Suppose a company has fixed costs of $1800 and variable costs per unit of 

15

16

*x* + 1110 dollars,

 where *x* is the total number of units produced. Suppose further that the selling price of its product is 

1200 − 

1

16

*x* dollars per unit. Find the maximum profit.

To find the maximum profit, we need to calculate the profit function based on the given cost and revenue functions. Profit $ P(x) $ is the difference between revenue and total cost:

\[
P(x) = R(x) - C(x)
\]

where $ R(x) $ is the revenue and $ C(x) $ is the total cost.

### 1. Revenue Function

The selling price per unit is given by:

\[
\text{Price per unit} = 1200 - \frac{1}{16}x
\]

So, the total revenue $ R(x) $ is the product of the selling price per unit and the number of units sold:

\[
R(x) = x \times \left( 1200 - \frac{1}{16}x \right) = 1200x - \frac{1}{16}x^2
\]

### 2. Cost Function

The total cost $ C(x) $ is the sum of fixed costs and variable costs. The fixed cost is given as \$1800, and the variable cost per unit is given by:

\[
\text{Variable cost per unit} = \frac{15}{16}x + 1110
\]

The total variable cost is:

\[
\text{Total variable cost} = x \times \left( \frac{15}{16}x + 1110 \right) = \frac{15}{16}x^2 + 1110x
\]

So, the total cost function is:

\[
C(x) = 1800 + \frac{15}{16}x^2 + 1110x
\]

### 3. Profit Function

Now, we can compute the profit function by subtracting the total cost from the revenue:

\[
P(x) = \left( 1200x - \frac{1}{16}x^2 \right) - \left( 1800 + \frac{15}{16}x^2 + 1110x \right)
\]

Simplifying:

\[
P(x) = 1200x - \frac{1}{16}x^2 - 1800 - \frac{15}{16}x^2 - 1110x
\]

\[
P(x) = (1200x - 1110x) - \left( \frac{1}{16}x^2 + \frac{15}{16}x^2 \right) - 1800
\]

\[
P(x) = 90x - \frac{16}{16}x^2 - 1800
\]

\[
P(x) = 90x - x^2 - 1800
\]

### 4. Maximizing Profit

To find the maximum profit, we need to find the critical points by taking the derivative of $ P(x) $ and setting it equal to zero:

\[
P'(x) = 90 - 2x
\]

Set $ P'(x) = 0 $ to find the critical point:

\[
90 - 2x = 0
\]

\[
2x = 90
\]

\[
x = 45
\]

So, the number of units that maximizes profit is $ x = 45 $.

### 5. Maximum Profit

To find the maximum profit, substitute $ x = 45 $ back into the profit function $ P(x) $:

\[
P(45) = 90(45) - (45)^2 - 1800
\]

\[
P(45) = 4050 - 2025 - 1800
\]

\[
P(45) = 4050 - 3825
\]

\[
P(45) = 225
\]

Therefore, the maximum profit is **\$225**.

---

Do you want more details or have any questions?

Here are 5 related questions:
1. How would the profit change if the fixed cost increased by 10%?
2. What happens to the profit function if the variable cost per unit is reduced?
3. How does the price elasticity affect profit maximization?
4. What if the selling price was fixed at \$1200 for all units sold?
5. How can the company reduce costs to increase its maximum profit?

**Tip:** Maximizing profit often involves balancing marginal cost and marginal revenue.

Suppose a company has fixed costs of $1800 and variable costs per unit of 15/16*x + 1110 dollars, where *x* is the total number of units produced. Suppose further that the selling price of its product is 1200 − 1/16*x dollars per unit. Find the maximum profit.

Solve the problem of maximizing profit for a company with fixed costs of $1800 and variable costs based on the number of units produced. This example uses quadratic equations, revenue and cost functions, and differentiation to find the critical points and maximum profit.

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