Problem 8 - A First Course in Mathematical Modeling

Question

An electric power-generating company charges different rates for residential and business users. (You might consider some reasons why this would be so.) The cost of producing the electricity is the same for all users and equals \(\$ 1000\) in fixed costs plus an additional \(\$ 200\) for each unit produced. If residential customers use \(x\) units of electricity, they pay \(p=1200-2 x\) dollars for each unit. On the other hand, commercial customers pay \(q=1000-y\) dollars for each of the \(y\) units of electricity they use. What price should the power company charge each type of customer to maximize profit? What is the maximum profit?

Step-by-Step Solution

Answer

Charge residential at $700 per unit, commercial at $600 per unit; maximum profit is $220,000.

Step 1: Define Profit Functions

To find the maximum profit, we first need to define the profit functions for both residential and commercial customers. The total revenue from residential customers is \( R(x) = x(1200 - 2x) \), and the revenue from commercial customers is \( R(y) = y(1000 - y) \). The cost of producing \( x+y \) units is given by \( C(x,y) = 1000 + 200(x + y) \). Thus, the total profit function \( P(x,y) \) can be defined as the revenue minus cost: \[ P(x,y) = x(1200 - 2x) + y(1000 - y) - (1000 + 200(x + y)) \]."

Step 2: Simplify the Profit Function

Let's simplify the expression for the profit function:\[P(x, y) = x(1200 - 2x) + y(1000 - y) - 1000 - 200x - 200y\]Expand the terms:\[P(x, y) = 1200x - 2x^2 + 1000y - y^2 - 1000 - 200x - 200y\]Combining like terms gives:\[P(x, y) = -2x^2 - y^2 + 1000x + 800y - 1000\]

Step 3: Find Critical Points for Maximum Profit

To find the maximum profit, take the partial derivatives of \( P(x, y) \) with respect to \( x \) and \( y \), and set them to zero:1. \( \frac{\partial P}{\partial x} = -4x + 1000 = 0 \), solving gives \( x = 250 \).2. \( \frac{\partial P}{\partial y} = -2y + 800 = 0 \), solving gives \( y = 400 \).These critical points represent the quantities \( x \) and \( y \) that maximize the profit function.

Step 4: Confirm Maximum via Second Derivative Test

Check the second derivative test to confirm that these critical points lead to a maximum:- \( \frac{\partial^2 P}{\partial x^2} = -4 \) (which is negative).- \( \frac{\partial^2 P}{\partial y^2} = -2 \) (which is negative).Since both second derivatives are negative, the function is concave down in each variable, confirming a local maximum.

Step 5: Calculate Maximum Profit and Prices

Finally, substitute \( x = 250 \) and \( y = 400 \) back into the price equations and profit function to find the maximum profit and corresponding prices:- For residential users: \( p = 1200 - 2(250) = 700 \).- For commercial users: \( q = 1000 - 400 = 600 \).The maximum profit is:\[P(250, 400) = -2(250)^2 - (400)^2 + 1000(250) + 800(400) - 1000 = 220,000\]

Key Concepts

Profit Maximization

Profit maximization is the process of determining the best output and pricing levels to achieve the highest profit. In this context, it involves setting optimal prices for residential and commercial customers to ensure the power company earns the most profit while meeting demand effectively.
The profit function is formed by subtracting the total costs from the total revenue. For the power company, the revenue comes from the selling electricity to two types of users, residential and commercial, each with a distinct price-demand relationship.
To maximize profit, the company calculates the output levels, or electricity units, for each customer type that lead to the highest possible profit. The calculation involves analyzing the profit function and finding the quantities where profits are at their peak.

Cost Structure

Understanding the cost structure is vital for determining the profitability of a business model. Fixed costs, in this case, are constant—it costs the power company \(1000 irrespective of how much electricity they produce.
Variable costs, on the other hand, depend on the units of electricity produced. Here, each unit of electricity incurs a cost of \)200.
Therefore, the total cost function for the power company becomes a sum of fixed and variable costs:

• Fixed costs: \(1000
• Variable costs: \)200 per unit of electricity

This results in a total production cost function of: \[C(x,y) = 1000 + 200(x + y)\]
This simplified equation is used to measure how changing the quantity of electricity impacts the total costs.

Price Optimization

Price optimization involves setting the prices to maximize revenue and profit. For the power company, the price per unit changes depending on the quantity used by the customers:

• Residential: \( p = 1200 - 2x \)
• Commercial: \( q = 1000 - y \)

These equations show that the price per unit decreases as more electricity is used, signifying a declining demand curve.
Optimization means balancing a higher price per unit with the number of units sold to maximize total revenue without exceeding customer price sensitivities.
By calculating the optimal prices for residential and commercial users when the profits are maximized, the power company determined that residential customers should be charged \\(700 per unit and commercial customers \\)600 per unit.

Differential Calculus

Differential calculus is a mathematical tool used to determine the rate at which quantities change. In profit maximization, it's used to find the points where the profit function peaks by analyzing its rate of change.

• First, partial derivatives of the profit function with respect to the quantities of electricity are taken: \( \frac{\partial P}{\partial x} \) and \( \frac{\partial P}{\partial y} \).
• Setting these derivatives to zero finds the critical points, suggesting maximum or minimum profit levels.
• Second, the second derivatives \( \frac{\partial^2 P}{\partial x^2} \) and \( \frac{\partial^2 P}{\partial y^2} \) help determine if these critical points suggest a maximum or a minimum. Negative values confirm a maximum.

This series of steps provides a systematic approach to discover the ideal sales quantities for maximum profitability.