| Exam Board | AQA |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Three-variable constraint reduction |
| Difficulty | Moderate -0.3 This is a standard D1 linear programming question requiring constraint formulation, variable substitution (z=y), graphical solution, and vertex testing. While multi-part with several steps, it follows the routine D1 template with no novel insights required—slightly easier than average due to the straightforward constraint reduction and typical graphical method application. |
| 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 | Marks | Guidance |
| \(6x + 4y + 3z \geq 420\) | B1 | Roses inequality |
| \(6x + 6y + 4z \geq 480\) | B1 | Carnations inequality |
| \(6x + 4y + 4z \leq 720\) | B1 | Dahlias inequality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(z = y\) into rose and carnation inequalities: \(6x + 4y + 3y \geq 420 \Rightarrow 6x + 7y \geq 420\) | M1 | Substituting \(z = y\) into at least one inequality |
| \(6x + 6y + 4y \geq 480 \Rightarrow 6x + 10y \geq 480 \Rightarrow 3x + 5y \geq 240\) and \(6x + 4y + 4y \leq 720 \Rightarrow 6x + 8y \leq 720 \Rightarrow 3x + 4y \leq 360\) | A1 | All three shown correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Three boundary lines drawn correctly | B3 | B1 each for \(6x+7y=420\), \(3x+5y=240\), \(3x+4y=360\) |
| Feasible region correctly identified and labelled | B1 | Correct region shaded/indicated |
| Objective function \(P = 4x + 5y\) stated or implied | B1 | Profit function with \(z=y\): \(4x+2.5y+2.5y\) |
| Direction of objective line indicated correctly | B1 | Arrow in correct direction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum at vertex of feasible region, intersection of \(6x+7y=420\) and \(3x+4y=360\) giving \(x=0\), \(y=60\) | M1 | Correct vertex identified from diagram |
| Maximum profit \(= £300\); \(x=0\) gold, \(y=60\) silver, \(z=60\) bronze bouquets | A1 | Correct values stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New objective: \(P = 2x + 6y + 6y = 2x + 12y\) (with \(z=y\)) | M1 | Correct objective function formed |
| Check vertices of feasible region | M1 | Evaluating objective at vertices |
| Maximum profit \(= £720\); \(x=0\), \(y=60\), \(z=60\) or appropriate vertex giving maximum | A1 | Correct maximum and bouquet numbers stated |
# Question 7:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6x + 4y + 3z \geq 420$ | B1 | Roses inequality |
| $6x + 6y + 4z \geq 480$ | B1 | Carnations inequality |
| $6x + 4y + 4z \leq 720$ | B1 | Dahlias inequality |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $z = y$ into rose and carnation inequalities: $6x + 4y + 3y \geq 420 \Rightarrow 6x + 7y \geq 420$ | M1 | Substituting $z = y$ into at least one inequality |
| $6x + 6y + 4y \geq 480 \Rightarrow 6x + 10y \geq 480 \Rightarrow 3x + 5y \geq 240$ and $6x + 4y + 4y \leq 720 \Rightarrow 6x + 8y \leq 720 \Rightarrow 3x + 4y \leq 360$ | A1 | All three shown correctly |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Three boundary lines drawn correctly | B3 | B1 each for $6x+7y=420$, $3x+5y=240$, $3x+4y=360$ |
| Feasible region correctly identified and labelled | B1 | Correct region shaded/indicated |
| Objective function $P = 4x + 5y$ stated or implied | B1 | Profit function with $z=y$: $4x+2.5y+2.5y$ |
| Direction of objective line indicated correctly | B1 | Arrow in correct direction |
## Part (b)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum at vertex of feasible region, intersection of $6x+7y=420$ and $3x+4y=360$ giving $x=0$, $y=60$ | M1 | Correct vertex identified from diagram |
| Maximum profit $= £300$; $x=0$ gold, $y=60$ silver, $z=60$ bronze bouquets | A1 | Correct values stated |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New objective: $P = 2x + 6y + 6y = 2x + 12y$ (with $z=y$) | M1 | Correct objective function formed |
| Check vertices of feasible region | M1 | Evaluating objective at vertices |
| Maximum profit $= £720$; $x=0$, $y=60$, $z=60$ or appropriate vertex giving maximum | A1 | Correct maximum and bouquet numbers stated |
These pages (22, 23, and 24) are **answer space pages** from the AQA June 2013 MD01 exam paper — they contain no mark scheme content whatsoever.
- Pages 22–23 are blank lined answer spaces for **Question 7**
- Page 24 is a blank "Do Not Write On This Page" page
To find the mark scheme for this paper (AQA June 2013 MD01 - Decision Mathematics 1), you would need to access the separate **mark scheme document**, which is available directly from **AQA's website** at aqa.org.uk under past papers for this qualification.7 Paul is a florist. Every day, he makes three types of floral bouquet: gold, silver and bronze.
Each gold bouquet has 6 roses, 6 carnations and 6 dahlias.\\
Each silver bouquet has 4 roses, 6 carnations and 4 dahlias.\\
Each bronze bouquet has 3 roses, 4 carnations and 4 dahlias.\\
Each day, Paul must use at least 420 roses and at least 480 carnations, but he can use at most 720 dahlias.
Each day, Paul makes $x$ gold bouquets, $y$ silver bouquets and $z$ bronze bouquets.
\begin{enumerate}[label=(\alph*)]
\item In addition to $x \geqslant 0 , y \geqslant 0$ and $z \geqslant 0$, find three inequalities in $x , y$ and $z$ that model the above constraints.
\item On a particular day, Paul makes the same number of silver bouquets as bronze bouquets.
\begin{enumerate}[label=(\roman*)]
\item Show that $x$ and $y$ must satisfy the following inequalities.
$$\begin{aligned}
& 6 x + 7 y \geqslant 420 \\
& 3 x + 5 y \geqslant 240 \\
& 3 x + 4 y \leqslant 360
\end{aligned}$$
\item Paul makes a profit of $\pounds 4$ on each gold bouquet sold, a profit of $\pounds 2.50$ on each silver bouquet sold and a profit of $\pounds 2.50$ on each bronze bouquet sold. Each day, Paul sells all the bouquets he makes. Paul wishes to maximise his daily profit, $\pounds P$.
Draw a suitable diagram, on the grid opposite, to enable this problem to be solved graphically, indicating the feasible region and the direction of the objective line.\\
(6 marks)
\item Use your diagram to find Paul's maximum daily profit and the number of each type of bouquet he must make to achieve this maximum.
\end{enumerate}\item On another day, Paul again makes the same number of silver bouquets as bronze bouquets, but he makes a profit of $\pounds 2$ on each gold bouquet sold, a profit of $\pounds 6$ on each silver bouquet sold and a profit of $\pounds 6$ on each bronze bouquet sold.
Find Paul's maximum daily profit, and the number of each type of bouquet he must make to achieve this maximum.\\
(3 marks)
Turn over -
\end{enumerate}
\hfill \mbox{\textit{AQA D1 2013 Q7 [16]}}