To find the maximum revenue, we need to express the revenue $R(p)$ as a function of the price $p$.

Revenue is calculated as:

$R(p) = p \times D(p)$

Given that the demand function is $D(p) = 40 - p$, the revenue function becomes:

$R(p) = p \times (40 - p)$

Simplifying this:

$R(p) = 40p - p^2$

Now, we need to find the value of $p$ that maximizes the revenue. To do this, we take the derivative of $R(p)$ with respect to $p$, and set it equal to zero to find the critical points.

1. Differentiate $R(p)$:

$R'(p) = 40 - 2p$

2. Set $R'(p) = 0$ to find the critical point:

$40 - 2p = 0$

$2p = 40$

$p = 20$

Now that we have $p = 20$, we check if this value gives a maximum by using the second derivative test.

3. The second derivative of $R(p)$ is:

$R''(p) = -2$

Since $R''(p) = -2$ is negative, this confirms that $p = 20$ gives a maximum.

4. To find the maximum revenue, substitute $p = 20$ back into the revenue function:

$R(20) = 20 \times (40 - 20) = 20 \times 20 = 400$

Thus, the maximum revenue is 400 units when the price is $p = 20$.

Would you like further details or have any questions?

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Here are 5 related questions to consider:

1. What happens to revenue if the price is set below or above the optimal price?
2. How would changes in demand (e.g., different demand functions) affect the maximum revenue?
3. What is the impact of fixed costs on the overall profit given this revenue function?
4. Can we determine the elasticity of demand from this function, and how would it influence pricing strategy?
5. How does the maximum revenue change if the company offers discounts?

Tip: The derivative is a powerful tool in economics and business to optimize functions like revenue, cost, and profit. Understanding how to apply it helps make informed decisions.