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Let's go through the given problem step by step:

The revenue $ R(p) $ as a function of the unit price $ p $ is given by:

\[
R(p) = -5p^2 + 15,000p
\]

### (a) At what prices $ p $ is revenue zero?

To find the prices where the revenue is zero, we set $ R(p) = 0 $ and solve for $ p $:

\[
0 = -5p^2 + 15,000p
\]

Factor out $ p $:

\[
p(-5p + 15,000) = 0
\]

This gives two solutions:

1. $ p = 0 $
2. $ -5p + 15,000 = 0 $

Solving the second equation:

\[
-5p = -15,000
\]
\[
p = 3,000
\]

So, the revenue is zero at $ p = 0 $ and $ p = 3,000 $.

### (b) For what range of prices will revenue exceed $1,000,000?

We need to find the values of $ p $ such that:

\[
R(p) > 1,000,000
\]

Substitute the revenue equation:

\[
-5p^2 + 15,000p > 1,000,000
\]

Rearrange the inequality:

\[
-5p^2 + 15,000p - 1,000,000 > 0
\]

Now solve this quadratic inequality. First, solve the equality to find the boundary points:

\[
-5p^2 + 15,000p - 1,000,000 = 0
\]

Divide through by -5 to simplify:

\[
p^2 - 3,000p + 200,000 = 0
\]

Now, use the quadratic formula:

\[
p = \frac{-(-3,000) \pm \sqrt{(-3,000)^2 - 4(1)(200,000)}}{2(1)}
\]
\[
p = \frac{3,000 \pm \sqrt{9,000,000 - 800,000}}{2}
\]
\[
p = \frac{3,000 \pm \sqrt{8,200,000}}{2}
\]
\[
p = \frac{3,000 \pm 2,863.56}{2}
\]

Thus, we have:

1. $ p = \frac{3,000 + 2,863.56}{2} = 2,931.78 $
2. $ p = \frac{3,000 - 2,863.56}{2} = 68.22 $

The revenue will exceed $1,000,000 when $ 68.22 < p < 2,931.78 $.

### Summary:
- (a) The revenue is zero at $ p = 0 $ and $ p = 3,000 $.
- (b) The revenue exceeds $1,000,000 for prices in the range $ 68.22 < p < 2,931.78 $.

Would you like more details on the solution, or do you have any questions?

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Here are five related questions to expand your understanding:
1. How would the revenue function change if the quadratic term had a different coefficient?
2. What does the shape of the revenue function represent in terms of the business's pricing strategy?
3. Can the maximum revenue be found from this function, and how would you calculate it?
4. What happens to the revenue as the price $ p $ becomes very large?
5. How does the discriminant in the quadratic formula help determine the range of prices where revenue exceeds a certain value?

**Tip:** When solving inequalities with quadratic functions, finding the roots helps establish the boundary points. Always check the direction of the inequality to determine the correct range of solutions.

Suppose that the manufacturer of a gas clothes dryer has found that when the unit price is p dollars, the revenue R (in dollars) is R(p) = -5p^2 + 15,000p. (a) At what prices p is revenue zero? (b) For what range of prices will revenue exceed $1,000,000?

Find the prices at which revenue for a gas clothes dryer manufacturer is zero and the range of prices where revenue exceeds $1,000,000. This problem explores algebraic concepts such as quadratic equations and revenue functions, suitable for Grades 9-12.

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