Final answer:
To find the optimal operating time for an evaporator to maximize product over 30 days, each operational cycle including the required 6-hour maintenance time is considered. The total operating hours in 30 days are calculated for each given option and the option with the highest value of total operating hours is chosen as the optimal operating time. The correct answer is option (d)
Step-by-step explanation:
To determine the optimal operating time between cleanings for a continuous evaporator to maximize the amount of product over a 30-day period, we need to consider the total operating time minus the downtime for maintenance. Let's define a cycle as the period of operating time followed by the downtime for cleaning, which is 6 hours. To maximize the product, we need to maximize the number of hours the evaporator operates in a 30-day cycle.
Let T be the total operating hours between cleanings.
The total time for each cycle would then be T hours of operation plus 6 hours of maintenance, making it T + 6 hours per cycle.
The number of cycles in 30 days can be determined by dividing 720 (the number of hours in 30 days) by T + 6.
The total operating time in 30 days would be the number of cycles multiplied by T.
To find the value of T that maximizes this total operating time, we would take the derivative of this function with respect to T, set it to zero, and solve for T.
However, since we have options (a through d) for T, we can calculate the total operating time for each option and choose the one yielding the highest value.
After calculating the total operating time for each option, we would choose the one with the highest value. So, the question is asking for a basic optimization problem that takes into account production time and maintenance time to find the maximum product yield over a month.