Problem 81
Question
A security firm currently has 5000 customers and charges \(\$ 20\) per month to monitor each customer's home for intruders. A marketing survey indicates that for each dollar the monthly fee is decreased, the firm will pick up an additional 500 customers. Let \(R(x)\) represent the revenue generated by the security firm when the monthly charge is \(x\) dollars. Find the value of \(x\) that results in the maximum monthly revenue.
Step-by-Step Solution
Verified Answer
The value of \(x\) that results in the maximum monthly revenue is 15, meaning the firm should charge $5 (20 - 15) per customer to maximize revenue.
1Step 1: Define the number of customers as a function of x
Since the price decrease increases the number of customers, we can write the number of customers, n, as a function of x where \(n(x) = 5000 + 500x\). This is because the firm starts with 5000 customers and gains 500 additional customers for each dollar decrease in the price.
2Step 2: Define the price per customer as a function of x
Since the price decreases per dollar, we can define the price, p, as a function of x where \(p(x) = 20 - x\). This is because the initial price is $20 and it decreases by $1 for each increase in x.
3Step 3: Formulate the revenue function R(x)
The revenue function is defined as the product of the price function and the customer function, hence \(R(x) = p(x) \cdot n(x) = (20 - x)(5000 + 500x) = 100000 - 20000x + 500x - 500x² = -500x² - 15000x + 100000\).
4Step 4: Find the maximum of R(x)
To find the maximum revenue, we need to find the maximum of the function R(x). This occurs at the vertex of the parabola represented by the quadratic function. For a function of the form \(ax² + bx + c\), the x-coordinate of the vertex is given by \(-b/(2a)\). Our a is -500 and b is -15000, hence the vertex is at \(x = -(-15000)/(2*-500) = 15\).
5Step 5: Verify the result
To verify our result, we can substitute x=15 into the revenue function and see if it gives a valid result: \(R(15) = -500*15² + 15000*15 + 100000 = 125000\) which is reasonable.
Key Concepts
Revenue FunctionQuadratic FunctionsVertex of a Parabola
Revenue Function
Understanding how a business can optimize its revenue is crucial for its success. A revenue function is a mathematical representation of how a company's revenue depends on its pricing strategy and sales volume. It's expressed as a function of the price charged for goods or services.
The scenario of the security firm in our exercise is a classic example. They initially have 5000 customers at a charge of \(20 each. The revenue function, denoted as R(x), explains the total revenue generated based on varying the price, x, as they adjust their monthly charges. Each decrease in price by \)1 attracts 500 more customers, leading to the revenue function R(x) = (20 - x)(5000 + 500x). This function helps businesses forecast the outcomes of pricing strategies.
The scenario of the security firm in our exercise is a classic example. They initially have 5000 customers at a charge of \(20 each. The revenue function, denoted as R(x), explains the total revenue generated based on varying the price, x, as they adjust their monthly charges. Each decrease in price by \)1 attracts 500 more customers, leading to the revenue function R(x) = (20 - x)(5000 + 500x). This function helps businesses forecast the outcomes of pricing strategies.
Quadratic Functions
Quadratic functions are one of the most fundamental concepts in algebra and play a key role in various applications like physics, engineering, and economics. These functions are written in the standard form ax² + bx + c, where a, b, and c are constants, and x is the variable.
In our case, after expanding and simplifying the revenue function, we get R(x) = -500x² + 15000x + 100000, a quadratic equation where a = -500, b = 15000, and c = 100000. The graph of this equation forms a parabola. Since a is negative, the parabola opens downwards, indicating that the firm's revenue will reach a maximum value for a specific charge per customer. The goal is to find this optimal price x that maximizes revenue.
In our case, after expanding and simplifying the revenue function, we get R(x) = -500x² + 15000x + 100000, a quadratic equation where a = -500, b = 15000, and c = 100000. The graph of this equation forms a parabola. Since a is negative, the parabola opens downwards, indicating that the firm's revenue will reach a maximum value for a specific charge per customer. The goal is to find this optimal price x that maximizes revenue.
Vertex of a Parabola
The vertex of a parabola is the point where the curve changes direction; in the case of a quadratic function representing revenue like ours, it's also where the function reaches its maximum or minimum value. For a downward-opening parabola, the vertex represents the maximum point.
Mathematically, the x-coordinate of the vertex is calculated using the formula \( x = \frac{-b}{2a} \) when the quadratic function is in standard form. Applying this to our revenue function, with a = -500 and b = 15000, we find that the vertex occurs at x = 15. This concludes that the security firm would generate maximum monthly revenue by charging \( 20 - 15 = \) \(5 less per month. Thus, the optimal new monthly charge would be \)15.
Mathematically, the x-coordinate of the vertex is calculated using the formula \( x = \frac{-b}{2a} \) when the quadratic function is in standard form. Applying this to our revenue function, with a = -500 and b = 15000, we find that the vertex occurs at x = 15. This concludes that the security firm would generate maximum monthly revenue by charging \( 20 - 15 = \) \(5 less per month. Thus, the optimal new monthly charge would be \)15.
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