Question

Problem 3: Johnson City Cooperage manufactures 30-gallon and 55-gallon fiber drums. Each 30-gallon drum takes 30 minutes to make, each 55-gallon drum takes 40 minutes to make, and the company has at most 10,000 minutes available each week. Also, workplace limitations and product demand indicate that the number of 55-gallon drums produced plus half the number of 30-gallon drums produced should be at Teast 160, and the number of 30-gallon drums should be at least twice the number of 55-gallon drums Johnson City Cooperage's manufacturing costs are $4 for each 30-gallon drum and S6 for each 55-gallon drum. How many of each drum should be made each week to satisfy the production constraints and minimize production cost? Find the minimum production cost. Summarize Problem Information: Determine Decision Variables: Determine Objective Function: Determine Constraints: Write a sentence or two containing the solution & and interpretation of the solution:

Answer 1

Answer: ---- Let Date: ---28/9/2019 x = number of 30-gallon drums y = number of 55-gallon drums Then the problem can be summarized as follows: Time to make(min utes) Cost($) 30-gallon drums 55-gallon drums Availability 30 40 10000 4 6 - We have to minimize the cost. Let the objective function be f. Therefore, the problem is as follows: Minimise f = 4x+6y subject to 30x +40 y 10000 ... (1) Also, y+2160 2y+x 320 .....(2) And

x22y x-2y 20 .....(3) All the variables are non-negative. Multiplying both sides of (2) by -1, we get -x-2y5-320 ---(4) Multiplying both sides of (3) by -1, we get -x+2y 50 .....(5) Therefore, the problem is as follows: Maximise - f =-4x-6y subject to 30x + 40 y 510000 -x-2y <-320 -x+2y=0 Introducing slack variables, we get 30x +40y+s, = 10000 ...... (6) -x-2y+s, =-320 ... (7)

-x+2y+s; = 0 ..... (8) And - f =-4x-6y 4x+6y- f = 0 ...... (9) Thus we have following matrix 10000 x y $ 30 40 1 -1 -2 0 -1 2 0 $ $; 0 0 1 0 0 1 -f 0 0 0 Since the last column contains a negative entry-320, we choose 1st column as pivot column to make it positive and entry with the smallest positive quotient as Pivot entry. x y s 30 40 1 -1 -2 0 -1 2 0 S2 0 1 0 S3 0 0 1 - f 0 0 0 10000 10000/30 = 333. 33 -320 -3207–1 = 320+ 0 07-1=0

| 4 6 0 0 0 1 0 Now we applying the row operations R, →-1R, R → R -30R, R→ R+R R. R-4R x y , SzSz - f 0 -20 1 30 0 1 2 0 -1 0 0 320 0 4 0 -1 1 0 320 0 -2 0 4 0 1 -1280 400 Since the last row contains negatives. We choose 2nd column with the negative entry as pivot column and the element with the smallest positive quotient as pivot entry. xy S SS - 0 -20 1 30 0 0 1 2 0 -1 0 0 400 320 4007-20 = -20 320/2 = 160

320/4 = 80+ 0 0 4 -2 0 0 1 4 1 0 0 1 320 -1280 Now we applying the row operations R R R. → R, +20R, R2 + R2 -2R R. R. +2R [x y s 8 8 -f 0 0 1 25 5 0 2000 1 0 0 -1/2 -1/ 2 0 160 0 1 0 -1/4 1/4 0 80 0 0 0 7/2 1/2 1 -1120 Since the upper portion in the last column has no negatives and the last row contains only positive or zero entries, therefore the solution is optimal. x=400

y = 400 Therefore, number of 30-gallon drums = 160 number of 55-gallon drums = 80 Also, minimum cost = $1120