Question

John's Lawn Mowing Service is a small business that acts as a price-taker (i.e., \(M R=P\) ). The prevailing market price of lawn mowing is \(\$ 20\) per acre. John's costs are given by total cost \(=0.1 q^{2}+10 q+50\) where \(q=\) the number of acres John chooses to cut a day. a. How many acres should John choose to cut to maximize profit? b. Calculate John's maximum daily profit. c. Graph these results, and label John's supply curve.

Short answer

Based on the given information: 1. John's revenue function is \(R(q) = 20q\). 2. His profit function is \(\pi(q) = 20q - (0.1q^2 + 10q + 50)\). 3. The first derivative of the profit function with respect to \(q\) is \(\frac{d\pi}{dq} = 10 - 0.2q\). 4. By setting the first derivative to zero, we find that John should choose to cut 50 acres per day to maximize his profit. 5. John's maximum daily profit is \(\$200\). 6. To graph the results, plot the revenue, cost, and profit functions, as well as John's supply curve derived from the marginal cost curve. The intersection points and maximum points help determine the break-even point and optimal production choice. In conclusion, John should cut 50 acres per day to maximize his daily profit of \$200. To represent this graphically, plot the revenue, cost, and profit functions along with the derived supply curve from the marginal cost curve.

Step-by-step solution

Determine the Revenue Function

First, we will determine John's revenue function. Since he is a price taker, the prevailing market price is \(P = \$20\). His revenue function is the price per acre multiplied by the number of acres he chooses to cut: \(R(q) = P \cdot q = 20q\)

Determine the Profit Function

Next, we will write down John's profit function which is the difference between his revenue and total cost. We are given the total cost function as \(C(q) = 0.1q^2 + 10q + 50\). So, the profit function will be: \(\pi(q) = R(q) - C(q) = 20q - (0.1q^2 + 10q + 50)\)

Find the First Derivative of the Profit Function

Now we need to find the first derivative of the profit function with respect to \(q\) to determine the optimal number of acres: \(\frac{d\pi}{dq} = \frac{d(20q - 0.1q^2 - 10q - 50)}{dq}\) Using calculus: \(\frac{d\pi}{dq} = 20 - 0.2q - 10\) Simplify further: \(\frac{d\pi}{dq} = 10 - 0.2q\)

Set the First Derivative to Zero and Solve for q

To find the optimal number of acres to maximize profit, set the first derivative to zero: \(10 - 0.2q = 0\) Now, solve for \(q\): \(q = \frac{10}{0.2} = 50\) Therefore, John should choose to cut 50 acres per day to maximize his profit.

Calculate John's Maximum Daily Profit

Now that we know the optimal number of acres, we will use it to calculate the maximum daily profit. Substitute the value of \(q\) in the profit function: \(\pi(50) = 20(50) - (0.1(50^2) + 10(50) + 50)\) \(\pi(50) = 1000 - (250 + 500 + 50)\) \(\pi(50) = 1000 - 800\) \(\pi(50) = 200\) Thus, John's maximum daily profit is \(\$200\).

Graphing the Results and John's Supply Curve

First, plot the revenue function \(R = 20q\) and the cost function \(C(q) = 0.1q^2 + 10q + 50\) on the same graph. The intersection point between these functions corresponds to the break-even point. Next, plot the profit function \(\pi(q) = 20q - (0.1q^2 + 10q + 50)\), its maximum point represents John's optimal production choice, which is 50 acres. To represent John's supply curve, plot the portion of the marginal cost curve, \(\frac{dC}{dq} = 0.2q + 10\), that lies above the minimum average variable cost. The supply curve is the portion of the marginal cost curve that producers are willing to supply at that price.

Key concepts

Price Taker

In economics, a price taker is an individual or company that has no control over the market price of its product. A price taker must accept the market price that is determined by the intersection of global supply and demand. For example, John's Lawn Mowing Service, as a small business, is a price taker in the lawn mowing market. This means that John's business cannot influence the pricing and must accept the prevailing rate of $20 per acre.
The characteristics of a price taker include:

• Small market share relative to the market size.
• Little influence on price setting.
• Typically occurs in perfectly competitive markets.

John must adjust his operations and focus on maximizing profit given the existing market price, which is defined as a crucial aspect of his business strategy.

Revenue Function

The revenue function represents the total income generated from selling goods or services at a given price. For John, the revenue can be calculated by using the formula: \( R(q) = P \cdot q \), where \( P \) is the price per acre and \( q \) is the number of acres mowed. Here, since John is a price taker, the price is fixed at $20 per acre, so his revenue function is:

• \( R(q) = 20q \)

This straightforward calculation allows John to easily determine his income based on the number of acres cut.
Key aspects of the revenue function in price-taking scenarios include:

• Linear relationship between quantity and revenue.
• Direct fluctuation with changes in quantity.
• The foundation for analyzing profit potential.

This understanding helps John to appreciate how quantity affects his earnings.

Profit Function

The profit function captures the difference between total revenue and total costs, shedding light on the actual earnings after expenses. For John's Lawn Mowing Service, the profit function can be expressed as:\
\[ \pi(q) = R(q) - C(q) \]
where \( R(q) = 20q \) and \( C(q) = 0.1q^2 + 10q + 50 \). Thus, the profit function becomes:

• \( \pi(q) = 20q - (0.1q^2 + 10q + 50) \)

This equation helps John to assess his operations strategically.
Key considerations regarding the profit function include:

• Differentiating to find maximum profit.
• Understanding intersections with cost for break-even analysis.
• Clarifying constraints related to production capacity.

By optimizing the number of acres mowed, John can achieve the maximum profit of $200, following a calculated strategy.

Marginal Cost

Marginal cost represents the additional cost of producing one more unit of output, crucial for decision-making in maximizing profit. For John, the marginal cost (MC) can be derived from the total cost function \( C(q) = 0.1q^2 + 10q + 50 \). By taking the derivative of the cost function, the marginal cost is:

• \( \frac{dC}{dq} = 0.2q + 10 \)

This represents the increase in cost per additional acre mowed.
Marginal cost plays a vital role in:

• Determining the supply curve, essential in competitive markets.
• Informing decisions about expanding or reducing output.
• Aligning production with market price to ensure profit maximization.

For John, understanding marginal cost aids in aligning production levels with profitability goals, ensuring efficient and economically sound operations.