Edexcel D1 2010 June — Question 7 11 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2010
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear Programming
TypeGraphical feasible region identification
DifficultyModerate -0.8 This is a standard textbook linear programming question requiring routine graphical methods: plotting lines, identifying feasible regions, and optimizing a linear objective function. All techniques are algorithmic with no novel insight required, making it easier than average for A-level, though the multi-part structure and 11 marks indicate reasonable length.
Spec7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables

\includegraphics{figure_6} Keith organises two types of children's activity, 'Sports Mad' and 'Circus Fun'. He needs to determine the number of times each type of activity is to be offered. Let \(x\) be the number of times he offers the 'Sports Mad' activity. Let \(y\) be the number of times he offers the 'Circus Fun' activity. Two constraints are $$x \leq 15$$ and $$y > 6$$ These constraints are shown on the graph in Figure 6, where the rejected regions are shaded out.
  1. Explain why \(y = 6\) is shown as a dotted line. [1] Two further constraints are $$3x \geq 2y$$ and $$5x + 4y \geq 80$$
  2. Add two lines and shading to Diagram 1 in the answer book to represent these inequalities. Hence determine the feasible region and label it R. [3] Each 'Sports Mad' activity costs £500. Each 'Circus Fun' activity costs £800. Keith wishes to minimise the total cost.
  3. Write down the objective function, C, in terms of \(x\) and \(y\). [2]
  4. Use your graph to determine the number of times each type of activity should be offered and the total cost. You must show sufficient working to make your method clear. [5]
(Total 11 marks)

Part (a)
AnswerMarks Guidance
AnswerMarks Guidance
To indicate the strict inequalityB1
Total: 1 mark
Part (b)
AnswerMarks Guidance
AnswerMarks Guidance
\(3x = 2y\) and \(5x + 4y = 80\) added to the diagram.<br>R correctly labelled.B1, B1 B1
Total: 3 marks
Part (c)
AnswerMarks Guidance
AnswerMarks Guidance
[Minimise C =] \(500x + 800y\)B1, B1
Total: 2 marks
Part (d)
AnswerMarks Guidance
AnswerMarks Guidance
Point testing or Profit line<br>Seeking integer solutions<br>(11, 7) at a cost of £ 11 100.M1 A1 M1 B1, B1 M1: Profit line [gradient accept reciprocal, minimum length line passes through (0, 2.5) (4, 0)] OR testing 2 points in their FR near two different vertices.<br>1A1: Correct profit line OR 2 points correctly tested in correct FR (my points)<br>e.g. \((7\frac{7}{11},10\frac{10}{11})=12 363\frac{7}{11}\) or \((7,11)=12 300\)<br>\((8,10)=12 000\) or \((8,11)=12 800\)<br>\((11\frac{1}{4},6)=10 400\) or \((11,6)=10 300\)<br>\((15,6)=12 300\) or \((15,7)=13 100\)<br>\((15,22\frac{1}{2})=25 500\) or \((15,22)=25 100\)<br>\((11,7)=11100\)<br><br>2M1: Seeking integer solution in correct FR (so therefore no \(y = 6\) points)<br>1B1: (11,7) CAO<br>2B1: £11 100 CAO
Total: 5 marks
Grand Total for Q7: 11 marks
## Part (a)

| Answer | Marks | Guidance |
|--------|-------|----------|
| To indicate the strict inequality | B1 | |

**Total: 1 mark**

## Part (b)

| Answer | Marks | Guidance |
|--------|-------|----------|
| $3x = 2y$ and $5x + 4y = 80$ added to the diagram.<br>R correctly labelled. | B1, B1 B1 | |

**Total: 3 marks**

## Part (c)

| Answer | Marks | Guidance |
|--------|-------|----------|
| [Minimise C =] $500x + 800y$ | B1, B1 | |

**Total: 2 marks**

## Part (d)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Point testing or Profit line<br>Seeking integer solutions<br>(11, 7) at a cost of £ 11 100. | M1 A1 M1 B1, B1 | M1: Profit line [gradient accept reciprocal, minimum length line passes through (0, 2.5) (4, 0)] OR testing 2 points in their FR near two different vertices.<br>1A1: Correct profit line OR 2 points correctly tested in correct FR (my points)<br>e.g. $(7\frac{7}{11},10\frac{10}{11})=12 363\frac{7}{11}$ or $(7,11)=12 300$<br>$(8,10)=12 000$ or $(8,11)=12 800$<br>$(11\frac{1}{4},6)=10 400$ or $(11,6)=10 300$<br>$(15,6)=12 300$ or $(15,7)=13 100$<br>$(15,22\frac{1}{2})=25 500$ or $(15,22)=25 100$<br>$(11,7)=11100$<br><br>2M1: Seeking integer solution in correct FR (so therefore no $y = 6$ points)<br>1B1: (11,7) CAO<br>2B1: £11 100 CAO |

**Total: 5 marks**

**Grand Total for Q7: 11 marks**

---
\includegraphics{figure_6}

Keith organises two types of children's activity, 'Sports Mad' and 'Circus Fun'.
He needs to determine the number of times each type of activity is to be offered.

Let $x$ be the number of times he offers the 'Sports Mad' activity. Let $y$ be the number of times he offers the 'Circus Fun' activity.

Two constraints are
$$x \leq 15$$
and $$y > 6$$

These constraints are shown on the graph in Figure 6, where the rejected regions are shaded out.

\begin{enumerate}[label=(\alph*)]
\item Explain why $y = 6$ is shown as a dotted line.
[1]

Two further constraints are
$$3x \geq 2y$$
and $$5x + 4y \geq 80$$

\item Add two lines and shading to Diagram 1 in the answer book to represent these inequalities.
Hence determine the feasible region and label it R.
[3]

Each 'Sports Mad' activity costs £500.
Each 'Circus Fun' activity costs £800.

Keith wishes to minimise the total cost.

\item Write down the objective function, C, in terms of $x$ and $y$.
[2]

\item Use your graph to determine the number of times each type of activity should be offered and the total cost. You must show sufficient working to make your method clear.
[5]
\end{enumerate}

(Total 11 marks)

\hfill \mbox{\textit{Edexcel D1 2010 Q7 [11]}}