| Exam Board | OCR |
|---|---|
| Module | Further Discrete AS (Further Discrete AS) |
| Year | 2020 |
| Session | November |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Three-variable constraint reduction |
| Difficulty | Moderate -0.3 This is a straightforward three-variable linear programming problem that reduces to two variables. Part (a) is trivial algebra (6-x-y). Part (b) requires basic constraint interpretation and substitution - standard textbook exercises. Part (c) involves graphing linear inequalities and finding a vertex, which is routine for Further Maths students. The problem requires no novel insight, just methodical application of standard LP techniques, making it slightly easier than average. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.06d Graphical solution: feasible region, two variables |
| Activity | Walking coast path | Visiting bird sanctuary | Visiting garden centre |
| Value | 5 points per hour | 3 points per hour | 2 points per hour |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | 6 – x – y |
| [1] | This expression, in any form | |
| 6 | (b) | (i) |
| Answer | Marks |
|---|---|
| 12 is constant so total value is max when 3x + y is max | M1 |
| Answer | Marks |
|---|---|
| [3] | 5x + 3y |
| Answer | Marks |
|---|---|
| (ii) | y ≥ 6 – x – y |
| ⇒ x + 2y ≥ 6 | M1 |
| Answer | Marks |
|---|---|
| [2] | y ≥ their expression from part (a) |
| Answer | Marks |
|---|---|
| (iii) | x ≤ 3.5 and x + 2y ≥ 6 |
| ⇒ y ≥ 1.25 (or y ≥ 0) | M1 |
| Answer | Marks |
|---|---|
| [2] | x is at most 3.5 (or x is at most 5) and x + 2y is at least 6 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (c) | x 0 3.5 3.5 0 |
| Answer | Marks |
|---|---|
| 1 hour visiting the garden centre | M1 |
| Answer | Marks |
|---|---|
| [7] | May have other lines shown, e.g. profit lines |
| Answer | Marks | Guidance |
|---|---|---|
| x | 0 | 3.5 |
| y | 3 | 1.25 |
| 3x+y | 3 | 11.75 |
Question 6:
6 | (a) | 6 – x – y | B1
[1] | This expression, in any form
6 | (b) | (i) | Total value = 5x + 3y + 2(6 – x – y)
= 3x + y + 12
12 is constant so total value is max when 3x + y is max | M1
M1
A1
[3] | 5x + 3y
+ 2(6 – x – y) or 2(their expression from part (a))
3x + y + 12 seen
or explaining why constant can be ignored
(ii) | y ≥ 6 – x – y
⇒ x + 2y ≥ 6 | M1
A1
[2] | y ≥ their expression from part (a)
Leading to given expression
(iii) | x ≤ 3.5 and x + 2y ≥ 6
⇒ y ≥ 1.25 (or y ≥ 0) | M1
A1
[2] | x is at most 3.5 (or x is at most 5) and x + 2y is at least 6
Leading to implication that y cannot be negative (as given)
Argued algebraically or in words
6 | (c) | x 0 3.5 3.5 0
y 3 1.25 1.5 5
3x+y 3 11.75 12 5
Tamsin should spend
3.5 hours walking the coast path
1.5 hours visiting the bird sanctuary
1 hour visiting the garden centre | M1
M1
M1
A1
M1
M1
A1
[7] | May have other lines shown, e.g. profit lines
Assume x is on the horizontal axis and y on the vertical
axis, unless labelled to the contrary
Vertical boundary x = 3.5
Line x + 2y = 6 through (0, 3) and (4, 1)
Line x + y = 5, through (0, 5) and (4, 1)
Shading to show this feasible region
Checking vertices or using a sliding profit line
May be implied from their solution provided consistent
with their feasible region
Interpretation of their x and y in context,
provided x ≤ 3.5 and both are ≥ 0
(6 – their x – their y) hours (≥ 0) visiting the garden centre
x | 0 | 3.5 | 3.5 | 0
y | 3 | 1.25 | 1.5 | 5
3x+y | 3 | 11.75 | 12 | 5
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recorded or monitored6 Tamsin is planning how to spend a day off. She will divide her time between walking the coast path, visiting a bird sanctuary and visiting the garden centre.
Tamsin has given a value to each hour spent doing each activity. She wants to decide how much time to spend on each activity to maximise the total value of the activities.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
Activity & Walking coast path & Visiting bird sanctuary & Visiting garden centre \\
\hline
Value & 5 points per hour & 3 points per hour & 2 points per hour \\
\hline
\end{tabular}
\end{center}
Tamsin's requirements are that she will spend:
\begin{itemize}
\item a total of exactly 6 hours on the three activities
\item at most 3.5 hours walking the coast path
\item at least as long at the bird sanctuary as at the garden centre
\item at least 1 hour at the garden centre.
\begin{enumerate}[label=(\alph*)]
\item If Tamsin spends $x$ hours walking the coast path and $y$ hours visiting the bird sanctuary, how many hours does she spend visiting the garden centre?
\end{itemize}
Tamsin models her problem as the linear programming formulation
Maximise $P = 3 x + y$\\
Subject to
$$\begin{aligned}
& x \leqslant 3.5 \\
& x + 2 y \geqslant 6 \\
& x + y \leqslant 5 \\
& x \geqslant 0
\end{aligned}$$
and
\item \begin{enumerate}[label=(\roman*)]
\item Explain why the maximum total value of the activities done is achieved when $3 x + y$ is maximised.
\item Show how the requirement that she spends at least as long at the bird sanctuary as at the garden centre leads to the constraint $x + 2 y \geqslant 6$.
\item Explain why there is no need to require that $y \geqslant 0$.
\end{enumerate}\item Represent the constraints graphically and hence find a solution to Tamsin's problem.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete AS 2020 Q6 [15]}}