Problem 3 - Applied Calculus

Question

Revenue is given by \(R(q)=450 q\) and cost is given by \(C(q)=10,000+3 q^{2} .\) At what quantity is profit maximized? What is the total profit at this production level?

Step-by-Step Solution

Answer

Profit is maximized at a quantity of 75 with a total profit of 6,875.

Step 1: Define the Profit Function

Profit is calculated as the difference between Revenue and Cost. The profit function, \( P(q) \), is given by \( P(q) = R(q) - C(q) \). Using the given functions, we have: \( P(q) = 450q - (10,000 + 3q^2) = 450q - 10,000 - 3q^2 \).

Step 2: Find the Derivative of the Profit Function

To find the quantity at which profit is maximized, take the derivative of the profit function with respect to \( q \). This gives us: \( P'(q) = \frac{d}{dq}[450q - 10,000 - 3q^2] = 450 - 6q \).

Step 3: Set the Derivative to Zero

To find the critical points, set the derivative equal to zero and solve for \( q \). So, \( 450 - 6q = 0 \). Solving this equation: \[6q = 450\] \[q = \frac{450}{6} = 75\].

Step 4: Check the Second Derivative for Maximum

Find the second derivative to confirm that this critical point is a maximum. The second derivative is: \( P''(q) = \frac{d}{dq}[-6q] = -6 \). Since this is negative, it confirms a maximum point.

Step 5: Calculate the Total Profit at the Maximum Quantity

Substitute \( q = 75 \) back into the original profit function to find the total profit: \( P(75) = 450 \times 75 - 10,000 - 3 \times 75^2 \). Calculate each term: \( 450 \times 75 = 33,750 \), \( 3 \times 75^2 = 16,875 \). Thus, \( P(75) = 33,750 - 10,000 - 16,875 = 6,875 \).

Key Concepts

Revenue Function

To understand profit maximization, we first look at the Revenue Function. Revenue is the income a company earns from selling its goods. It's typically expressed as a function of the quantity of units sold, denoted by \( q \). In our problem, the revenue function is given by \( R(q) = 450q \). Here, \( 450 \) is the amount earned per unit sold.

• Revenue function calculates total earnings.
• The coefficient (450) is the price per unit.
• \( q \) represents the quantity sold.

The simplicity of this linear function allows us to easily determine how revenue changes as sales increase. Each additional unit sold adds another 450 to the revenue, indicating positive linear growth.

Cost Function

The Cost Function represents the total expense incurred in producing a certain quantity of products. In our exercise, this function is given by \( C(q) = 10,000 + 3q^2 \).

• The term \( 10,000 \) is a fixed cost, a constant expense regardless of the number of goods produced.
• The term \( 3q^2 \) is a variable cost, increasing quadratically with the quantity \( q \).
• This reflects economies, or diseconomies, of scale in production as \( q \) changes.

Fixed costs are unavoidable, like rent or salaries, while variable costs depend on production levels. Understanding these components helps businesses plan how to cover costs and maintain long-term profitability.

Derivative Calculation

In profit maximization, finding the derivative is crucial. The derivative of a function gives the rate at which it changes. For our profit function, \( P(q) = 450q - 10,000 - 3q^2 \), the derivative with respect to \( q \) is \( P'(q) = 450 - 6q \).

• The derivative represents the slope of the profit curve at any point \( q \).
• Setting \( P'(q) = 0 \) finds the critical points; where profit neither increases nor decreases sharply.
• This helps identify potentially maximum or minimum profit quantities.

By solving \( 450 - 6q = 0 \), we find \( q = 75 \). This step is vital for knowing the production quantity that influences profit levels significantly.

Second Derivative Test

After finding the critical quantity, we confirm it as a maximum point using the Second Derivative Test. The second derivative of a function demonstrates how the slope of the first derivative is changing. For the profit function's second derivative, we have \( P''(q) = -6 \).

• A negative second derivative (\( -6 \)) indicates a concave down curve.
• This confirms a maximum point, as the slope of the profit curve is decreasing beyond \( q = 75 \).
• This test is essential to ensure that the quantity calculated truly yields the highest profit.

Therefore, using the second derivative helps decisively determine the optimal production level for maximizing profit, giving businesses confidence in their strategic decisions.