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.