Suppose a competitive firm has as its total cost function:

TC=25+3q2TC=25+3q2


Suppose the firm's output can be sold (in integer units) at $54 per unit.

**Use calculus and formulas to find a solution (don't just build a table in a spreadsheet as in the previous lesson).**

**Hint 1:** The first derivative of the total profit function, which is cumulative, is the marginal profit function, which is incremental. The lecture and formula summary explain how to compute the derivative.

Set the marginal profit equal to zero to define an equation for the optimal quantity q.

**Hint 2:** When computing the total profit for a candidate quantity, use the total profit function you define (rather than summing the marginal profits using the marginal profit function).

How many integer units should the firm produce to maximize profit?

Please specify your answer as an integer. **In the case of equal profit from rounding up and down for a non-integer initial solution quantity, proceed with the higher quantity.**

**Answer (2 points)**

Save your answer

What is the total profit at the optimal integer output level?

Please specify your answer as an integer.

To solve this problem, we need to find the quantity $ q $ that maximizes the firm's profit.

### Step 1: Define the total cost and total revenue functions

The total cost (TC) function is given by:

\[
TC(q) = 25 + 3q^2
\]

The firm's output can be sold at $54 per unit. Therefore, the total revenue (TR) function is:

\[
TR(q) = 54q
\]

### Step 2: Define the total profit function

The total profit (π) is the difference between total revenue (TR) and total cost (TC):

\[
\pi(q) = TR(q) - TC(q) = 54q - (25 + 3q^2)
\]
\[
\pi(q) = 54q - 25 - 3q^2
\]

### Step 3: Find the derivative of the total profit function

To find the quantity $ q $ that maximizes profit, we need to compute the derivative of the total profit function (which is the marginal profit) and set it equal to zero:

\[
\frac{d\pi}{dq} = \frac{d}{dq} \left( 54q - 25 - 3q^2 \right)
\]
\[
\frac{d\pi}{dq} = 54 - 6q
\]

### Step 4: Set the marginal profit equal to zero

To find the critical points, set the derivative of the profit function equal to zero:

\[
54 - 6q = 0
\]
\[
6q = 54
\]
\[
q = 9
\]

So, the firm should produce 9 units to maximize profit.

### Step 5: Check the second derivative to confirm it's a maximum

The second derivative of the profit function is:

\[
\frac{d^2\pi}{dq^2} = -6
\]

Since the second derivative is negative, this confirms that $ q = 9 $ is a point of maximum profit.

### Step 6: Calculate the total profit at $ q = 9 $

Now, let's calculate the total profit at $ q = 9 $:

\[
\pi(9) = 54(9) - 25 - 3(9)^2
\]
\[
\pi(9) = 486 - 25 - 243
\]
\[
\pi(9) = 218
\]

Thus, the total profit at the optimal output level of 9 units is 218.

### Final Answer:

- The firm should produce **9 units** to maximize profit.
- The total profit at the optimal output level is **218**.

Would you like further details or have any questions?

---

Here are 5 related questions:
1. How does a change in the unit price affect the optimal quantity?
2. What happens to the optimal quantity if the total cost function changes?
3. Can marginal profit be used to evaluate non-integer output levels effectively?
4. How does the fixed cost component (like 25 in this case) affect profit maximization?
5. What are the economic implications of a negative second derivative in profit maximization?

**Tip:** The second derivative test is a great tool to confirm whether a critical point is a maximum or minimum.

Suppose a competitive firm has as its total cost function: TC = 25 + 3q^2. Suppose the firm's output can be sold (in integer units) at $54 per unit. Use calculus and formulas to find a solution. How many integer units should the firm produce to maximize profit? What is the total profit at the optimal integer output level?

Learn how to solve a profit maximization problem for a firm with the total cost function TC = 25 + 3q^2 and revenue based on output sold at $54 per unit. We use calculus, including differentiation, to find the optimal integer quantity (q = 9) for maximum profit, resulting in a total profit of $218.

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