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Let's break down the problem step by step:

### Part a:
We are given the equation for the number of cushions sold:
\[
c = 230 - Hp
\]
where \( p \) is the price of each cushion, and \( H \) is a constant. The problem states that Lynn expects to sell 10 more cushions for every £1 she lowers the price. This means \( \frac{dc}{dp} = -10 \).

From the equation \( c = 230 - Hp \), the derivative of \( c \) with respect to \( p \) is:
\[
\frac{dc}{dp} = -H
\]
Equating \( -H \) to \( -10 \) (since she expects to sell 10 more cushions per £1 decrease):
\[
H = 10
\]
So, the value of \( H \) is **10**.

---

### Part b:
We are given the revenue equation:
\[
r = p(230 - Hp)
\]
We need to rearrange this into the form \( A - B(p - C)^2 \), where \( A \), \( B \), and \( C \) are constants.

Substituting \( H = 10 \) from part a into the revenue equation:
\[
r = p(230 - 10p)
\]
Expanding this:
\[
r = 230p - 10p^2
\]

Now, we complete the square on the quadratic term:
\[
r = -10(p^2 - 23p)
\]
To complete the square inside the parentheses:
\[
r = -10 \left( \left( p - \frac{23}{2} \right)^2 - \frac{529}{4} \right)
\]
Simplifying this:
\[
r = -10 \left( p - \frac{23}{2} \right)^2 + \frac{5290}{4}
\]
So, the equation is in the form:
\[
r = A - B(p - C)^2
\]
where:
- \( A = \frac{5290}{4} \),
- \( B = 10 \),
- \( C = \frac{23}{2} \).

---

### Part c:
We need to find how Lynn can increase her revenue by £122.50 through lowering her prices and find the optimal price.

From part b, the maximum revenue occurs when \( p = C = \frac{23}{2} = 11.5 \) (as this minimizes the squared term). 

To find the maximum revenue:
\[
r_{\text{max}} = \frac{5290}{4} = 1322.50
\]
The initial revenue at a price of £15 is:
\[
r = 15(230 - 10 \times 15) = 15 \times 80 = 1200
\]
So, the increase in revenue is:
\[
1322.50 - 1200 = 122.50
\]
Thus, lowering the price to £11.50 increases the revenue by **£122.50**, and the **optimum price is £11.50**.

---

Would you like more details on any of these steps or have questions?

Here are some follow-up questions to expand understanding:
1. What is the significance of completing the square when maximizing revenue?
2. How does changing the price affect the revenue in this model?
3. Why is the derivative \(\frac{dc}{dp} = -10\) used to determine \(H\)?
4. Could Lynn reduce the price further to increase sales even more?
5. How does quadratic behavior in the revenue equation affect price decisions?

**Tip**: Completing the square is a key technique in optimization problems involving quadratic equations.

Lynn is selling cushions as part of an enterprise project. The number of cushions sold can be modeled by the equation c = 230 - Hp. Determine the value of H, and rearrange the revenue equation r = p(230 - Hp) into a specific form to find how Lynn can maximize her revenue.

Learn how to maximize revenue by rearranging and optimizing the quadratic equation in the context of a cushion-selling enterprise. Solve for the optimum price that Lynn should charge to increase her revenue by £122.50. Suitable for students in Grades 9-12.

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