Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

After being in business for \(t\) years, a manufacturer of cars is producing \(120+2 t+3 t^{2}\) units per year. The sales price in dollars per unit has risen according to the formula \(6000+700 t\). Write a formula for the manufacturer's yearly revenue \(R(t)\) after \(t\) years.

Step-by-Step Solution

Answer

The revenue formula is \( R(t) = 2100t^3 + 19400t^2 + 85200t + 720000 \).

Step 1: Understanding Revenue Formula

Revenue is the total income earned by selling goods or services. In this case, it's calculated as the product of the number of units produced and the price per unit. We need to find the revenue function based on the units produced and the sales price.

Step 2: Identify the Units Produced Function

The problem states that the units produced each year is given by the formula: \(120 + 2t + 3t^2\). This represents the number of cars produced after \(t\) years.

Step 3: Identify the Sales Price Function

The sales price per unit (per car) is given by the formula \(6000 + 700t\), which depends on the number of years \(t\). This represents the price at which each car is sold after \(t\) years.

Step 4: Compute the Revenue Function

The revenue \(R(t)\) is calculated by multiplying the number of units produced with the sales price per unit: \[ R(t) = (120 + 2t + 3t^2) \times (6000 + 700t) \].

Step 5: Simplify the Revenue Function

Expand the equation obtained in the previous step to get a polynomial expression for \(R(t)\):\[ R(t) = (120 + 2t + 3t^2)(6000 + 700t) \].Expanding this, do:1. Multiply \(120\) by both terms of \(6000 + 700t\).2. Multiply \(2t\) by both terms of \(6000 + 700t\).3. Multiply \(3t^2\) by both terms of \(6000 + 700t\).Summing these gives us:\[ R(t) = 720000 + 84000t + 1200t + 1400t^2 + 18000t^2 + 2100t^3 = 720000 + 85200t + 19400t^2 + 2100t^3 \].Thus, the formula for the manufacturer's yearly revenue is \( R(t) = 2100t^3 + 19400t^2 + 85200t + 720000 \).

Key Concepts

Polynomial Revenue Function

A polynomial revenue function is a mathematical expression used to calculate the total income a company generates. It usually involves variables such as sales price and quantity sold. In our car manufacturer's case, the revenue function is a polynomial involving the variable \(t\) for time in years. The function here arises by multiplying the units produced each year by the sales price. The result, after simplification, is a third-degree polynomial:

• The term \(2100t^3\) comes from compounding changes in units over time and increasing price.
• The terms \(19400t^2\), \(85200t\), and \(720000\) are derived from multiplying each part of the production formula \(120 + 2t + 3t^2\) with parts of the sales price formula \(6000 + 700t\).

Each term in this polynomial displays how the compound effects of production rates and price adjustments contribute to revenue.

Units Produced Function

In this problem, the units produced function is represented by \(120 + 2t + 3t^2\). This quadratic equation provides insight into how many units (cars) are manufactured each year based on time \(t\).

• The constant term \(120\) indicates the initial production rate when the business started.
• The term \(2t\) suggests a linear increase in production as time passes, each year bettering the count by 2 units from the base production.
• The quadratic term \(3t^2\) exhibits accelerated growth over time; every year production gains even more cars as the company becomes more experienced and efficient.

By analyzing this function, businesses can understand the growth pattern in their production processes over time.

Sales Price Function

The sales price function \(6000 + 700t\) represents how the price per unit (per car) changes each year. It's a linear function depending solely on time \(t\).

• Initially, each car is priced at \(\\(6000\).
• As the business continues, the price of each car rises by \(\\)700\) per year. This increase could reflect adjustments due to enhanced features, inflation, or market demand changes.

This function is critical as it helps businesses project revenue, adjust pricing strategies, and understand market dynamics.

Calculus in Business Applications

Calculus is an essential tool in business, particularly when optimizing processes like production and pricing to maximize revenue. Here, calculus helps by providing functions that describe dynamic systems and project future outcomes over time.

• By using derivatives, businesses can determine the rate of change in units produced and pricing.
• Polynomials, like our revenue function \(R(t) = 2100t^3 + 19400t^2 + 85200t + 720000\), give complex models of revenue over time, considering both linear and nonlinear growth impacts.
• Through integration, businesses calculate total revenues over specific periods or understand cumulative growth.

These applications highlight how calculus aids in making informed strategic decisions to enhance profitability and efficiency.