To find the price that maximizes revenue, we use the demand equation:

$p = -\frac{1}{3}x + 100$

From part (c), the quantity $x$ that maximizes revenue is found to be $x = 150$.

Now, substituting $x = 150$ into the demand equation:

$p = -\frac{1}{3} (150) + 100$

$p = -50 + 100$

$p = 50$

Thus, the company should charge $50 to maximize revenue.

Would you like further clarification on any part? Here are some related questions:

1. How is the revenue function derived from the demand equation?
2. Why does revenue maximize at $x = 150$?
3. What would happen to revenue if the price were set higher or lower than $50?
4. Can we verify the maximum revenue using calculus (derivative method)?
5. How does this type of problem relate to real-world pricing strategies?

Tip: The maximum revenue occurs at the vertex of the quadratic revenue function, which can be found using the vertex formula $x = -\frac{b}{2a}$.