John's Lawn Mowing Service is a small business that acts as a price-taker (i.e., MR = P). The prevailing market price of lawn mowing is $20 per acre. John's costs are given by total cost = 0.1q^2 + 10q + 50, where q = the number of acres John chooses to cut a day. Calculate John's maximum daily profit.
user
Sure, I can help you with that.

In this case, as John's Lawn Mowing Service is a price-taker (i.e., MR = P), he has no control over the market price. Therefore, he cannot choose to charge any other price than the prevailing $20 per acre.

Now, let's identify the revenue function and the total cost function.

Revenue function:
For John, revenue (R) is equal to the number of acres he cuts (q) times the market price (P), which is $20 per acre. So, R = P * q = $20 * q = 20q.

Total cost function:
You've already given the total cost function as:

Total cost (TC) = 0.1q^2 + 10q + 50

Profit function:
Profit (π) is equal to the difference between total revenue (TR) and total cost (TC). So, π = TR - TC.

Substituting the revenue and cost functions, we get:

π = 20q - [0.1q^2 + 10q + 50]

Simplifying, we get:

π = 10q - 0.1q^2 - 50

This is a linear equation in slope-intercept form with a negative coefficient for the quadratic term. This means that it intersects the x-axis at two points, one at the x-intercept and one at infinity.

So, to find the x-intercept, make the coefficient of q^2 equal to zero.

0.1q^2 + 10q + 50 = 0

Factor the coefficient of q to get:

0.1q^2 + 10q + 50 = (0.1q + 50)(q + 2) = 0

Set each factor equal to zero and solve for q, using the Zero-Factor Theorem.

0.1q + 50 = 0 or q = -500

The solution is not realistic, as cutting 500 acres of lawn is not possible, so we don't consider this solution.

Also, you cannot set q equal to zero, as that would mean no output, hence zero revenue.

Therefore, the only remaining solution is the output level where marginal cost (MC) is equal to marginal revenue (MR), which gives the level of output where profit is maximized.

MC and MR are obtained by finding the first derivative of the cost and revenue functions respectively.

MC = dTC/dq = 0.1q + 10

MR = dR/dq = 20q

Setting them equal, we get:

0.1q + 10 = 20q

Solving for q, we get:

q = 50 acres

So, John's Lawn Mowing Service should aim to cut lawns across 50 acres each day for optimal short-term profits.

Substituting this into the profit function, we get:

π = 10(50) - 0.1(50)^2 - 50 = $150 - 2500 + 50 = -1350

$-1350 is the maximum profit John can make, excluding any expenses such as overhead, labor, fuel, etc. It's negative, meaning he's actually incurring a loss.

Therefore, your maximum daily profit is actually -$1350, which is not sustainable for a business. John might need to rethink his pricing strategies to survive.
assistant
John's Lawn Mowing Service is a small business that acts as a price-taker (i.e., MR = P). The prevailing market price of lawn mowing is $20 per acre. John's costs are given by total cost = 0.1q^2 + 10q + 50, where q = the number of acres John chooses to cut a day. Calculate John's maximum daily profit.
user
To find John's maximum daily profit, we need to find his profit functionfirst, then find the quantity that maximizes profit. The profit function is given by:

Profit = Revenue - Cost

Revenue is calculated by multiplying the price by the quantity (Revenue = P * q). Given that the market price is $20 per acre, the revenue function is:

Revenue = 20q

John's total cost function is given as:

Total Cost = 0.1q^2 + 10q + 50

Now we can write the profit function as:

Profit = Revenue - Total Cost
       = 20q - (0.1q^2 + 10q + 50)

Now let's simplify the profit function:

Profit = 20q - 0.1q^2 - 10q - 50
       = -0.1q^2 + 10q - 50

To find the quantity that maximizes profit, we need take the derivative of the profit function with respect to q and set it equal to zero:

d(Profit)/dq = -0.2q + 10

Now set the derivative equal to zero to find the optimal quantity:

-0.2q + 10 = 0

Add 0.2q to both sides:

0.2q = 10

Now, divide both sides by 0.2 to solve for q:

q = 10 / 0.2
q = 50 acres

Now that we have the optimal quantity, we can calculate the maximum daily profit by plugging this value of q back into the profit function:

Profit = -0.1(50)^2 + 10(50) - 50
       = -250 + 500 - 50
       = $200

John's maximum daily profit is $200 when he chooses to cut 50 acres per day.
assistant