A sports manufacturer produces two products: footballs and baseballs. These products can be produced either during the morning shift or the evening shift. The cost of manufacturing the football and the baseball in the morning shift is $20 each, and the cost of manufacturing the football and the baseball in the evening shift is $25 each. The amounts of labor, leather, inner plastic lining, and demand requirements are given as follows: Resource Football Baseball Labor (hours/unit) 0.75 2 Leather (pounds/unit) 7 15 Inner plastic lining (pounds/unit) 0.5 2 Total demand (units) 1500 1200 Based on the information about the company, we know that the maximum labor hours available in the morning shift and evening shift are 5,000 hours and 2,000 hours, respectively, per month. The maximum amount of leather available for the morning shift is 15,000 pounds per month and 14,000 pounds per month for the evening shift. The maximum amount of inner plastic lining available for the morning shift is 2,000 pounds per month and 1,500 pounds per month for the evening shift.

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nwandukelechi nwandukelechi · Answer 1 · 2020-02-06T10:01:16+01:00

Step-by-step explanation:

From the above illlustration,

Let x, be the number of footballs produced in the morning shift,

y, the number of baseball in the morning shift,

z, the number of football in the evening shift,

t, the number of baseball in the evening shift.

Minimizing the objective function,

min {20(x+y) + 25(z + t)}

Therefore, since the number of labor hours is for both shifts(morning and evening shifts), we add the following constraints:

0.75x + 2y ≤ 5000

0.75z + 2t ≤ 2000

Remember, the amount of leather available in the shifts is also limited. The following constraints are got:

7x + 15y ≤ 15000

7z + 15t ≤1 4000

Also, adding the constraints for the use of inner plastic lining, we have:

0.5x + 2y ≤ 2000

0.5z + 2t ≤ 1500

Modelling their demands through the following constraints:

x + z ≥ 1500

y + t ≥ 1200

Also, we are producing whole number of baseballs or footballs but we only, so

x, y, z, t ∈Z.

Finally,

min20(x + y) + 25(z + t)