American Vending (AV) supplies vended food to a large university. Because students often kick the machines out of anger and frustration, management has a constant repair problem. The machines break down on an average of three per hour, and the breakdowns are distributed in a Poisson manner. Downtime costs the company $25 per hour per machine, and each maintenance worker gets $16 per hour. One worker can service machines at an average rate of five per hour distributed exponentially. What is the optimal maintenance crew size for servicing the machines

case 2 with two workers is the optimal decision.

Step-by-step explanation:

Case 1—One worker:A= 3/hour Poisson, ¡x =5/hour exponential The average number of machines in the system isL = - 3. = 4 = lJr machines' ix-A 5 - 3 2 2Downtime cost is $25 X 1.5 = $37.50 per hour; repair cost is $4.00 per hour; and total cost per hour for 1worker is $37.50 + $4.00

= $41.50.Downtime (1.5 X $25) = $37.50 Labor (1 worker X $4) = 4.00

$41.50

Case 2—Two workers: K = 3, pl= 7L= r= = 0.75 machine1 p. -A 7 - 3Downtime (0.75 X $25) = S J 8.75Labor (2 workers X S4.00) = 8.00S26.75Case III—Three workers:A= 3, p= 8L= ——r = 5- ^= § = 0.60 machinepi -A 8 - 3 5Downtime (0.60 X $25) = $15.00 Labor (3 workers X $4) = 12.00 $27.00

Comparing the costs for one, two, three workers, we see that case 2 with two workers is the optimal decision.

3.how is solving radical equations different than solving linear equations?

step-by-step explanation: