Chapter 11: Problem 1

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

Expert verified

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.