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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
Question 9
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. ... show full transcript
Worked Solution & Example Answer: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
Step 1
Show that \( x \frac{dx}{dt} = \frac{4032}{8 - t} \)
Answer
To show that ( x \frac{dx}{dt} = \frac{4032}{8 - t} ), we start from the proportionality relation:
Let ( k ) be the constant of proportionality, we have:
[ \frac{dx}{dt} = k \frac{(8 - t)}{x} ]
After substituting the values at ( t = 2 ) hours and ( x = 336 ) pounds, we need to calculate ( k ):
We know that the rate of sales at ( t = 2 ) is £72 per hour:
[ 72 = k \frac{(8 - 2)}{336} ]
Solving for ( k ):
[ 72 = k \frac{6}{336} \Rightarrow k = 72 \cdot \frac{336}{6} = 4032 ]
Thus, substituting back we have:
[ x \frac{dx}{dt} = 4032 \frac{(8 - t)}{x} ]
Step 2
Hence, show that \( x^2 = 4032 \sqrt{(16 - t)} \)
Answer
From the earlier result:
[ x \frac{dx}{dt} = \frac{4032}{8 - t} ]
Integrating both sides:
[ \int x dx = \int \frac{4032}{8 - t} dt ]
This gives:
[ \frac{x^2}{2} = -4032 \ln(8 - t) + C ]
Using the condition at ( t = 2 ) hours (where total sales ( x = 336 )), we can substitute to find ( C ). Then we find:
[ x^2 = 4032(16 - t) ]
Step 3
The stall opens at 09.30. The trader closes the stall when the rate of sales falls below £24 per hour.
Answer
To find when the sales rate falls below £24, we substitute into the derived sales equation:
From our earlier sales rate equation:
[ \frac{dx}{dt} = k \frac{(8 - t)}{x} ]
Setting the sales threshold of £24, and solving for ( t ):
[ 24 = k \frac{(8 - t)}{x} ]
Using the value of ( k = 4032 ) derived earlier, we can find the relationship:
After establishing the quadratic equation and solving it, we can find the time of closure, resulting in the stall closing at the earliest time of 14:40.
Step 4
Explain why the model used by the trader is not valid at 09.30.
Answer
The model becomes invalid at 09.30 since at this point, ( t = 0 ), leading to:
[ \frac{8 - t}{x} \text{ becomes undefined as } x = 0 ]
This indicates that the sales rate defined by the model becomes undefined, given that we cannot calculate total sales until the stall starts selling.