| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Graphical optimization with objective line |
| Difficulty | Easy -1.2 This is a standard D1 linear programming question following a very familiar template: formulate inequalities from word problems, plot constraints, identify feasible region, use objective line method, and find optimal vertex. All steps are routine textbook exercises with no novel problem-solving required. The only mild challenge is calculating the exact intersection point in part (e), but this is straightforward simultaneous equations. Well below average difficulty for A-level maths overall. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b 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 |
| \(20x + 65y \leq 520\) or equivalently \(4x + 13y \leq 104\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Line 1 drawn correctly | B1 | |
| Line 2 drawn correctly | B1 | |
| Line 3 drawn correctly | B1 | |
| Region \(R\) correctly identified | DB1 (R) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. \(P = 2x + 3y\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Drawing an objective line – accept reciprocal gradient | M1 | Drawing either the correct objective line or their objective line (based on answer to (c)) or the reciprocal of the correct objective line or the reciprocal of their objective line – if their line is shorter than the length equivalent to that of the line from \((0,1)\) to \((1.5, 0)\) then M0. Line must be correct to within one small square if extended from axis to axis |
| Correct objective line minimum length equivalent to \((0,1)\) to \((1.5, 0)\) | A1 | Drawing the correct objective line – same condition that the line must be correct to within one small square if extended from axis to axis |
| V correctly labelled | A1 | Correct V labelled clearly on their graph – dependent on scoring at least B1B1B1B0 in (b) and the two previous marks in this part |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V\left(\dfrac{624}{59}, \dfrac{280}{59}\right)\) | M1 A1 | Simultaneous equations being used to find V. Must have scored at least B1B1B0B0 in (b) and candidates must have labelled one of their vertices as V. Must be solving either the pair of simultaneous equations \(20x+65y=520\) and \(7x+8y=112\) or \(7x+8y=112\) and \(-x+24y=24\). Correct exact coordinates as either \(\left(\dfrac{624}{59}, \dfrac{280}{59}\right)\) or \(\left(10\dfrac{34}{59}, 4\dfrac{44}{59}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Testing integer solutions around V, \(x=11\) and \(y=4\) is optimal integer solution, so they should buy 11 standard containers and 4 deluxe containers | M1 A1 | Testing any two of \((11,4)\) or \((9,5)\) or \((10,5)\) or \((10,4)\) or \((11,5)\) in a correct objective function or the correct pair of inequalities. f1A1: CSO (all previous 12 marks must have been awarded) – must have tested \((11,4)\) in the correct objective function or correct pair of inequalities – accept \(x=11\) and \(y=4\) or stated as a pair of coordinates |
| Cost is £480 | B1 | CAO – this mark is not dependent on any previous mark and condone lack of units |
# Question 8:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $20x + 65y \leq 520$ or equivalently $4x + 13y \leq 104$ | B1 | oe |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Line 1 drawn correctly | B1 | |
| Line 2 drawn correctly | B1 | |
| Line 3 drawn correctly | B1 | |
| Region $R$ correctly identified | DB1 (R) | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. $P = 2x + 3y$ | B1 | |
## Question 8:
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Drawing an objective line – accept reciprocal gradient | M1 | Drawing either the correct objective line **or** their objective line (based on answer to (c)) **or** the reciprocal of the correct objective line **or** the reciprocal of their objective line – if their line is shorter than the length equivalent to that of the line from $(0,1)$ to $(1.5, 0)$ then M0. Line must be correct to within one small square if extended from axis to axis |
| Correct objective line minimum length equivalent to $(0,1)$ to $(1.5, 0)$ | A1 | Drawing the correct objective line – same condition that the line must be correct to within one small square if extended from axis to axis |
| V correctly labelled | A1 | Correct V labelled clearly on their graph – **dependent on scoring at least B1B1B1B0 in (b) and the two previous marks in this part** |
**(3 marks)**
---
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V\left(\dfrac{624}{59}, \dfrac{280}{59}\right)$ | M1 A1 | Simultaneous equations being used to find V. Must have scored at least B1B1B0B0 in (b) and candidates must have labelled one of their vertices as V. Must be solving either the pair of simultaneous equations $20x+65y=520$ and $7x+8y=112$ **or** $7x+8y=112$ and $-x+24y=24$. Correct exact coordinates as either $\left(\dfrac{624}{59}, \dfrac{280}{59}\right)$ or $\left(10\dfrac{34}{59}, 4\dfrac{44}{59}\right)$ |
**(2 marks)**
---
### Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Testing integer solutions around V, $x=11$ and $y=4$ is optimal integer solution, so they should buy 11 standard containers and 4 deluxe containers | M1 A1 | Testing any two of $(11,4)$ or $(9,5)$ or $(10,5)$ or $(10,4)$ or $(11,5)$ in a correct objective function **or** the correct pair of inequalities. f1A1: CSO (**all previous 12 marks must have been awarded**) – must have tested $(11,4)$ in the correct objective function **or** correct pair of inequalities – accept $x=11$ and $y=4$ or stated as a pair of coordinates |
| Cost is £480 | B1 | CAO – this mark is not dependent on any previous mark and condone lack of units |
**(3 marks)**
**Total: 14 marks**8. Charlie needs to buy storage containers.
There are two different types of storage container available, standard and deluxe.
Standard containers cost $\pounds 20$ and deluxe containers cost $\pounds 65$. Let $x$ be the number of standard containers and $y$ be the number of deluxe containers.
The maximum budget available is $\pounds 520$
\begin{enumerate}[label=(\alph*)]
\item Write down an inequality, in terms of $x$ and $y$, to model this constraint.
Three further constraints are:
$$\begin{aligned}
x & \geqslant 2 \\
- x + 24 y & \geqslant 24 \\
7 x + 8 y & \leqslant 112
\end{aligned}$$
\item Add lines and shading to Diagram 1 in the answer book to represent all four constraints. Hence determine the feasible region and label it R .
The capacity of a deluxe container is $50 \%$ greater than the capacity of a standard container. Charlie wishes to maximise the total capacity.
\item State an objective function, in terms of $x$ and $y$.
\item Use the objective line method to find the optimal vertex, V, of the feasible region. You must make your objective line clear and label the optimal vertex V.
\item Calculate the exact coordinates of vertex V.
\item Determine the number of each type of container that Charlie should buy. You must make your method clear and calculate the cost of purchasing the storage containers.
Write your name here
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Final output $\_\_\_\_$\\
(b)\\
6.
\begin{figure}[h]
\begin{center}
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\caption{Figure 5\\[0pt]
[The total weight of the network is 384]}
\end{center}
\end{figure}
\includegraphics[max width=\textwidth, alt={}, center]{22ff916a-4ba8-4e0c-9c53-e82b0aff0b98-24_2651_1940_118_121}\\
\includegraphics[max width=\textwidth, alt={}, center]{22ff916a-4ba8-4e0c-9c53-e82b0aff0b98-25_2261_50_312_36}
\section*{Q uestion 7 continued}
(c) $\_\_\_\_$\\
(d)
\section*{Diagram 2}
(Total 12 marks)\\
□
8.
\begin{center}
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\end{center}
Diagram 1
\section*{Q uestion 8 continued}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{22ff916a-4ba8-4e0c-9c53-e82b0aff0b98-28_2646_1833_116_118}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2016 Q8 [14]}}