Question

A power company faces demands during both peak and off-peak times. If a price of p₁ dollars per kilowatt-hour is charged during the peak time, customers will demand 60 – 0.5 p₁ kwh of power. If a price of p₂ dollars is charged during the off-peak time, then customers will demand 40 – p₂ kwh. The power company must have sufficient capacity to meet demand during both the peak and off-peak times. It costs $ 10 per day to maintain each kilowatt-hour of capacity. Determine how the power company can maximize daily revenues less operating costs.

Answer 1

To maximize daily revenues less operating costs, the power company needs to determine the price that will maximize demand during both peak and off-peak times. This can be done by finding the values of p₁ and p₂ that will maximize the function f(p₁, p₂) = (p₁ * DP + p₂ * DO) - 10 * (DP + DO), where DP = 60 - 0.5p₁ and DO = 40 - p₂.

Step-by-step explanation:

To maximize daily revenues fewer operating costs, the power company needs to determine the price that will maximize demand during both peak and off-peak times. Let's denote the demand during the peak time as DP and the demand during the off-peak time as DO. According to the given information, DP = 60 - 0.5p₁ and DO = 40 - p₂.

The power company's revenue during the peak time is given by RP = p₁ * DP, and the revenue during the off-peak time is RO = p₂ * DO. We can calculate the total revenue as TR = RP + RO. We can also calculate the operating costs as OC = 10 * (DP + DO).

To maximize the daily revenues fewer operating costs, we need to find the values of p₁ and p₂ that will maximize TR - OC. This can be done by finding the maximum value of the function f(p₁, p₂) = (p₁ * DP + p₂ * DO) - 10 * (DP + DO).

By solving the system of equations DP = 60 - 0.5p₁ and DO = 40 - p₂, we can find the values of p₁ and p₂ that maximize f(p₁, p₂). Then, we can substitute these values back into the equations for DP and DO to find the corresponding demands, revenues, and operating costs.