Problem 66 - College Algebra and Calculus: An Applied Approach

Question

The revenue \(R\) and cost \(C\) for a product are given by \(R=x(75-0.0005 x)\) and \(C=30 x+250,000\), where \(R\) and \(C\) are measured in dollars and \(x\) represents the number of units sold (see figure). (a) How many units must be sold to obtain a profit of at least \(\$ 750,000 ?\) (b) The demand equation for the product is \(p=75-0.0005 x\) where \(p\) is the price per unit. What prices will produce a profit of at least \(\$ 750,000 ?\) (c) As the number of units increases, the revenue eventually decreases. After this point, at what number of units is the revenue approximately equal to the cost? How should this affect the company's decision about the level of production?

Step-by-Step Solution

Answer

To obtain a profit of at least $750,000, the company must sell a certain number of units which will be obtained from Step 2. Prices that will similarly yield this profit can be calculated from the demand equation using the \(x\) value. The gained insights on the number of units that make revenue equal to cost will guide the company's production decision.

Step 1: Find the profit function

The profit (\(P\)) can be found by subtracting the cost (\(C\)) from the revenue (\(R\)). So, \(P = R - C = x(75 - 0.0005x) - (30x + 250000)\).

Step 2: Solve for \(x\) to obtain a profit of at least $750,000

We need to solve the inequality \(P \ge 750000\). Substituting the profit function we obtained, we get: \(x(75 - 0.0005x) - (30x + 250000) \ge 750000\). Solve this inequality for \(x\).

Step 3: Find the price that will yield a profit of at least $750,000

Substitute the \(x\) value obtained in Step 2 into the demand equation \(p = 75 - 0.0005x\) and solve for \(p\).

Step 4: Determine the number of units that make revenue equal to cost

We need to solve for \(x\) when \(R = C\), that is, \(x(75 - 0.0005x) = (30x + 250000)\). Solve this equation for \(x\).

Step 5: Observe the effect on the company's decision

The value obtained for \(x\) in step 4 will represent the number of units at which profit max out before starting to decrease. If production continues beyond this level, the company will start to take losses; as such, the company should not produce this number of units.

Key Concepts

Revenue Function

A revenue function is used to determine how much money a business brings in through selling its goods or services. In this exercise, the revenue function is expressed as \[ R = x(75 - 0.0005x) \]where \( R \) is the total revenue in dollars, and \( x \) is the number of units sold. The expression inside the parentheses, \(75 - 0.0005x\), represents the price per unit as defined by the demand equation. This means the revenue can be seen as the number of units times the price per unit.

• The term \( 75 \) indicates the starting price.
• The term \(-0.0005x\) shows how the price per unit decreases as more units are sold. This reflects potential market saturation or competitive pricing dynamics.

Thus, understanding the revenue function helps businesses predict how changes in sales volume affect overall earnings. For maximizing profit, it’s key to find the number of units where revenue is still increasing but not yet driving down prices too far.

Cost Function

The cost function details the expenses a business incurs to produce a certain number of units. Here, the cost function is structured as:\[ C = 30x + 250,000 \]where \( C \) stands for the total cost in dollars and \( x \) is the number of units.

• The term \(30x\) represents the variable cost, which increases with each additional unit. This component typically includes materials, labor, and other unit-specific expenses.
• The term \(250,000\) is the fixed cost, which remains constant regardless of production output. This involves costs like salaries of permanent staff, rent, and utilities.

By examining the cost function, businesses can determine the total expenses that need to be offset by revenue to achieve profitability. Grasping this relationship aids in cost management and decision-making about production levels.

Demand Equation

The demand equation captures the relationship between price and quantity. For this problem, the demand equation is:\[ p = 75 - 0.0005x \]where \( p \) represents the price per unit, and \( x \) is the number of units sold. In essence, it reflects the law of demand: as more units are sold, the price should decrease.

• The equation starts at a maximum price of \(75\) dollars per unit when no units are sold, illustrating the initial price point.
• The \(-0.0005x\) component signifies the price decrease per unit, showcasing a classical linear demand scenario.

Understanding the demand equation allows businesses to set optimal pricing strategies that align with production levels to maximize profits. It also helps predict how shifts in quantity can affect pricing and revenue.

Inequality Solving

Inequality solving is a mathematical technique used to find a range of values rather than a specific number. In this exercise, inequalities help determine how many units need to be sold—or at what price—to achieve a desired profit level.To achieve at least a \(\$ 750,000\) profit, the inequality used is:\[ x(75 - 0.0005x) - (30x + 250,000) \geq 750,000 \]This inequality ensures that the profit equation yields a value of at least 750,000 dollars. Solving this involves:

• Rearranging terms to isolate \( x \).
• Utilizing mathematical operations to simplify and solve for \( x \) values.

Inequality solving is vital since it provides solutions to real-world constraints or conditions that businesses face. By finding the unit levels that fit profitability constraints, a company can plan production and pricing effectively.