Calculus Applications: Cost, Revenue, and Profit Functions

Price-Demand Equation and Its Analysis

Understanding the Price-Demand Relationship

The price-demand equation models the relationship between the price of a product and the quantity demanded by consumers. In this scenario, the equation is given by:

• Equation: $x = 10,000 - 1,000p$

• Where x is the number of headphones demanded, and p is the price per set in dollars.

Solving for price as a function of demand:

• Rearrange the equation to solve for p:

$x = 10,000 - 1,000p$ $1,000p = 10,000 - x$ $p = \frac{10,000 - x}{1,000}$

• Domain: The domain is the set of all possible values of x for which the price is non-negative and demand is realistic.

• Since $p \geq 0$, $10,000 - x \geq 0 \implies x \leq 10,000$

• Also, $x \geq 0$ (cannot have negative demand).

• Domain: $0 \leq x \leq 10,000$

Cost Function and Marginal Cost

Understanding the Cost Function

The cost function represents the total cost of producing x units. Here, it is given by:

• Cost Function: $C(x) = 7,000 + 2x$

• Where $7,000$ is the fixed cost and $2x$ is the variable cost per unit.

Marginal Cost Function:

• The marginal cost is the derivative of the cost function with respect to x:

$C'(x) = \frac{d}{dx}[7,000 + 2x] = 2$

• Interpretation: The marginal cost is $2, meaning each additional unit produced increases the total cost by $2.

Revenue Function

Defining and Finding the Revenue Function

The revenue function calculates the total income from selling x units at price p:

• Revenue Function: $R(x) = x \cdot p$

• Substitute $p$ from the price-demand equation:

$R(x) = x \cdot \left(\frac{10,000 - x}{1,000}\right) = \frac{10,000x - x^2}{1,000}$

• Domain: $0 \leq x \leq 10,000$ (same as the demand domain)

Intersection of Cost and Revenue Functions

Break-Even Analysis

The intersection points of the cost and revenue functions represent the break-even points, where total revenue equals total cost (no profit or loss).

• Set $C(x) = R(x)$:

$7,000 + 2x = \frac{10,000x - x^2}{1,000}$

• Multiply both sides by 1,000 to clear denominators:

$7,000,000 + 2,000x = 10,000x - x^2$ $7,000,000 + 2,000x - 10,000x + x^2 = 0$ $x^2 - 8,000x + 7,000,000 = 0$

• This is a quadratic equation in x. The solutions give the break-even quantities.

• Interpretation: At these values of x, the business neither makes a profit nor a loss.

Profit Function

Defining and Analyzing the Profit Function

The profit function is the difference between revenue and cost:

• Profit Function: $P(x) = R(x) - C(x)$

• Substitute the given functions:

$P(x) = \frac{10,000x - x^2}{1,000} - (7,000 + 2x)$ $= \frac{10,000x - x^2 - 7,000,000 - 2,000x}{1,000}$ $= \frac{8,000x - x^2 - 7,000,000}{1,000}$

• Domain: $0 \leq x \leq 10,000$

Marginal Analysis

Estimating Profit Using Marginal Analysis

Marginal analysis uses derivatives to estimate the change in profit for producing one additional unit. The marginal profit at $x = 1,000$ is:

• First, find the derivative of the profit function:

$P(x) = \frac{8,000x - x^2 - 7,000,000}{1,000}$ $P'(x) = \frac{8,000 - 2x}{1,000}$

• At $x = 1,000$:

$P'(1,000) = \frac{8,000 - 2,000}{1,000} = \frac{6,000}{1,000} = 6$

• Interpretation: The estimated profit from producing the 1,001st unit is approximately $6.

Summary Table: Key Functions

Function	Formula	Domain	Interpretation
Price-Demand	$p = \frac{10,000 - x}{1,000}$	$0 \leq x \leq 10,000$	Price as a function of demand
Cost	$C(x) = 7,000 + 2x$	All $x \geq 0$	Total production cost
Revenue	$R(x) = \frac{10,000x - x^2}{1,000}$	$0 \leq x \leq 10,000$	Total sales revenue
Profit	$P(x) = \frac{8,000x - x^2 - 7,000,000}{1,000}$	$0 \leq x \leq 10,000$	Net profit
Marginal Cost	$C'(x) = 2$	All $x$	Cost of producing one more unit
Marginal Profit	$P'(x) = \frac{8,000 - 2x}{1,000}$	$0 \leq x \leq 10,000$	Profit from one additional unit

Example Application

• Suppose the company wants to know the price at which 5,000 units will be sold:

$p = \frac{10,000 - 5,000}{1,000} = \frac{5,000}{1,000} = 5$ So, the price per set should be $5 to sell 5,000 units.

• To find the break-even points, solve $x^2 - 8,000x + 7,000,000 = 0$:

$x = \frac{8,000 \pm \sqrt{(8,000)^2 - 4 \cdot 1 \cdot 7,000,000}}{2}$ $x = \frac{8,000 \pm \sqrt{64,000,000 - 28,000,000}}{2}$ $x = \frac{8,000 \pm \sqrt{36,000,000}}{2}$ $x = \frac{8,000 \pm 6,000}{2}$ $x = 7,000$ or $x = 1,000$

• Interpretation: The company breaks even at 1,000 and 7,000 units sold.