Question

Gator Brews manufactures three types of brews: A, B, and C. These brews are produced via twa production processes: 1 and 2. Running process 1 for an hour costs $400 and yields three hundred gallons of brew A, one hundred gallons of brew B, and one hundred gallons of brew C. Running process 2 for an hour costs $100 and yields one hundred gallons of brew A, one hundred gallons of brew B. To meet customer demands, at least 1000 gallons of brew A, 500 gallons of brew B, and 300 gallons of brew C must be produced daily. Gator Brews wants to minimize the cost of running both processes. Set up and answer this question in worksheet "Part 6" using the simplex method in solver. (15 points) X1-hrs of process 1 X2 - hrs of process 2 MIN Z = 400x1 +100X2 s.t. Gallons of A 300 x1 + 100 x2 1000 Gallons of B 100 x1 + 100 x2 2500 Gallons of C 100 x1 + 0x2 2 300

Answer 1

Answer:

z(min)  = 1400

x = 3

y = 2

Step-by-step explanation:

Formulation:

Let´s call

x number of running hours of process 1

y number of running hours of process 2

Then Objective Function to minimize is:

z  =  400*x  + 100*y

Constrains:

Subject to

Demand constraint ( by type of brews)

300*x  + 100y ≥ 1000

100*x + 100*y ≥ 500

100*x ≥ 300

General constraint

y≥0     Note that we already have  x≥= (in the third constraint)

By using the on-line slver we find:

z (min) = 1400 $

x = 3

y = 2