91Ó°ÊÓ

The **Revenue Function** is crucial when dealing with demand and pricing. It helps us understand how changes in price affect total revenue. Generally, revenue \(R\(p\)\) is calculated as the product of price \(p\) and quantity demanded \(q\). In this exercise, the demand function is \[ q = 500 - 2p \], and the revenue function is thus: \[ R(p) = p(500 - 2p) = 500p - 2p^2 \]. This quadratic equation will help us analyze revenue behavior as price changes. \[ \[\[\begin{align*} \text{When } p = 0, & \ R(0) = 0 \ \text{When } p = 250, & \ R(250) = 0 \end{align*}\]\] \] For prices between 0 and 250, the revenue varies, and we'll delve deeper using derivatives to understand this variation better.

To find the **Critical Points** of the revenue function, we need to take its derivative and set it to zero. The derivative tells us the rate of change of revenue with respect to price. For \(R(p) = 500p - 2p^2\), differentiate it with respect to \(p\) to get: \[ \frac{dR}{dp} = 500 - 4p \] Setting the derivative to zero to find critical points: \[ 500 - 4p = 0 \] \[ 4p = 500 \] \[ p = 125 \] This means at **p = 125**, a critical point is found. At this price, the revenue function has a critical point, meaning it could be a maximum or minimum. Since it's a quadratic function opening downwards, \(p = 125\) is where the revenue is maximized. Beyond these calculations, examining the sign of the derivative helps in confirming that this point provides the maximum revenue.

The **Derivative Analysis** is key to understanding the behavior of the revenue function over different price intervals. By differentiating \[ R(p) = 500p - 2p^2 \], we found \[ \frac{dR}{dp} = 500 - 4p \]. We obtained a critical point at \(p = 125 \).
Now let's interpret it further:

• For \: 0 < p < 125 \, \(\frac {dR}{dp}\) > 0 (Revenue is increasing)
• For \: p > 125 \, \(\frac {dR}{dp}\) < 0 (Revenue is decreasing)

Therefore, **Revenue increases** when \(\0 < p < 125\,\) and then **Revenue decreases** beyond \(\p > 125\).
This tells us that the revenue function reaches its peak, or in mathematical terms, its maximum at the critical point **p = 125**. Understanding these intervals helps businesses strategically price their products to maximize revenue.