| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Vertex testing for optimization |
| Difficulty | Moderate -0.5 This is a standard D1 linear programming question requiring routine techniques: reading inequalities from a graph, finding intersection points, and testing vertices with given objective functions. While multi-part, each step follows textbook procedures with no novel problem-solving required, making it slightly easier than average A-level difficulty. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2y \leq x + 12\); \(5y \geq 2x\); \(5x + 2y \leq 60\); \(x \geq 0\) | B1, B1, B1, B1 | (4) |
| \((0,0), (0,6), (8,10), \left(\frac{300}{29}, \frac{120}{29}\right)\) | B1 M1 A1 | (3) |
| At \((0, 0)\) \(P = 0\); At \((0, 6)\) \(P = 6\); At \((8, 10)\) \(P = 26\); At \(\left(\frac{300}{29}, \frac{120}{29}\right)\) \(P = \frac{720}{29}\) therefore \((8,10)\) is the optimal vertex | M1 A1 A1 | (3) |
| At \((0, 0)\) \(P = 0\); At \((0, 6)\) \(P = 6\); At \((8, 10)\) \(P = 30\); At \(\left(\frac{300}{29}, \frac{120}{29}\right)\) \(P = 30\) therefore the set of points on the line \(5x + 2y = 60\) for which \(8 \leq x \leq \frac{300}{29}\) gives \(P\) its maximum value | M1 A1 A1 | (3) |
| 13 marks |
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2y \leq x + 12$; $5y \geq 2x$; $5x + 2y \leq 60$; $x \geq 0$ | B1, B1, B1, B1 | (4) |
| $(0,0), (0,6), (8,10), \left(\frac{300}{29}, \frac{120}{29}\right)$ | B1 M1 A1 | (3) |
| **At $(0, 0)$** $P = 0$; **At $(0, 6)$** $P = 6$; **At $(8, 10)$** $P = 26$; **At** $\left(\frac{300}{29}, \frac{120}{29}\right)$ $P = \frac{720}{29}$ therefore $(8,10)$ is the optimal vertex | M1 A1 A1 | (3) |
| **At $(0, 0)$** $P = 0$; **At $(0, 6)$** $P = 6$; **At $(8, 10)$** $P = 30$; **At** $\left(\frac{300}{29}, \frac{120}{29}\right)$ $P = 30$ therefore the set of points on the line $5x + 2y = 60$ for which $8 \leq x \leq \frac{300}{29}$ gives $P$ its maximum value | M1 A1 A1 | (3) |
| | | **13 marks** |
**Notes for Question 3:**
- a1B1: Either $5y \geq 2x$ or $5x + 2y \leq 60$ (accept strict inequality for this mark)
- a2B1: Both $5y \geq 2x$ **and** $5x + 2y \leq 60$ correct
- a3B1: Correct equation of the line $2y = x + 12$ (aef) – accept any inequality for this mark
- a4B1: Both $2y \leq x + 12$ **and** $x \geq 0$
- b1B1: The three coordinates $(0, 0), (0, 6)$ and $(8, 10)$ correct
- b1M1: Using simultaneous equations to find fourth vertex – must be a correct method to solve simultaneous equations and must arrive at $x = \ldots$ and $y = \ldots$ but allow slips/errors. This mark can be awarded for the correct exact coordinates stated with no working
- b1A1: CAO for $\left(\frac{300}{29}, \frac{120}{29}\right)$ or stated in terms of $x$ and $y$
- c1M1: Point testing at least two of their vertices using the correct objective function (objective line is M0)
- c1A1: Point testing three of the correct vertices correctly
- c2A1: All four correct vertices tested correctly and correct conclusion that $(8, 10)$ is the optimal vertex
- d1M1: Either point testing (at least two vertices) using the correct objective function or using the objective line method (by considering an objective line with the correct gradient)
- d1A1: Recognising that the points on the line $5x + 2y = 60$ **or** points in the interval $8 \leq x \leq \frac{300}{29}$ (or $\frac{120}{29} \leq y \leq 10$) give $P$ its maximum value. Stating/recognising that the gradient of the objective is the same as the gradient of $5x + 2y = 60$ scores M1A1. Just stating the two points for which $P = 30$ is A0A0
- d2A1: Stating that all points on $5x + 2y = 60$ when $8 \leq x \leq \frac{300}{29}$ (or $\frac{120}{29} \leq y \leq 10$) give $P$ its maximum value
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{049de386-42a9-4f16-8be3-9324382e4988-04_1684_1492_194_283}
\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. The equations of two of the lines have been given.
\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 $P$, where $P = k x + y$.
\item For the case $k = 2$, use point testing to find the optimal vertex of the feasible region.
\item For the case $k = 2.5$, find the set of points for which $P$ takes its maximum value.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2016 Q3 [13]}}