Answer:
$21357
Explanation:
The table below gives the number of minutes required for each item
Machine time Craftsman time
Item X 13 20
Item Y 19 29
The company has 40 hours of machine time available in the next working week but only 35 hours of craftsman time. Machine time is costed at $100 per hour worked and craftsman time is costed at $20 per hour worked. Both machine and craftsman idle times incur no costs. The revenue received for each item produced (all production is sold) is $200 for X and $300 for Y. The company has a specific contract to produce 10 items of X per week for a particular customer. Formulate the problem of deciding how much to produce per week as a linear program hence make the decision.
Let be the number of items of and be the number of items of .
We need to maximize:
200 + 300 − 100(ℎ ) − 20( )
subject to:
13 + 19≤ 40(60) ℎ
20 + 29≤ 35(60)
≥ 10
,≥ 0
The object function is now:
[tex]200x + 300y - 100(\frac{13x+19y}{60}) -20(\frac{20x+29y}{60})\\197.1667x+295.8667y\\subject to:\\13x+19y\leq 2400\\20x+29y\leq 2100\\x\geq 10\\x,y\geq 0[/tex]
the maximum occurs at the intersection of =10 and 20 + 29≤ 2100.
Solving simultaneously, we have that =10 and =65.52. Substituting in the objective function:
197.1667(10) + 295.8667(65.52)=$21357
