https://file.aiquickdraw.com/mathbot/file/1726433463402q0ni41if.jpg

Let's analyze the mathematical model given in the image. The equation provided is:

\[
P = 100 - 6.25(x - 9)^2
\]

where:
- \(P\) represents the annual profit in thousands of pounds.
- \(x\) is the selling price of the toy in pounds.

The questions are:

### Part (a)
**Why is £15 not a sensible selling price for the toy?**

To answer this, we need to calculate the profit at \(x = 15\). Substituting \(x = 15\) into the profit equation:

\[
P = 100 - 6.25(15 - 9)^2 = 100 - 6.25(6)^2 = 100 - 6.25 \times 36 = 100 - 225 = -125
\]

The profit at £15 is -125 (in thousands), meaning a loss of £125,000. Therefore, £15 is not a sensible price since the company would incur a significant loss.

---

### Part (b)
**Given the company made an annual profit of more than £800,000, find the least possible selling price for the toy.**

Since the profit is measured in thousands of pounds, £800,000 translates to \(P = 800\) in thousands of pounds. We set the equation equal to 80:

\[
80 = 100 - 6.25(x - 9)^2
\]

Now, solving for \(x\):

\[
80 - 100 = -6.25(x - 9)^2
\]
\[
-20 = -6.25(x - 9)^2
\]
\[
(x - 9)^2 = \frac{20}{6.25} = 3.2
\]
\[
x - 9 = \pm \sqrt{3.2} \approx \pm 1.79
\]
\[
x = 9 \pm 1.79
\]

Thus, the two possible values for \(x\) are approximately:

\[
x = 9 + 1.79 = 10.79 \quad \text{and} \quad x = 9 - 1.79 = 7.21
\]

So, the least possible selling price is approximately **£7.21**.

---

### Part (c)
**(i) The maximum possible annual profit,**
**(ii) The selling price that maximises the annual profit.**

The equation is in the form \(P = a - b(x - h)^2\), which represents a parabola that opens downwards. The vertex form of this equation indicates that the maximum profit occurs at the vertex \(x = 9\).

- **(ii)** The selling price that maximizes the profit is \(x = 9\) pounds.

To find the maximum possible profit, substitute \(x = 9\) into the equation:

\[
P = 100 - 6.25(9 - 9)^2 = 100 - 0 = 100
\]

- **(i)** The maximum possible profit is **£100,000** (since \(P = 100\) in thousands).

---

Would you like further details on any of these calculations, or help with additional questions? Here are 5 related questions to expand on this:

1. What is the domain and range of the profit function \(P(x)\)?
2. How would the model change if production costs were included?
3. What does the value \(9\) in the equation represent in the context of the company’s operations?
4. How sensitive is the profit to changes in the selling price around \(x = 9\)?
5. How can the company use this model to determine break-even points?

**Tip:** When analyzing quadratic models, always check the vertex, as it provides valuable information about maxima and minima!

A company makes a particular type of children's toy. The annual profit made by the company is modeled by the equation P = 100 - 6.25(x - 9)^2, where P is the profit measured in thousands of pounds and x is the selling price of the toy in pounds. Using the model, (a) explain why £15 is not a sensible selling price for the toy. (b) Find the least possible selling price for the toy given that the company made an annual profit of more than £800,000. (c) State the maximum possible annual profit and the selling price that maximizes the annual profit.

Learn how to maximize profit using quadratic equations. We explore why £15 is not a sensible selling price, find the least possible price with an annual profit of over £800,000, and determine the maximum profit and the price that achieves it.

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