| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2018 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Reverse engineering constraints from graph |
| Difficulty | Moderate -0.3 This is a standard D1 linear programming question requiring reading constraints from a graph, finding vertices, and optimizing. Part (d) requires understanding objective function gradients but follows a well-practiced technique. Slightly easier than average A-level due to being a routine textbook-style question with predictable parts. |
| Spec | 7.06b Slack variables: converting inequalities to equations7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y \leq 3x\) | B3,2,1,0 | a1B1: either two equations correct or one correct inequality (condone strict inequality) |
| \(3y \geq x\) | a2B1: two correct inequalities | |
| \(5x + 3y \leq 15\) | a3B1: CAO all three inequalities correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0,0)\), \(\left(\dfrac{15}{14}, \dfrac{45}{14}\right)\), \(\left(\dfrac{5}{2}, \dfrac{5}{6}\right)\) | B1 M1 A1 | b1B1: \((0,0)\). b1M1: using simultaneous equations to get other two vertices. b1A1: CAO of \(\left(\frac{15}{14},\frac{45}{14}\right)\) and \(\left(\frac{5}{2},\frac{5}{6}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \((0,0)\), \(P = 0\) | M1 | Point testing at least two vertices using correct objective function \(P = 2x + 3y\) |
| \(\left(\dfrac{15}{14}, \dfrac{45}{14}\right)\), \(P = \dfrac{165}{14}\), therefore \(\left(\dfrac{15}{14}, \dfrac{45}{14}\right)\) is the optimal vertex | A1 | Point testing two correct vertices correctly |
| \(\left(\dfrac{5}{2}, \dfrac{5}{6}\right)\), \(P = \dfrac{15}{2}\) | A1 | All three correct exact vertices tested correctly with correct conclusion; \(P = \frac{165}{14}\) or \(11\frac{11}{14}\) must be clear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\!\left(\dfrac{15}{14}\right) + k\!\left(\dfrac{45}{14}\right) \geq 2\!\left(\dfrac{5}{2}\right) + k\!\left(\dfrac{5}{6}\right)\) | M1 | Setting up a linear inequality (any inequality sign) or equation involving new objective \(Q = 2x + ky\) and candidate's two non-zero vertices from (b) |
| \(k \geq \dfrac{6}{5}\) | A1 | CAO; allow strict inequality; correct answer with no working scores M1A0 |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y \leq 3x$ | B3,2,1,0 | a1B1: either two equations correct or one correct inequality (condone strict inequality) |
| $3y \geq x$ | | a2B1: two correct inequalities |
| $5x + 3y \leq 15$ | | a3B1: CAO all three inequalities correct |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0,0)$, $\left(\dfrac{15}{14}, \dfrac{45}{14}\right)$, $\left(\dfrac{5}{2}, \dfrac{5}{6}\right)$ | B1 M1 A1 | b1B1: $(0,0)$. b1M1: using simultaneous equations to get other two vertices. b1A1: CAO of $\left(\frac{15}{14},\frac{45}{14}\right)$ and $\left(\frac{5}{2},\frac{5}{6}\right)$ |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $(0,0)$, $P = 0$ | M1 | Point testing at least two vertices using correct objective function $P = 2x + 3y$ |
| $\left(\dfrac{15}{14}, \dfrac{45}{14}\right)$, $P = \dfrac{165}{14}$, therefore $\left(\dfrac{15}{14}, \dfrac{45}{14}\right)$ is the optimal vertex | A1 | Point testing two correct vertices correctly |
| $\left(\dfrac{5}{2}, \dfrac{5}{6}\right)$, $P = \dfrac{15}{2}$ | A1 | All three correct exact vertices tested correctly with correct conclusion; $P = \frac{165}{14}$ or $11\frac{11}{14}$ must be clear |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\!\left(\dfrac{15}{14}\right) + k\!\left(\dfrac{45}{14}\right) \geq 2\!\left(\dfrac{5}{2}\right) + k\!\left(\dfrac{5}{6}\right)$ | M1 | Setting up a linear inequality (any inequality sign) or equation involving new objective $Q = 2x + ky$ and candidate's two non-zero vertices from (b) |
| $k \geq \dfrac{6}{5}$ | A1 | CAO; allow strict inequality; correct answer with no working scores M1A0 |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e0c89aba-9d2e-469b-8635-d513df0b65a4-05_1198_908_226_584}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows the constraints of a linear programming problem in $x$ and $y$, where $R$ is the feasible region.
\begin{enumerate}[label=(\alph*)]
\item Determine the inequalities that define the feasible region.
\item Find the exact coordinates of the vertices of the feasible region.
The objective is to maximise $P = 2 x + 3 y$.
\item Use point testing at each vertex to find the optimal vertex, $V$, of the feasible region and state the corresponding value of $P$ at $V$.\\
(3)
The objective is changed to maximise $Q = 2 x + k y$, where $k$ is a constant.
\item Find the range of values of $k$ for which the vertex identified in (c) is still optimal.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2018 Q4 [11]}}