Suppose a company has fixed costs of $49,400 and a variable cost per unit of 1/3x + 333 dollars, where x is the total number of units produced. Suppose further that the selling price of its product is 2259- 2/3x dollars per unit. Find the maximum revenue.

To solve this problem and find the maximum revenue, we need to model the revenue function and maximize it.

### Step 1: Define the Revenue Function
Revenue is defined as the selling price per unit multiplied by the number of units sold. The selling price per unit is given as:

\[
P(x) = 2259 - \frac{2}{3}x
\]

So, the revenue function $R(x)$ in terms of $x$ (the number of units sold) is:

\[
R(x) = x \cdot P(x) = x \cdot \left(2259 - \frac{2}{3}x\right)
\]

This simplifies to:

\[
R(x) = 2259x - \frac{2}{3}x^2
\]

### Step 2: Maximize the Revenue Function
To find the value of $x$ that maximizes the revenue, we need to take the derivative of the revenue function $R(x)$, set it equal to zero, and solve for $x$.

1. Differentiate $R(x)$:

\[
\frac{dR}{dx} = 2259 - \frac{4}{3}x
\]

2. Set the derivative equal to zero and solve for $x$:

\[
2259 - \frac{4}{3}x = 0
\]

\[
\frac{4}{3}x = 2259
\]

\[
x = \frac{2259 \times 3}{4} = 1694.25
\]

### Step 3: Verify It's a Maximum
To confirm that this value of $x = 1694.25$ is a maximum, we check the second derivative of $R(x)$:

\[
\frac{d^2R}{dx^2} = -\frac{4}{3}
\]

Since the second derivative is negative, the revenue function is concave down, indicating that $x = 1694.25$ is indeed a maximum.

### Step 4: Calculate the Maximum Revenue
Now that we have the value of $x$, we substitute it back into the revenue function $R(x)$ to find the maximum revenue:

\[
R(1694.25) = 2259(1694.25) - \frac{2}{3}(1694.25)^2
\]

First, calculate each part:
\[
2259 \times 1694.25 = 3,826,920.75
\]
\[
\frac{2}{3} \times (1694.25)^2 = \frac{2}{3} \times 2860701.06 = 1,907,134.04
\]

Now subtract:

\[
R(1694.25) = 3,826,920.75 - 1,907,134.04 = 1,919,786.71
\]

Thus, the maximum revenue is approximately **$1,919,786.71**.

---

Do you want more details on any step? Here are 5 questions to deepen your understanding:

1. How does the second derivative test confirm a maximum or minimum?
2. Why do we set the derivative equal to zero to find the maximum point?
3. How would the result change if the selling price had a different form?
4. Can you explain why revenue is modeled by multiplying price and quantity?
5. What would the implications be if the cost per unit were higher?

**Tip**: When dealing with optimization problems, always check the second derivative to confirm whether you have found a maximum or minimum point.

Suppose a company has fixed costs of $49,400 and a variable cost per unit of 1/3x + 333 dollars, where x is the total number of units produced. Suppose further that the selling price of its product is 2259 - 2/3x dollars per unit. Find the maximum revenue.

Explore how to maximize the revenue of a company with fixed costs and a quadratic revenue function. This detailed solution involves the use of derivatives and the second derivative test to find the maximum revenue. Suitable for Grades 11-12 students learning optimization techniques.

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