To maximize revenue, we first need to express the revenue function in terms of $x$.

Step 1: Define the Revenue Function

Revenue $R$ is given by:

$R = p \cdot x$

Substituting $p = -\frac{1}{3}x^2 + 100$:

$R(x) = x \left( -\frac{1}{3}x^2 + 100 \right)$

$R(x) = -\frac{1}{3}x^3 + 100x$

Step 2: Find the Critical Points

To find the maximum revenue, take the derivative of $R(x)$ and set it equal to zero.

$R'(x) = -\frac{3}{3}x^2 + 100$

$R'(x) = -x^2 + 100$

Setting $R'(x) = 0$:

$-x^2 + 100 = 0$

$x^2 = 100$

$x = 10 \quad \text{or} \quad x = -10$

Since $x$ must be in the range $0 \leq x \leq 30$, we take $x = 10$.

Step 3: Find the Price

Substituting $x = 10$ into the demand equation:

$p = -\frac{1}{3}(10)^2 + 100$

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

$p = 100 - 33.33$

$p = 66.67$

Conclusion:

The company should charge $66.67 to maximize revenue.

Would you like a step-by-step breakdown of any part? Here are some related questions:

1. How do we confirm that this is a maximum using the second derivative test?
2. What is the maximum revenue at this price?
3. How does price elasticity affect revenue in this case?
4. How would the revenue change if the demand function were different?
5. What happens to revenue when the price is set too high or too low?

Tip: The revenue function is always maximized where marginal revenue = 0, which comes from differentiating the revenue equation! 🚀