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The given revenue function is:

\[
R(x) = x \cdot p = 1764x - 0.12x^3
\]

To solve parts A and B, we'll need to calculate the first and second derivatives of the revenue function, as these are used to determine local minima and concavity.

### Step 1: First derivative (R'(x))

The first derivative helps to find the critical points:

\[
R'(x) = \frac{d}{dx} \left(1764x - 0.12x^3 \right) = 1764 - 3(0.12)x^2
\]

Simplify:

\[
R'(x) = 1764 - 0.36x^2
\]

Set the first derivative equal to zero to find the critical points:

\[
1764 - 0.36x^2 = 0
\]

Solving for \(x\):

\[
0.36x^2 = 1764 \quad \Rightarrow \quad x^2 = \frac{1764}{0.36} = 4900
\]
\[
x = \sqrt{4900} = 70
\]

So, \(x = 70\) is a critical point. This is where the function may have a local minimum or maximum.

### Step 2: Second derivative (R''(x))

Now, we calculate the second derivative to find where the graph is concave up or concave down:

\[
R''(x) = \frac{d}{dx}(1764 - 0.36x^2) = -0.72x
\]

To find concavity, we analyze the sign of \(R''(x)\):
- If \(R''(x) > 0\), the graph is concave upward.
- If \(R''(x) < 0\), the graph is concave downward.

At \(x = 70\), \(R''(70) = -0.72(70) = -50.4\), which is negative, indicating concave downward at \(x = 70\).

To determine when the graph is concave upward, solve \(R''(x) > 0\):

\[
-0.72x > 0 \quad \Rightarrow \quad x < 0
\]

However, since \(x\) must be between 0 and 95 (given the problem's constraints), there is **no interval** where the graph is concave upward.

### Conclusion:
- **Part A**: The local minimum occurs at \(x = 70\).
- **Part B**: The graph is **never concave upward**.

Would you like further clarification on any of these steps?

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Here are five related questions:
1. How do you confirm if \(x = 70\) is a local minimum or maximum?
2. What is the significance of the second derivative in determining concavity?
3. Can a revenue function like this have multiple critical points?
4. How would the solution change if the revenue function had a different cubic coefficient?
5. Why is concavity important in understanding revenue models?

**Tip:** Always check the signs of the first and second derivatives to understand the behavior of functions, especially when dealing with maxima, minima, and concavity.

The marketing research department of a computer company used a large city to test market the firm's new laptop. The department found the relationship between price p (dollars per unit) and the demand x (units per week) was given by p = 1764 - 0.12x^2, for 0 < x < 95. Using this equation, the weekly revenue function becomes R(x) = 1764x - 0.12x^3. Find the local minimum and determine the interval(s) where the revenue function is concave upward.

Learn how to find the local minimum and analyze the concavity of the revenue function R(x) = 1764x - 0.12x^3. This solution uses first and second derivatives to determine critical points and intervals of concavity. Suitable for high school calculus students.

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