The revenue $R(x)$ for a company is calculated as the selling price per unit multiplied by the number of units sold. The problem gives the selling price per unit as:

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

Thus, the total revenue function is:

$R(x) = x \times \left(1200 - \frac{1}{16}x\right)$

Expanding this, we get:

$R(x) = 1200x - \frac{1}{16}x^2$

Now, to find the maximum revenue, we need to find the critical points by taking the derivative of $R(x)$ and setting it equal to zero.

Step 1: Differentiate the revenue function

$R'(x) = \frac{d}{dx} \left(1200x - \frac{1}{16}x^2\right)$

$R'(x) = 1200 - \frac{1}{8}x$

Step 2: Set $R'(x) = 0$ to find the critical points

$1200 - \frac{1}{8}x = 0$

Solving for $x$:

$\frac{1}{8}x = 1200$

$x = 1200 \times 8 = 9600$

Step 3: Verify that this critical point is a maximum by using the second derivative test

The second derivative of $R(x)$ is:

$R''(x) = \frac{d}{dx} \left(1200 - \frac{1}{8}x\right) = -\frac{1}{8}$

Since $R''(x) = -\frac{1}{8}$ is negative, the function is concave down at $x = 9600$, indicating that this is a maximum point.

Step 4: Calculate the maximum revenue

Substitute $x = 9600$ into the revenue function:

$R(9600) = 1200(9600) - \frac{1}{16}(9600^2)$

Now calculating:

$R(9600) = 1200 \times 9600 - \frac{1}{16} \times (9600)^2$

$R(9600) = 11520000 - \frac{1}{16} \times 92160000$

$R(9600) = 11520000 - 5760000$

$R(9600) = 5760000$

Thus, the maximum revenue is $5,760,000.

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Would you like further details or have any questions?

Related questions:

1. How do we calculate the total profit instead of just revenue?
2. What would be the effect on revenue if the selling price changed?
3. How does variable cost impact the maximum revenue?
4. Can we find the break-even point using this information?
5. What is the role of fixed costs in determining the overall profitability?

Tip: In problems involving revenue, profit, or cost functions, always identify key points (maximums/minimums) by differentiating and solving for critical points.