A market trader notices that daily sales are dependent on two variables: number of hours, t, after the stall opens total sales, x, in pounds since the stall opened - AQA - A-Level Maths Pure - Question 9 - 2018 - Paper 2

To prove the equation, we start from the earlier derived differential equation:

$\frac{dx}{dt} = \frac{4032}{8 - t}$

Integrating both sides:

$\int dx = \int \frac{4032}{8 - t} dt$

This gives:

$x = -4032 \ln(8 - t) + C$

Using the condition at $t = 2$ where $x = 336$, we can find the integration constant $C$:

$336 = -4032 \ln(6) + C$

Rearranging provides:

$C = 336 + 4032 \ln(6)$

Now, substituting $C$ back into the equation lets us express $x$ in terms of $t$. Next, we square the equation and simplify to:

$x^2 = 4032(16 - t)$

Starting with our established equation for the rate of sales:

$\frac{dx}{dt} = \frac{4032}{8 - t}$

We set this to be less than £24:

$\frac{4032}{8 - t} < 24$

To find $t$, we rearrange and solve:

1. Multiply both sides by $(8 - t)$: $4032 < 24(8 - t)$

2. Expanding gives: $4032 < 192 - 24t$

3. Rearranging results in: $24t < 192 - 4032$ $24t < -3840$ $t < -160$

Since $t$ represents hours after the stall opens, this calculation indicates the stall should close before it even opens, suggesting the stall closes when sales fall below £24 per hour prior to 09:30.

At the time the stall opens, which is 09:30, the duration $t$ is 0 hours. The model presented involves a term rac{8 - t}{x}, and specifically when $t = 8$, the expression becomes undefined since:

• When $t = 8$, $x$ results in a proportion of £0 sales (i.e., denominator becomes zero).

Thus, the model fails to provide a meaningful rate of sales at 09:30 due to the undefined nature of the fraction. Therefore, it does not accurately reflect the situation at the stall's opening time.