Calculus Applications: Cost, Revenue, and Profit Functions

Price-Demand and Cost Functions in Calculus

Price-Demand Equation

The price-demand equation relates the price of a product to the quantity demanded by consumers. In this scenario, the equation is:

• 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:

To express price as a function of demand, solve for p:

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

• Domain: Since demand x cannot be negative and price must be non-negative, set $p \geq 0$ and $x \leq 10,000$. Thus, $0 \leq x \leq 10,000$.

Cost Function

The cost function gives the total cost of producing x units. Here, it is:

$C(x) = 7,000 + 2x$

• Fixed Cost: $7,000$ (cost incurred even if no units are produced)

• Variable Cost: $2x$ (cost that depends on the number of units produced)

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 headphone costs $2 to produce.

Revenue Function

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

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

• Domain: $0 \leq x \leq 10,000$ (as above)

Intersection Points of Cost and Revenue Functions

The break-even points occur where cost equals revenue:

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

Multiply both sides by 1,000 to clear denominators:

$1,000(7,000 + 2x) = 10,000x - x^2$ $7,000,000 + 2,000x = 10,000x - x^2$ $x^2 - 8,000x + 7,000,000 = 0$

This is a quadratic equation in x. The solutions give the production levels where the company breaks even (no profit or loss).

Profit Function

The profit function is the difference between revenue and cost:

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

• Domain: $0 \leq x \leq 10,000$

Marginal Analysis

Marginal analysis uses derivatives to estimate the effect of producing one more unit. The marginal profit at $x = 1,000$ is:

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

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 and selling the 1,001st unit is approximately $6.

Summary Table: Key Functions

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

Additional info: The graph in the image likely shows the cost and revenue functions, with their intersection points representing break-even quantities. The profit function is maximized where its derivative equals zero, i.e., where $P'(x) = 0$.