Hasil Terjemahan Ke Inggris.docx

1. A company has a factory that will produce two types of products, namely silk and wool fabrics. To produce these two products necessary raw materials silk, wool yarn and labor. To produce the required 2kg damask silk yarn and 2 hours of labor required to produce wool fabrics silk yarn 3kg, 2kg wool yarn and 1 hour labor. Maximum supply of silk yarn is 60kg / day, the wool 30kg / day and labor 40jam / day. Both kinds of products provide benefits of𝑅𝑅40,000,000.00 for silk and Rp30.000.000,00 for wool fabrics. Determine the number of units of each type of product to be produced each day so that the benefits can be maximized. Discussion: For example: 𝑅 = crape 𝑅 = woolen cloth Model mtk Fabrics silk

fabrics wool

Max provision

2

3

60

Silk

-

2

30

Wool

2

1

40

Power keja

𝑅(𝑥, 𝑥) = 40𝑅 + 30𝑅 2𝑅 + 3𝑅 ≤ 60 2𝑅 ≤ 30

... .benang silk wool... .benang

2𝑅 + 𝑅 ≤ 40

... .tenaga work

(𝑅, 𝑅) 𝑅 𝑅 0 40 (0.40) (20.0) 20 0 2𝑅 + 3𝑅 = 60 𝑅 0 30 𝑅 =

(in millions)

𝑅 20 0 15

(𝑅, 𝑅) (0.20) (30.0)

2𝑅 = 30

2𝑅 + 𝑅 = 40

point C 𝑅 = 15 → 2𝑅 + 3 (15) = 60 → 2𝑅 = 60 − 45 → 𝑅 = 7.5 𝑅((7.5),15) Point D 2𝑅 + 3𝑅 = 60 2𝑅 + 𝑅 = 40 2𝑅 = 20 𝑅 = 10 𝑅 = 10 → 2𝑅 + (10) = 40 2𝑅 = 30 → 𝑅 = 15

so 𝑅 (15.10)

𝑅 (0,0) → 𝑅(0,0) = 40(0) + 30(0) = 0 𝑅 (0.15) → 𝑅(0,15) = 40(0) + 30(15) = 0 + 450 = 450 𝑅((7.5),15) → 𝑅((7.5),15) = 40(7.5) + 30(15) = 300 + 450 = 750 𝑅(15.10) → 𝑅(15.10) = 40(15) + 30(10) = 600 + 300 = 900 𝑅 (20.0) → 𝑅(20,0) = 40(20) + 30(0) = 800 + 0 = 800

2. Prove that 49𝑅 − 36𝑅 divisible by 13 Discussion: Use mathematical induction Un / 𝑅 = 1 49 1 − 361 = 13 True can be divided 13 ➢ Hypothesis: Un / 𝑅 = 𝑅 49𝑅 − 36𝑅 right can be divided by 13 ➢ Induction Un / 𝑅 = 𝑅 + 1 49𝑅 + 1 − 36𝑅 + 1 = 49.49𝑅 − 36.36𝑅 = (36 + 13). 49𝑅 − 36.36𝑅 = 36.49𝑅 + 13: 49𝑅 − 36.36𝑅 = 36(49𝑅 − 36 𝑥) + 13: 49𝑅 Because 49𝑅 − 36𝑅 correctly can be split 13 then 36(49𝑅 − 36 𝑥) is true can be split 13 . Then for 13: 49𝑅 can also be shared13. So 36(49𝑅 − 36 𝑥) + 13: 49𝑅 can be split13. 3. a total of 𝑅 of the managers of an organization will be divided into four commissions under the following conditions: (i) Every member belonging to the right two commissions , (ii) Each of the two commissions have exactly the same members. Determine the value of n. Discussion: (i)

each member incorporated into exactly two commissions

(ii)

every two commissions have exactly the same members

as there are 4 commission the number of pairs that can be made is commission 4 𝑅2 = 6

because there are many couples commission 6 then the number of members is at least 6 for less than 6 then there is a member who belong to more than 2 commissions. If there are more than 6 then there will be an entry into a commission but do not fit into the other three commissions. This is contrary to the statements (i) as a result many members there are 6 people. Illustration: Suppose: The commission is a, b, c and d 𝑅𝑅 stated i-th member with 1 ≤ 𝑅 ≤ 6 Commissiona commission b commissionc d commission 𝑅1

𝑅1

𝑅2

𝑅3

𝑅2

𝑅4

𝑅4

𝑅5

𝑅3

𝑅5

𝑅6

𝑅6