Question

Blue Ridge Power Company has a maximum capacity of 2.5 million kilowatt-hour (kwh) of electric power available on a daily basis. The demand (in millions of kwh) for power from its customers for high demand hours and low demand hours is determined by the following formulas: High demand = 5.8 - 0.06Ph +0.005PL Low demand = 3.0 - 0.11PI + 0.008Ph The variable Pl equals the price per kilowatt-hour during low demand hours, and Ph is the price per kilowatt-hour during high demand hours. Suppose the daily demand for electric power is the summation of the power from high demand hours and power from low demand hours. What would be the appropriate constraint of electric power?

- 5.8 -0.06Ph +0.005PI + 3.0 -0.11PI + 0.008Ph = 2.5 O
- (5.8 -0.06Ph +0.005PI) – (3.0 -0.11PI + 0.008Ph) = 2.5 O
- (5.8 -0.06Ph +0.005PI)Ph + (3.0 - 0.11P1 + 0.008ph)PI = 2.5 O
- 3.0 -0.11P1 + 0.008Ph = 2.5

Answer 1

The appropriate constraint of electric power can be represented by the equation:

- (5.8 -0.06Ph +0.005PI) – (3.0 -0.11PI + 0.008Ph) = 2.5. The correct answer is option 2.

Step-by-step explanation:

The student has asked for the appropriate constraint for electric power when combining high demand and low demand hours for Blue Ridge Power Company that has a maximum capacity of 2.5 million kilowatt-hour (kWh) per day. The formulas for high demand and low demand in terms of the price per kilowatt-hour during high (Ph) and low (Pl) demand hours are given.

Considering the total demand should not exceed the maximum capacity, the correct constraint equation that represents the combined power demands equalling to the maximum capacity of 2.5 million kWh is:

- (5.8 - 0.06Ph + 0.005Pl) + (3.0 - 0.11Pl + 0.008Ph) = 2.5

Here, we sum the expressions for high demand and low demand, and this total demand should equal the maximum capacity available.