Problem 40 - Precalculus : Building Concepts and Connections

Question

The price \(s\) (in dollars) of a product is given by \(s(q)=100-0.1 q, 0 \leq q \leq 1000,\) where \(q\) is the number of units sold per day. It costs \(\$ 10,000\) per day to operate the factory and an additional \(\$ 12\) for each unit produced. (a) Find the daily revenue function, \(R(q)\) (b) Find the daily cost function, \(C(q)\) (c) The profit function is given by \(P(q)=R(q)-C(q)\) For what values of \(q\) will the profit be greater than or equal to zero?

Step-by-Step Solution

Answer

Revenue function is \(R(q) = 100q - 0.1q^2\), the cost function is \(C(q) = $10,000 + $12 \times q\), and the profit function is \(P(q) = 0.1q^2 + 88q - $10,000\). The values of 'q' where profit is ≥ 0 can be found by solving the quadratic equation \(0.1q^2 + 88q - $10,000 = 0\).

Step 1: Determine the daily revenue function R(q)

The daily revenue function, \(R(q)\), can be obtained by multiplying the price function, \(s(q)\), with the number of units sold, 'q'. This yields, \(R(q) = s(q) \times q = (100 - 0.1q) \times q = 100q - 0.1q^2\).

Step 2: Determine the daily cost function C(q)

The daily cost function, \(C(q)\), is a summation of the constant daily operating cost ($10,000) and per unit production cost ($12) times units produced (q). Therefore, \(C(q) = \$10,000 + \$12 \times q\).

Step 3: Calculate the profit function P(q)

The profit function, \(P(q)\), is the difference between the revenue function, \(R(q)\), and cost function, \(C(q)\). Hence, \(P(q) = R(q) - C(q) = (100q - 0.1q^2) - (\$10,000 + \$12 \times q) = 100q - 0.1q^2 - \$10,000 - 12q = 0.1q^2 + 88q - \$10,000 \)

Step 4: Find the values of q for which profit is ≥ 0

To get these values, the profit function has to be set equal to zero and solved for 'q'. So, \(0.1q^2 + 88q - \$10,000 = 0\). Solving this quadratic equation will give the range of 'q' for which the profit is zero or higher.

Key Concepts

Revenue Function

In the context of business mathematics, understanding the revenue function is pivotal. The revenue function, often denoted as \( R(q) \), represents the total income generated from selling \( q \) units of a product or service. It is found by multiplying the price per unit, \( s(q) \), by the number of units sold, \( q \). In this exercise, the price of the product is modeled by the linear equation \( s(q) = 100 - 0.1q \). This implies that as more items are sold, the price per item decreases, a concept commonly known as price markdown or demand-driven pricing. Consequently, the revenue function is written as:

• \( R(q) = s(q) \times q \)
• Substituting the known values, we get \( R(q) = (100 - 0.1q) \times q \)
• Simplifying, \( R(q) = 100q - 0.1q^2 \)

This is a quadratic function, hinting that there may be a maximum revenue point.

Cost Function

The cost function, \( C(q) \), is a crucial tool for determining the expenses associated with producing a given number of units. In this scenario, costs are comprised of both fixed and variable components.

• Fixed Costs: These are costs that do not change with the level of production, such as the daily operating cost of \(10,000.
• Variable Costs: Incurred on a per-unit basis, here it is \)12 per unit produced.

Thus, the cost function is:\[C(q) = 10,000 + 12q\]This linear equation reflects the direct relationship between the number of units produced and the costs incurred. By carefully estimating the cost function, businesses can forecast expenses and adjust pricing or production levels accordingly to ensure profitability.

Quadratic Equation

Solving quadratic equations is fundamental in finding the roots or solutions where certain conditions are met, such as profit being zero or positive. In the exercise, we arrive at a quadratic equation when determining the values for \( q \) that make the profit function non-negative.The profit function, \( P(q) \), derived in previous steps is:\[P(q) = -0.1q^2 + 88q - 10,000\]To find when the profit is greater than or equal to zero, we set:\[-0.1q^2 + 88q - 10,000 \geq 0\]This becomes a quadratic inequality. Solving it involves finding the roots of the corresponding equation \( 0.1q^2 - 88q + 10,000 = 0 \) using the quadratic formula:\[q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 0.1\), \(b = -88\), and \(c = 10,000\). This step is critical for identifying feasible production levels that maintain a viable profit.

Precalculus

Precalculus serves as a foundational mathematical discipline that lays the groundwork for calculus. It typically includes topics such as functions, algebra, and trigonometry - all of which are evident in the mathematical modeling within this exercise.Understanding precalculus allows you to:

• Comprehend how to develop and analyze functions like the revenue and cost functions.
• Manipulate algebraic expressions and solve equations, as shown in the steps of creating \( R(q) \) and \( C(q) \).
• Solve quadratic equations and inequalities, a necessary skill for determining production levels in business scenarios.

These concepts facilitate the navigation of complex business problems, enabling students to apply mathematical thinking to maximize profit in practical applications. By mastering precalculus, you'll be well-prepared to approach calculus with confidence, where these skills are further refined.