Problem 8 - Calculus: Early Transcendentals

Question

Marketing tells you that if you set the price of an item at $$\$ 10$$ then you will be unable to sell it, but that you can sell 500 items for each dollar below $$\$10$$ that you set the price. Suppose your fixed costs total $$\$ 3000,$$ and your marginal cost is $$\$ 2$$$ per item. What is the most profit you can make?

Step-by-Step Solution

Answer

The maximum profit is \$5000.

Step 1: Determine the Price Function

Mark the price of each item as \( P \) dollars. According to the problem, at \( P = 10 \) dollars, the quantity sold \( Q = 0 \). The number of items sold increases by 500 for each dollar decrease in price.Let \( x \) be the amount of reduction from \$10. Then \( P = 10 - x \).Since for each dollar below 10, 500 more items are sold, the quantity can be expressed as \( Q = 500x \).

Step 2: Express Revenue as a Function of x

Revenue is calculated by multiplying the number of items sold \( Q \) by the price at which each item is sold \( P \).The revenue function \( R(x) \) is:\[R(x) = Q \times P = (500x) \times (10 - x) = 5000x - 500x^2\]

Step 3: Determine the Cost Function

Total cost \( C(x) \) comprises the fixed costs and variable costs (marginal cost multiplied by the quantity sold).Given a fixed cost of \\(3000 and a marginal cost of \\)2 per item:\[C(x) = 3000 + 2Q = 3000 + 2(500x) = 3000 + 1000x\]

Step 4: Determine the Profit Function

Profit \( \Pi(x) \) is calculated by subtracting costs from revenue.The profit function \( \Pi(x) \) is:\[\Pi(x) = R(x) - C(x) = (5000x - 500x^2) - (3000 + 1000x) = -500x^2 + 4000x - 3000\]

Step 5: Find the Value of x that Maximizes Profit

To maximize profit, take the derivative of the profit function \( \Pi(x) \) with respect to \( x \), and set it equal to zero to find the critical points.Compute \( \frac{d\Pi}{dx} = -1000x + 4000 \).Set the derivative to zero:\[-1000x + 4000 = 0 \1000x = 4000 \x = 4\]

Step 6: Calculate Maximum Profit

Substitute \( x = 4 \) back into the profit function to get the maximum profit.\[\Pi(4) = -500(4)^2 + 4000(4) - 3000 \= -500(16) + 16000 - 3000 \= -8000 + 16000 - 3000 \= 5000\]

Key Concepts

Revenue Function

Revenue is the amount of money a company brings in from selling its goods or services. In this exercise, revenue is computed by multiplying the number of items sold by the price at which they are sold.However, both the quantity and the price are functions of a variable, making it important to construct what's called a revenue function.In this particular example, the revenue function, denoted as \( R(x) \), is derived based on an item's price decrease.For every dollar the price reduces, an additional 500 items can be sold.
Thus, the formula becomes:

• Items sold, \( Q = 500x \): where \( x \) is the number of dollars below \( \$10 \)
• Revenue function, \( R(x) = 500x \times (10 - x) = 5000x - 500x^2 \)

The expression \( 5000x - 500x^2 \) captures how revenue changes as the price decreases.

Cost Function

Costs involve everything that a company spends to produce and sell its products.In this exercise, the total cost relies on two parts: fixed costs and variable costs.Fixed costs remain the same regardless of the number of items sold, while variable costs increase as more items are sold.

The given fixed cost is \( \\(3000 \), and the variable cost, or marginal cost, is \( \\)2 \) per item.To calculate the cost function \( C(x) \):

• Fixed cost is constant at \( \$3000 \).
• Variable cost is \( 2 \times Q = 2 \times 500x = 1000x \).

This results in the cost function: \( C(x) = 3000 + 1000x \). The cost function helps determine how much money the company will need to reach different levels of production.

Price Function

The price function explains how the selling price of an item changes in response to changes in other variables, such as quantity sold.In the context of this exercise, it's crucial to determine a price function to establish a relationship between the price and the quantity sold.

According to the problem statement, at a price of \( \\(10 \), no items are sold. For each dollar reduction in price, sales increase by 500 units.This can be captured by:

• Letting \( x \) be the dollar reduction from \( \\)10 \).
• Thus, the price function becomes \( P = 10 - x \).
• The corresponding quantity sold is \( Q = 500x \).

The price function \( P = 10 - x \) helps form the basis for constructing the revenue function.

Derivative

A derivative is a calculus concept that measures how a function changes as its input changes. When applied to business problems, derivatives help understand how different variables affect outcomes, like profit.To maximize or minimize any function, derivatives play a pivotal role.

To determine the maximum profit, one must find the derivative of the profit function \( \Pi(x) \) and set it to zero to locate critical points:

• The profit function derived was \( \Pi(x) = -500x^2 + 4000x - 3000 \).
• The derivative of the profit function is \( \frac{d\Pi}{dx} = -1000x + 4000 \).
• Setting the derivative to zero gives us: \(-1000x + 4000 = 0 \).

Solving this equation helps in identifying the price point that might maximize profits.

Critical Points

Critical points are values of variables that provide opportunities to find either maxima or minima of a function.In the context of profit maximization, identifying these points is essential to know the price or quantity level that optimizes profit outcomes.

After finding the derivative of the profit function in this problem, setting it to zero helps find the critical points:

• Solve \(-1000x + 4000 = 0\).
• This simplifies to \( x = 4 \).

At \( x = 4 \), substituting back into the profit function reveals: \( \Pi(4) = -8000 + 16000 - 3000 = 5000 \).With \( x = 4 \), the company maximizes its profit at \( \$5000 \), showing a prominent use of calculus in business problems.