Edexcel D1 2015 January — Question 6 12 marks

Exam BoardEdexcel
ModuleD1 (Decision Mathematics 1)
Year2015
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear Programming
TypeGraphical optimization with objective line
DifficultyModerate -0.8 This is a standard D1 linear programming question following a familiar template: formulate constraints, sketch feasible region, use objective line method. Parts (a)-(d) are routine textbook exercises requiring only recall of standard techniques. Part (e) adds a simple algebraic step but no conceptual challenge. Easier than average A-level maths overall.
Spec7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables

6. Jonathan is going to make hats to sell at a fete. He can make red hats and green hats. Jonathan can use linear programming to determine the number of each colour of hat that he should make. Let \(x\) be the number of red hats he makes and \(y\) be the number of green hats he makes.
One of the constraints is that there must be at least 30 hats.
  1. Write down an inequality, in terms of \(x\) and \(y\), to model this constraint. Two further constraints are $$\begin{aligned} & 2 y + x \geqslant 40 \\ & 2 y - x \geqslant - 30 \end{aligned}$$
  2. Write down two more constraints which apply.
  3. Represent all these constraints on Diagram 1 in the answer book. Hence determine, and label, the feasible region R . The cost of making a green hat is three times the cost of making a red hat. Jonathan wishes to minimise the total cost.
  4. Use the objective line (ruler) method to determine the number of red hats and number of green hats that Jonathan should make. You must clearly draw and label your objective line. Given that the minimum total cost of making the hats is \(\pounds 107.50\)
  5. determine the cost of making one green hat and the cost of making one red hat. You must make your method clear.

AnswerMarks Guidance
AnswerMarks Guidance
\(x + y \geq 30\)B1 (1)
\(x \geq 0, y \geq 0\)B1 (1)
Four lines correctly drawnB1 B1 B1 B1 (4)
Objective line drawn (\(m = -3\) or \(m = -\frac{1}{3}\)). (34,3) so 34 red hats and 3 green hatsM1 A1 A1 (3)
\(34r + 3g = 107.5\) \(g = 3r\) Leading to \(r = 2.50\) and \(g = 7.50\) So a red hat costs £2.50 and a green hat costs £7.50B1ft B1 DB1 (3)
12 marks
Notes for Question 6:
- a1B1: CAO (\(x + y \geq 30\)).
- b1B1: CAO (accept \(x, y \geq 0\) or \(x\) and \(y\) are non-negative) – do not accept strict inequalities.
In (c) lines must pass through one small square of the points stated:
- \(x + y = 30\) passes through (0, 30), (15, 15), (30, 0)
- \(2y + x = 40\) passes through (0, 20), (20, 10), (40, 0)
- \(2y – x = –30\) passes through (30, 0), (50, 10), (60, 15)
- c1B1: \(x + y = 30\) drawn correctly.
- c2B1: \(2y + x = 40\) drawn correctly.
- c3B1: \(2y – x = –30\) drawn correctly.
- c4B1: Region, R, correctly labelled – not just implied by shading - must have scored all three previous marks in this part. Condone lack of shading for \(x \geq 0\).
- d1M1: Drawing the correct objective line or its reciprocal (\(m = –3\) or \(m = –\frac{1}{3}\)). Line must be correct to within one small square if extended from axis to axis.
- d1A1: Correct point identified – accept as a coordinate (34, 3).
- d2A1: Correct answer – 34 red hats and 3 green hats (or 34r and 3g).
- e1B1ft: A 'correct' equation involving their optimal point from (d) (accept any values even if non-integer) and 107.50.
- e2B1: CAO on the relationship between the costs of green hats and red hats (\(g = 3r\)) – this mark may be implied e.g. \(34r + 3(3r) = 107.5\) would score the first two marks in this part.
- e3DB1: CAO – this mark is independent on having the correct optimal point (34, 3) in (d).
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + y \geq 30$ | B1 (1) | |
| $x \geq 0, y \geq 0$ | B1 (1) | |
| Four lines correctly drawn | B1 B1 B1 B1 (4) | |
| Objective line drawn ($m = -3$ or $m = -\frac{1}{3}$). (34,3) so 34 red hats and 3 green hats | M1 A1 A1 (3) | |
| $34r + 3g = 107.5$ $g = 3r$ Leading to $r = 2.50$ and $g = 7.50$ So a red hat costs £2.50 and a green hat costs £7.50 | B1ft B1 DB1 (3) | |
| | 12 marks | |

**Notes for Question 6:**

- a1B1: CAO ($x + y \geq 30$).
- b1B1: CAO (accept $x, y \geq 0$ or $x$ and $y$ are non-negative) – do not accept strict inequalities.

**In (c) lines must pass through one small square of the points stated:**

- $x + y = 30$ passes through (0, 30), (15, 15), (30, 0)
- $2y + x = 40$ passes through (0, 20), (20, 10), (40, 0)
- $2y – x = –30$ passes through (30, 0), (50, 10), (60, 15)

- c1B1: $x + y = 30$ drawn correctly.
- c2B1: $2y + x = 40$ drawn correctly.
- c3B1: $2y – x = –30$ drawn correctly.
- c4B1: Region, R, correctly labelled – not just implied by shading - must have scored all three previous marks in this part. Condone lack of shading for $x \geq 0$.
- d1M1: Drawing the correct objective line or its reciprocal ($m = –3$ or $m = –\frac{1}{3}$). Line must be correct to within one small square if extended from axis to axis.
- d1A1: Correct point identified – accept as a coordinate (34, 3).
- d2A1: Correct answer – 34 red hats and 3 green hats (or 34r and 3g).
- e1B1ft: A 'correct' equation involving their optimal point from (d) (accept any values even if non-integer) and 107.50.
- e2B1: CAO on the relationship between the costs of green hats and red hats ($g = 3r$) – this mark may be implied e.g. $34r + 3(3r) = 107.5$ would score the first two marks in this part.
- e3DB1: CAO – this mark is independent on having the correct optimal point (34, 3) in (d).

---
6. Jonathan is going to make hats to sell at a fete. He can make red hats and green hats. Jonathan can use linear programming to determine the number of each colour of hat that he should make.

Let $x$ be the number of red hats he makes and $y$ be the number of green hats he makes.\\
One of the constraints is that there must be at least 30 hats.
\begin{enumerate}[label=(\alph*)]
\item Write down an inequality, in terms of $x$ and $y$, to model this constraint.

Two further constraints are

$$\begin{aligned}
& 2 y + x \geqslant 40 \\
& 2 y - x \geqslant - 30
\end{aligned}$$
\item Write down two more constraints which apply.
\item Represent all these constraints on Diagram 1 in the answer book. Hence determine, and label, the feasible region R .

The cost of making a green hat is three times the cost of making a red hat. Jonathan wishes to minimise the total cost.
\item Use the objective line (ruler) method to determine the number of red hats and number of green hats that Jonathan should make. You must clearly draw and label your objective line.

Given that the minimum total cost of making the hats is $\pounds 107.50$
\item determine the cost of making one green hat and the cost of making one red hat. You must make your method clear.
\end{enumerate}

\hfill \mbox{\textit{Edexcel D1 2015 Q6 [12]}}