Problem 55 - Calculus Early Transcendentals: Pearson New International Edition

Question

The manufacturer of Zbars estimates that 100 units per month can be sold if the unit price is \(\$ 250\) and that sales will increase by 10 units for each \(\$ 5\) decrease in price. Write an expression for the price \(p(n)\) and the revenue \(R(n)\) if \(n\) units are sold in one month, \(n \geq 100\)

Step-by-Step Solution

Answer

Price: \(p(n) = 250 - 0.5(n - 100)\); Revenue: \(R(n) = n(300 - 0.5n)\).

Step 1: Understand the relationship

The problem states that the sales will increase if the price decreases. With a decrease of \(\$5\) in price, sales increase by 10 units.

Step 2: Write the expression for price

Initially, for 100 units, the price is \(\\(250\). For each additional 10 units, you reduce the price by \(\\)5\). Thus, the price function for \(n\) units can be expressed as: \(p(n) = 250 - 0.5(n - 100)\).

Step 3: Write the expression for revenue

Revenue is the product of the number of units sold and the price per unit. Revenue function is given by: \(R(n) = n \times p(n)\). Substitute \(p(n)\) from Step 2 into the revenue equation to get \(R(n) = n \times (250 - 0.5(n - 100))\).

Step 4: Simplify the revenue expression

Substitute the expression for \(p(n)\) into the revenue formula: \(R(n) = n \times \left(250 - 0.5 \times (n - 100)\right)\). Simplify further to get the final expression: \(R(n) = n \times (300 - 0.5n)\).

Key Concepts

Linear Functions

A linear function is a type of mathematical function that creates a straight line when graphed on a coordinate plane. It is usually written in the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. In our exercise, the price function \(p(n) = 250 - 0.5(n - 100)\) represents a linear relationship between the number of units sold \(n\) and the price per unit \(p\).

• The slope here, \(-0.5\), tells us how much the price decreases for each additional unit sold beyond 100.
• The term \(250\) represents the initial unit price when \(n = 100\).

To identify patterns in word problems, understanding linear functions helps you predict how one variable affects another, especially when variables change at constant rates. This concept is vital across various applications in economics, science, and everyday decision-making. When you graph the function, notice how every step down in price corresponds with an exact increase in the number of units, illustrating the predictable and proportional change characteristic of linear functions.

Revenue Optimization

Revenue optimization involves finding the quantity at which revenue is maximized. Revenue itself is the total income generated from selling goods or services. It's calculated as the product of the quantity sold and the price per unit. In this exercise, we focus on the revenue function derived: \(R(n) = n \times (300 - 0.5n)\). To optimize revenue, one would find the value of \(n\) that maximizes this function. This process often involves finding the vertex of a parabola, which represents the maximum or minimum value of a quadratic function.

• Understanding how changes in pricing influence revenue is crucial.
• Analyzing derived revenue functions can guide pricing strategies and business decisions.

In the real world, businesses rely on revenue optimization to maximize profits and meet financial goals, making it an indispensable concept for future entrepreneurs and business strategists.

Algebraic Expressions

Algebraic expressions are crucial tools in mathematics, representing numbers and operations in a generalized form using variables, constants, and operators. In the context of our exercise, both the price function \(p(n)\) and the revenue function \(R(n)\) are algebraic expressions.An understanding of these expressions allows you to:

• Substitute specific values, making calculations manageable.
• Manipulate them to simplify problems or solve for unknowns.

In the exercise, the algebraic expression for revenue \(R(n) = n\times(300 - 0.5n)\) gives insight into the dependence of revenue on both price and quantity sold. Utilizing algebraic expressions helps to break down complex problems into solvable pieces, enabling clearer communication of mathematical ideas. For students, mastering algebraic expressions is a gateway to tackling more advanced topics in mathematics and related fields.