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Let's break down the problem step by step.

### Given Information:
- The manufacturer can sell **500 metres** of fabric per week if the price is **$10 per metre**.
- For each **20 cents reduction** in price, the sales will increase by **25 metres**.
- Let **x** be the number of reductions, where each reduction is **20 cents**.

Now, let's solve the subparts:

### a) Expression for the cost per metre:
If the initial price is $10 and the price is reduced by **20 cents for each x**, the expression for the cost per metre is:

\[
\text{Cost per metre} = 10 - 0.2x
\]

### b) Expression for the number of metres sold:
For each reduction of **20 cents**, the manufacturer sells **25 more metres**. Hence, the number of metres sold after **x** reductions is:

\[
\text{Metres sold} = 500 + 25x
\]

### c) Expression for the total revenue (income):
The total revenue is the product of the price per metre and the number of metres sold. So, the total revenue $R(x)$ is:

\[
R(x) = (\text{Cost per metre}) \times (\text{Metres sold})
\]

Substituting the expressions from parts (a) and (b):

\[
R(x) = (10 - 0.2x)(500 + 25x)
\]

Expanding this expression:

\[
R(x) = 10(500 + 25x) - 0.2x(500 + 25x)
\]

Simplifying further:

\[
R(x) = 5000 + 250x - 100x - 5x^2
\]

So, the total revenue expression is:

\[
R(x) = 5000 + 150x - 5x^2
\]

### d) The value of $x$ that maximizes the total revenue:
To find the value of $x$ that maximizes the total revenue, we need to find the vertex of the quadratic function $R(x) = 5000 + 150x - 5x^2$. The formula to find the vertex of a quadratic function $ax^2 + bx + c$ is:

\[
x = -\frac{b}{2a}
\]

In this case, $a = -5$ and $b = 150$. Substituting these values:

\[
x = -\frac{150}{2(-5)} = \frac{150}{10} = 15
\]

Thus, $x = 15$ is the number of reductions that maximizes the total revenue.

### Verification that this gives a maximum:
Since the coefficient of $x^2$ in the quadratic expression is negative ($-5$), the parabola opens downwards, confirming that $x = 15$ corresponds to a maximum point.

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Would you like more details or further explanations? Here are some related questions:

1. How would the revenue change if the price was increased instead of reduced?
2. What would happen if the price could not be reduced beyond a certain limit?
3. How does the maximum revenue compare to the initial revenue when no price reduction is applied?
4. Could the manufacturer raise sales by increasing the price, assuming the market behaves differently?
5. How would the formula change if the sales increase was not linear with price reductions?

**Tip:** Maximizing revenue often involves finding a balance between price and sales volume. This relationship can be modeled with quadratic functions, and knowing how to find the vertex of a parabola helps in identifying optimal pricing strategies.

The manufacturer of a certain fabric estimates that she can sell 500 m of the fabric each week if the price is $10 per metre. However, market research indicates that each 20 cents per metre price reduction will increase sales by 25 metres. If the manufacturer reduces the cost per metre by x lots of 20 cents, find: a) an expression for the cost per metre, b) an expression for the number of metres sold, c) an expression for the total revenue (income), d) the value of x that makes this total revenue a maximum and explain how you know that your value of x will give a maximum.

Learn how to model and solve a fabric pricing problem using linear equations and quadratic functions. This solution explores how price reductions increase sales, leading to a quadratic revenue function that is maximized by finding the optimal price reduction. Suitable for Grades 9-12.

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