Problem 22 - Applied Calculus

Question

The demand equation for a quantity \(q\) of a product at price \(p,\) in dollars, is \(p=-5 q+4000 .\) Companies producing the product report the cost, \(C,\) in dollars, to produce a quantity \(q\) is \(C=6 q+5\) dollars. (a) Express a company's profit, in dollars, as a function of \(q\) (b) What production level earns the company the largest profit? (c) What is the largest profit possible?

Step-by-Step Solution

Answer

Maximal profit of $798,015 occurs at production level of 400 units.

Step 1: Express Revenue

First, we express the revenue function. The revenue, \( R \), from selling \( q \) units is the product of the price per unit, \( p \), and the quantity, \( q \). Thus, \( R = p \times q = (-5q + 4000)q = -5q^2 + 4000q \).

Step 2: Express Profit

Profit, \( P \), is the difference between revenue and cost. Given the revenue \( R = -5q^2 + 4000q \) and cost \( C = 6q + 5 \), the profit function is \( P(q) = R - C = (-5q^2 + 4000q) - (6q + 5) = -5q^2 + 3994q - 5 \).

Step 3: Find Maximum Profit using Calculus

To find the production level that maximizes profit, differentiate the profit function \( P(q) = -5q^2 + 3994q - 5 \) with respect to \( q \) and set the derivative equal to zero. The derivative is \( P'(q) = -10q + 3994 \). Setting \( P'(q) = 0 \) gives \(-10q + 3994 = 0 \), which solves to \( q = 399.4 \). Since \( q \) must be whole, consider \( q = 399 \) and \( q = 400 \).

Step 4: Calculate Profit at Critical Points

Substitute \( q = 399 \) and \( q = 400 \) into \( P(q) \) to find the actual profits. For \( q = 399 \), \( P(399) = -5(399)^2 + 3994(399) - 5 = 798008.5\).For \( q = 400 \), \( P(400) = -5(400)^2 + 3994(400) - 5 = 798015 \).

Step 5: Determine Largest Profit

Comparing the calculated profits, the profit at \( q = 400 \) is greater. Hence, the largest profit occurs at \( q = 400 \), resulting in a profit of \( 798015 \) dollars.

Key Concepts

Demand Equation

The demand equation is a crucial concept that businesses use to understand how much of a product consumers are willing to purchase at a given price. In our example, the demand equation is given by:
\[ p = -5q + 4000 \]
This equation tells us the relationship between the price \( p \) and the quantity demanded \( q \). Here, if you increase the quantity \( q \), the price \( p \) decreases, suggesting an inverse relationship typical of most demand equations.

• The slope of the demand curve is \(-5\), indicating the price drop for each additional unit sold.
• The intercept is 4000, reflecting the price if no units were sold.

Understanding the demand equation allows businesses to predict how changes in price can affect the quantity sold, which is fundamental for strategic planning and pricing.

Revenue Function

The revenue function represents the total income a company receives from selling its products. For any business, revenue is calculated by multiplying the price per unit by the quantity sold. With our demand equation \( p = -5q + 4000 \), the revenue function becomes:
\[ R(q) = p \times q = (-5q + 4000)q = -5q^2 + 4000q \]
This function is quadratic, indicating the revenue varies with the square of the quantity. Key points include:

• The function has a negative \( q^2 \) term, creating a parabolic shape that opens downwards. This suggests there is a peak revenue before it starts declining as quantity increases further.
• The importance of identifying the peak is that it helps in determining the optimal sales volume that maximizes revenue.

For businesses, the revenue function is essential in understanding potential earnings and setting strategies to enhance sales performance.

Cost Function

A cost function details the total cost a company incurs to produce a specific quantity of goods. Understanding costs is fundamental to assessing profitability. In this scenario, the cost function is:
\[ C(q) = 6q + 5 \]
This linear equation suggests that:

• The term \( 6q \) represents the variable cost that changes with production levels.
• The number 5 is the fixed cost, covering expenses that do not change with the production amount.

Understanding the cost function is key to analyzing how costs contribute to net profit or loss at different production levels, thereby influencing business decisions.

Calculus

Calculus is vital for finding maxima or minima in functions, which is crucial for optimizing business processes such as profit maximization. To determine the optimal production level for maximum profit, we use:
Differentiation, which involves finding the derivative of the profit function \( P(q) \), derived from the revenue and cost functions. The profit function given is:
\[ P(q) = -5q^2 + 3994q - 5 \]
To find the maximum profit point, we differentiate and solve for when the derivative \( P'(q) \) equals zero:
\[ P'(q) = -10q + 3994 = 0 \]
Solving this gives the critical point \( q = 399.4 \), rounded to the nearest whole number \( q = 400 \) for practical purposes.

• After finding the critical point, substitute it back into the profit function to confirm it yields the maximum profit.
• This application of calculus helps businesses optimize production levels effectively.