = 3𝑥 + 4𝑦

7𝑥 + 6𝑦 = 429𝑥 + 5𝑦 = 45

𝐴6019

,6319�

3x + 4y3, 0 9

5, 0 15

60/19, 63/19 432/19

0, 7 28

0, 2 8

𝑚𝑎𝑥 = 28 ; 𝑚𝑖𝑛 = 8

𝐺. 𝑆𝑜𝑎𝑙 𝑐𝑒𝑟𝑖𝑡𝑎

1) 𝐽𝑒𝑛𝑖 𝑚𝑒𝑛𝑗𝑢𝑎𝑙 𝑘𝑢𝑒 𝐴 & 𝐵.

𝐾𝑢𝑒 𝐴 𝑑𝑖𝑏𝑒𝑙𝑖 𝑅𝑝 4.000, − & 𝑑𝑖𝑗𝑢𝑎𝑙 𝑅𝑝 7.500, −

𝐾𝑢𝑒 𝐵 𝑑𝑖𝑏𝑒𝑙𝑖 𝑅𝑝 2.000, − & 𝑑𝑖𝑗𝑢𝑎𝑙 𝑅𝑝 4.000, −

𝑀𝑜𝑑𝑎𝑙𝑛𝑦𝑎 ℎ𝑎𝑛𝑦𝑎 𝑅𝑝 400.000, − &

𝑖𝑎 𝑚𝑎𝑘𝑠𝑖𝑚𝑢𝑚 𝑚𝑒𝑛𝑗𝑢𝑎𝑙 140 𝑘𝑢𝑒.

𝐻𝑖𝑡𝑢𝑛𝑔 𝑘𝑒𝑢𝑛𝑡𝑢𝑛𝑔𝑎𝑛 𝑚𝑎𝑘𝑠 𝑛𝑦𝑎.

𝐴 𝐵𝐵𝑒𝑙𝑖𝐽𝑢𝑎𝑙

𝑈𝑛𝑡𝑢𝑛𝑔

40007500𝟑𝟓00

20004000𝟐𝟎00

400.000140

→→ 𝑥 + 𝑦 ≤ 140

2𝑥 + 𝑦 ≤ 200

𝑥 = 60𝑦 = 80

� (60, 80)

100, 0

60, 80

0, 140

𝟑𝟓𝑥 + 𝟐𝟎𝑦

3500

2100 + 1600

2800

𝑢𝑛𝑡𝑢𝑛𝑔 𝑚𝑎𝑥 = 𝑅𝑝 370.000,−

𝒛

2) 𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑘𝑖𝑛 𝑡𝑎𝑠 𝑠𝑒ℎ𝑎𝑟𝑔𝑎 𝑅𝑝 200.000, −𝑝𝑒𝑟𝑙𝑢 𝑏𝑎ℎ𝑎𝑛 3 𝑘𝑔 & 𝑤𝑎𝑘𝑡𝑢 18 𝑗𝑎𝑚.

𝑈𝑛𝑡𝑢𝑘 𝑏𝑖𝑘𝑖𝑛 𝑠𝑒𝑝𝑎𝑡𝑢 𝑠𝑒ℎ𝑎𝑟𝑔𝑎 𝑅𝑝 300.000, −𝑝𝑒𝑟𝑙𝑢 𝑏𝑎ℎ𝑎𝑛 2 𝑘𝑔 & 𝑤𝑎𝑘𝑡𝑢 24 𝑗𝑎𝑚.

𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑚𝑎𝑘𝑠 𝑝𝑒𝑛𝑗𝑢𝑎𝑙𝑎𝑛 𝑗𝑖𝑘𝑎 𝑘𝑒𝑟𝑗𝑎720 𝑗𝑎𝑚 𝑗𝑖𝑘𝑎 𝑠𝑡𝑜𝑘 𝑏𝑎ℎ𝑎𝑛 90 𝑘𝑔.

𝑡𝑎𝑠 𝑠𝑒𝑝𝑎𝑡𝑢

𝑏𝑎ℎ𝑎𝑛

𝑤𝑎𝑘𝑡𝑢𝑗𝑢𝑎𝑙

3

18𝟐𝟎𝟎 𝑟𝑏

2

24𝟑𝟎𝟎 𝑟𝑏

90

720 →→ 3𝑥 + 2𝑦 ≤ 90

3𝑥 + 4𝑦 ≤ 120

𝑦 = 15𝑥 = 20

� (20, 15)

30, 0

20, 15

0, 30

𝟐𝟎𝟎𝑥 + 𝟑𝟎𝟎𝑦

6000

4000 + 4500

9000

𝑗𝑢𝑎𝑙 𝑚𝑎𝑥 = 𝑅𝑝 9 𝑗𝑢𝑡𝑎, −

𝒛

3) 𝐷𝑟. 𝐽𝑜 𝑚𝑒𝑛𝑦𝑎𝑟𝑎𝑛𝑘𝑎𝑛 𝐴𝑛𝑒 𝑢𝑛𝑡𝑢𝑘 𝑚𝑖𝑛𝑢𝑚

𝑚𝑖𝑛𝑖𝑚𝑎𝑙 10 𝑢𝑛𝑖𝑡 𝑣𝑖𝑡. 𝐴 & 8 𝑣𝑖𝑡. 𝐵 𝑑𝑎𝑙𝑎𝑚

𝑡𝑎𝑏𝑙𝑒𝑡 & 𝑘𝑎𝑝𝑠𝑢𝑙.

𝑇𝑖𝑎𝑝 𝑡𝑎𝑏𝑙𝑒𝑡 𝑏𝑒𝑟𝑖𝑠𝑖 2 𝑣𝑖𝑡. 𝐴 & 1 𝑣𝑖𝑡. 𝐵

𝐾𝑎𝑝𝑠𝑢𝑙 𝑏𝑒𝑟𝑖𝑠𝑖 1 𝑣𝑖𝑡. 𝐴 & 2 𝑣𝑖𝑡. 𝐵

𝐻𝑎𝑟𝑔𝑎 𝑡𝑎𝑏𝑙𝑒𝑡 𝑅𝑝 800/𝑏𝑢𝑎ℎ & 𝑘𝑎𝑝𝑠𝑢𝑙 600/𝑏𝑢𝑎ℎ

𝑇𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑏𝑎𝑛𝑦𝑎𝑘𝑛𝑦𝑎 𝑡𝑎𝑏𝑙𝑒𝑡 & 𝑘𝑎𝑝𝑠𝑢𝑙 𝑦𝑔

𝑚𝑒𝑠𝑡𝑖 𝑑𝑖𝑏𝑒𝑙𝑖 𝑎𝑔𝑎𝑟 𝑖𝑟𝑖𝑡 𝑏𝑖𝑎𝑦𝑎.

𝐴 𝐵

𝑡𝑎𝑏𝑙𝑒𝑡

𝑘𝑎𝑝𝑠𝑢𝑙

2

110

1

28

800

600

2𝑥 + 𝑦 ≥ 10𝑥 + 2𝑦 ≥ 8

(4, 2)

𝒛

8x + 6y8, 0 644, 2 44

0, 10 60

𝑖𝑎 ℎ𝑎𝑟𝑢𝑠 𝑏𝑒𝑙𝑖 4 𝑡𝑎𝑏𝑙𝑒𝑡 & 2 𝑘𝑎𝑝𝑠𝑢𝑙

𝑚𝑎𝑡𝑒𝑟𝑖 𝑠𝑒𝑙𝑒𝑠𝑎𝑖 . . . . . .

𝐾𝑒𝑟𝑗𝑎𝑘𝑎𝑛 𝑈𝐾 2.6 ℎ𝑎𝑙𝑚 60

2) 3)

𝑈𝐾 2.6 ℎ𝑎𝑙𝑚 602) 𝑆𝑢𝑎𝑡𝑢 𝑝𝑎𝑏𝑟𝑖𝑘 𝑏𝑖𝑘𝑖𝑛 2 𝑗𝑒𝑛𝑖𝑠 𝑚𝑖𝑛𝑢𝑚𝑎𝑛.𝑀𝑖𝑛𝑢𝑚𝑎𝑛 𝐴 𝑏𝑢𝑡𝑢ℎ 45 𝑚𝑒𝑛𝑖𝑡 𝑑𝑖 𝑚𝑒𝑠𝑖𝑛 𝐼 &

40 𝑚𝑒𝑛𝑖𝑡 𝑑𝑖 𝑚𝑒𝑠𝑖𝑛 𝐼𝐼

𝐾𝑒𝑢𝑛𝑡𝑢𝑛𝑔𝑎𝑛 𝑚𝑖𝑛𝑢𝑚𝑎𝑛 𝐴 𝑅𝑝 6.500, − &𝑚𝑖𝑛𝑢𝑚𝑎𝑛 𝐵 𝑅𝑝 8.000, −

𝐽𝑖𝑘𝑎 𝑎𝑑𝑎 40 𝑗𝑎𝑚 𝑘𝑒𝑟𝑗𝑎 𝑚𝑒𝑠𝑖𝑛 𝐼 & 3𝟒 𝑗𝑎𝑚𝑚𝑒𝑠𝑖𝑛 𝐼𝐼, 𝑡𝑒𝑛𝑡𝑢𝑘𝑎𝑛 𝑘𝑒𝑢𝑛𝑡𝑢𝑛𝑔𝑎𝑛 𝑚𝑎𝑘𝑠.

𝑀𝑖𝑛𝑢𝑚𝑎𝑛 𝐵 𝑏𝑢𝑡𝑢ℎ 60 𝑚𝑒𝑛𝑖𝑡 𝑑𝑖 𝑚𝑒𝑠𝑖𝑛 𝐼 &50 𝑚𝑒𝑛𝑖𝑡 𝑑𝑖 𝑚𝑒𝑠𝑖𝑛 𝐼𝐼

!" 2.6 ℎ'() 602) ,-'.- /'0123 02324 2 56427 )24-)'4.824-)'4 9 0-.-ℎ 45 )642. <2 )6724 = &

40 )642. <2 )6724 ==

"6-4.-4?'4 )24-)'4 9 @/ 6.500, − &)24-)'4 C @/ 8.000, −

E23' '<' 40 5') 3615' )6724 = & 3G 5'))6724 ==, .64.-3'4 36-4.-4?'4 )'37.

824-)'4 C 0-.-ℎ 60 )642. <2 )6724 = &50 )642. <2 )6724 ==

𝐴 𝐵

𝐼

𝐼𝐼

45

406500

60

508000

2400

2040𝒛

→→ 3𝑥 + 4𝑦 ≤ 160

4𝑥 + 5𝑦 ≤ 204

𝑥 = 16𝑦 = 28

� (16, 28)

6500x + 8000y51, 0 331.500

16, 28 328.0000, 40 320.000

𝑢𝑛𝑡𝑢𝑛𝑔 𝑚𝑎𝑘𝑠𝑅𝑝 331.500,−

𝑈𝐾 2.6 ℎ𝑎𝑙𝑚 603) 𝐴𝑑𝑎 2 𝑚𝑎𝑐𝑎𝑚 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 𝑏𝑒𝑟𝑛𝑢𝑡𝑟𝑖𝑠𝑖.𝑇𝑖𝑎𝑝 𝑜𝑛𝑠 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 𝐼 𝑚𝑒𝑛𝑔𝑎𝑛𝑑𝑢𝑛𝑔 2 𝑣𝑖𝑡. 𝐴

& 1 𝑣𝑖𝑡. 𝐵 𝑠𝑒ℎ𝑎𝑟𝑔𝑎 𝑅𝑝 5.500, −

𝐵𝑒𝑟𝑎𝑝𝑎 𝑜𝑛𝑠 𝑡𝑖𝑎𝑝 𝑗𝑒𝑛𝑖𝑠 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 𝑦𝑔

𝑇𝑖𝑎𝑝 𝑜𝑛𝑠 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 𝐼𝐼 𝑚𝑒𝑛𝑔𝑎𝑛𝑑𝑢𝑛𝑔 5 𝑣𝑖𝑡. 𝐴& 4 𝑣𝑖𝑡. 𝐵 𝑠𝑒ℎ𝑎𝑟𝑔𝑎 𝑅𝑝 8.000, −

𝑚𝑒𝑠𝑡𝑖 𝑑𝑖𝑏𝑒𝑙𝑖 𝑢𝑛𝑡𝑢𝑘 𝑚𝑒𝑚𝑒𝑛𝑢ℎ𝑖 𝑘𝑒𝑏𝑢𝑡𝑢ℎ𝑎𝑛𝑚𝑖𝑛𝑖𝑚𝑢𝑚 25 𝑣𝑖𝑡. 𝐴 & 15 𝑣𝑖𝑡. 𝐵

𝐴 𝐵

𝐼

𝐼𝐼

2

525

1

415

5.500

8.000𝒛

2𝑥 + 5𝑦 ≥ 25𝑥 + 4𝑦 ≥ 15

253,53

𝟓𝟓𝒙 + 𝟖𝟎𝒚

15, 0 82.500

0, 5 𝟒𝟎. 𝟎𝟎𝟎253,53

1775003

𝑚𝑖𝑛𝑖𝑚𝑢𝑚 = 𝟒𝟎. 𝟎00,−

𝑏𝑒𝑙𝑖 0 𝑜𝑛𝑠 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 𝐼& 5 𝑜𝑛𝑠 𝑚𝑎𝑘𝑎𝑛𝑎𝑛 𝐼𝐼