| Exam Board | Edexcel |
|---|---|
| Module | FD1 (Further Decision 1) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Parametric objective analysis |
| Difficulty | Challenging +1.2 This is a Further Maths Decision module question requiring identification of constraints from a graph, evaluation of an objective function at vertices (standard linear programming), and then parametric analysis to find when a vertex remains optimal. The parametric part (c) requires understanding of gradient conditions but is a well-established technique in FD1. More challenging than core A-level due to the Further Maths content and multi-part reasoning, but follows standard linear programming methodology without requiring novel insight. |
| Spec | 7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Any two correct inequalities from: \(2y \leq 5x\), \(y \geq x+1\), \(6x+5y \leq 30\) | B2,1,0 | Accept strict inequalities; accept equivalent inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Vertices: \(\left(\frac{2}{3}, \frac{5}{3}\right)\), \(\left(\frac{60}{37}, \frac{150}{37}\right)\), \(\left(\frac{25}{11}, \frac{36}{11}\right)\) | B1, B1 | One correct vertex (B1); all three correct (B1). Must be exact |
| \(\left(\frac{2}{3}, \frac{5}{3}\right) \to P = \frac{11}{3}\) | M1 | Testing all three vertices in correct objective function |
| \(\left(\frac{60}{37}, \frac{150}{37}\right) \to P = \frac{330}{37}\) | ||
| \(\left(\frac{25}{11}, \frac{36}{11}\right) \to P = \frac{111}{11}\), so optimal vertex is \(\left(\frac{25}{11}, \frac{36}{11}\right)\) | A1 | Correct three values of \(P\) and correct optimal vertex stated or clearly indicated on graph |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3\!\left(\frac{25}{11}\right) + \frac{36a}{11} > 3\!\left(\frac{60}{37}\right) + \frac{150a}{37}\) | M1 | Optimal point from (b) evaluated in \(Q\) compared to \(\left(\frac{60}{37},\frac{150}{37}\right)\) in \(Q\) with correct inequality |
| \(\Rightarrow a < \frac{5}{2}\) | A1 | CAO |
| \(3\!\left(\frac{25}{11}\right) + \frac{36a}{11} > 3\!\left(\frac{2}{3}\right) + \frac{5a}{3}\) | M1 | Optimal point from (b) evaluated in \(Q\) compared to \(\left(\frac{2}{3},\frac{5}{3}\right)\) in \(Q\) with correct inequality |
| \(\Rightarrow a > -3\) | A1 | CAO |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Any two correct inequalities from: $2y \leq 5x$, $y \geq x+1$, $6x+5y \leq 30$ | B2,1,0 | Accept strict inequalities; accept equivalent inequalities |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Vertices: $\left(\frac{2}{3}, \frac{5}{3}\right)$, $\left(\frac{60}{37}, \frac{150}{37}\right)$, $\left(\frac{25}{11}, \frac{36}{11}\right)$ | B1, B1 | One correct vertex (B1); all three correct (B1). Must be exact |
| $\left(\frac{2}{3}, \frac{5}{3}\right) \to P = \frac{11}{3}$ | M1 | Testing all three vertices in correct objective function |
| $\left(\frac{60}{37}, \frac{150}{37}\right) \to P = \frac{330}{37}$ | | |
| $\left(\frac{25}{11}, \frac{36}{11}\right) \to P = \frac{111}{11}$, so optimal vertex is $\left(\frac{25}{11}, \frac{36}{11}\right)$ | A1 | Correct three values of $P$ and correct optimal vertex stated or clearly indicated on graph |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $3\!\left(\frac{25}{11}\right) + \frac{36a}{11} > 3\!\left(\frac{60}{37}\right) + \frac{150a}{37}$ | M1 | Optimal point from (b) evaluated in $Q$ compared to $\left(\frac{60}{37},\frac{150}{37}\right)$ in $Q$ with correct inequality |
| $\Rightarrow a < \frac{5}{2}$ | A1 | CAO |
| $3\!\left(\frac{25}{11}\right) + \frac{36a}{11} > 3\!\left(\frac{2}{3}\right) + \frac{5a}{3}$ | M1 | Optimal point from (b) evaluated in $Q$ compared to $\left(\frac{2}{3},\frac{5}{3}\right)$ in $Q$ with correct inequality |
| $\Rightarrow a > -3$ | A1 | CAO |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bd357978-6464-43fd-854f-4188b5408e91-06_1171_1758_269_150}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the constraints of a linear programming problem in $x$ and $y$, where $R$ is the feasible region.
\begin{enumerate}[label=(\alph*)]
\item Write down the inequalities that define $R$.
The objective is to maximise $P$, where $P = 3 x + y$
\item Obtain the exact value of $P$ at each of the three vertices of $R$ and hence find the optimal vertex, $V$.
The objective is changed to maximise $Q$, where $Q = 3 x + a y$. Given that $a$ is a constant and the optimal vertex is still $V$,
\item find the range of possible values of $a$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD1 2020 Q4 [10]}}