A bus company models the extra weekly revenue, in pounds, from two proposed fare changes. Scheme 1 gives \(R_1 = -x^2 + 18x\). Scheme 2 gives \(R_2 = -x^2 +...
Question 1 Report
A bus company models the extra weekly revenue, in pounds, from two proposed fare changes. Scheme 1 gives \(R_1 = -x^2 + 18x\). Scheme 2 gives \(R_2 = -x^2 + 14x + 20\), where \(x\) is the size of the fare change in pence.
- Write \(R_1\) in the form \(-(x - a)^2 + b\), and state the maximum value of \(R_1\). (2)
- Write \(R_2\) in the form \(-(x - c)^2 + d\), and state the maximum value of \(R_2\). (2)
- State, with a reason, which scheme gives the higher maximum revenue. (2)
Answer Details
This question rewrites two quadratic revenue models in completed-square form to find and compare their maximum values.
Completing the square on \(R_1 = -x^2+18x\):
\[R_1 = -(x^2-18x) = -[(x-9)^2-81] = -(x-9)^2+81\]Since \(-(x-9)^2 \le 0\), the maximum value of \(R_1\) is \(81\), occurring at \(x=9\) [2 marks].
Completing the square on \(R_2 = -x^2+14x+20\):
\[R_2 = -(x^2-14x)+20 = -[(x-7)^2-49]+20 = -(x-7)^2+69\]The maximum value of \(R_2\) is \(69\), occurring at \(x=7\) [2 marks].
Since \(81 \gt 69\), Scheme 1 gives the higher maximum revenue [2 marks].
In completed-square form \(-(x-a)^2+b\) for a downward parabola, \(b\) is always the maximum value, reached exactly when \(x=a\); comparing the two \(b\) values is all that is needed to compare the two schemes' best-case revenue.
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