| Exam Board | AQA |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Three-variable constraint reduction |
| Difficulty | Moderate -0.8 This is a standard linear programming question with routine constraint formulation and graphical solution. Part (b) involves straightforward algebraic substitution to reduce from 3 to 2 variables, and parts (c)-(e) follow the typical D1 graphical method. While multi-step, it requires only textbook techniques with no novel insight or problem-solving beyond applying the standard algorithm. |
| 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 | Mark | Guidance |
| \(400x + 400y + 600z \leqslant 130000\) (wheat) | B1 | Or equivalent simplified form |
| \(200x + 500y + 200z \leqslant 70000\) (maize) | B1 | |
| \(400x + 100y + 200z \leqslant 72000\) (barley) | B1 | |
| \(z \geqslant 75\) (Supreme constraint) | B1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(z = \frac{x+y+z}{2}\), so \(x + y = z\) | M1 | Show substitution of \(z = x+y\) into inequalities |
| Correct simplification to given inequalities | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P = 50x + 100y + 150z\) with \(z = x+y\) | M1 | |
| \(P = 200x + 250y\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct objective line drawn | M1 | |
| \(x = 0\), \(y = 100\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Maximum profit \(= £25000\) | B1 | |
| Basic \(= 0\), Premium \(= 100\), Supreme \(= 100\) tonnes | B1 | 2 marks |
## Question 9:
**(a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $400x + 400y + 600z \leqslant 130000$ (wheat) | B1 | Or equivalent simplified form |
| $200x + 500y + 200z \leqslant 70000$ (maize) | B1 | |
| $400x + 100y + 200z \leqslant 72000$ (barley) | B1 | |
| $z \geqslant 75$ (Supreme constraint) | B1 | 4 marks |
**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $z = \frac{x+y+z}{2}$, so $x + y = z$ | M1 | Show substitution of $z = x+y$ into inequalities |
| Correct simplification to given inequalities | A1 | 2 marks |
**(d)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P = 50x + 100y + 150z$ with $z = x+y$ | M1 | |
| $P = 200x + 250y$ | A1 | 2 marks |
**(e)(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct objective line drawn | M1 | |
| $x = 0$, $y = 100$ | A1 | 2 marks |
**(e)(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Maximum profit $= £25000$ | B1 | |
| Basic $= 0$, Premium $= 100$, Supreme $= 100$ tonnes | B1 | 2 marks |
These pages (22, 23, and 24) are **answer space pages** from an AQA exam paper (P/Jun15/MD01). They contain:
- Blank lined answer spaces for **Question 9**
- A final blank page stating "There are no questions printed on this page"
**There is no mark scheme content on these pages.** These are student answer booklet pages, not mark scheme pages. They contain no questions, worked solutions, mark allocations, or examiner guidance.
To find the mark scheme for this paper (AQA MD01, June 2015), you would need to locate the separate official mark scheme document published by AQA.9 A company producing chicken food makes three products, Basic, Premium and Supreme, from wheat, maize and barley.
A tonne $( 1000 \mathrm {~kg} )$ of Basic uses 400 kg of wheat, 200 kg of maize and 400 kg of barley.\\
A tonne of Premium uses 400 kg of wheat, 500 kg of maize and 100 kg of barley.\\
A tonne of Supreme uses 600 kg of wheat, 200 kg of maize and 200 kg of barley.\\
The company has 130 tonnes of wheat, 70 tonnes of maize and 72 tonnes of barley available.
The company must make at least 75 tonnes of Supreme.\\
The company makes $\pounds 50$ profit per tonne of Basic, $\pounds 100$ per tonne of Premium and $\pounds 150$ per tonne of Supreme.
They plan to make $x$ tonnes of Basic, $y$ tonnes of Premium and $z$ tonnes of Supreme.
\begin{enumerate}[label=(\alph*)]
\item Write down four inequalities representing the constraints (in addition to $x , y \geqslant 0$ ).\\[0pt]
[4 marks]
\item The company want exactly half the production to be Supreme.
Show that the constraints in part (a) become
$$\begin{aligned}
x + y & \leqslant 130 \\
4 x + 7 y & \leqslant 700 \\
2 x + y & \leqslant 240 \\
x + y & \geqslant 75 \\
x & \geqslant 0 \\
y & \geqslant 0
\end{aligned}$$
\item On the grid opposite, illustrate all the constraints and label the feasible region.
\item Write an expression for $P$, the profit for the whole production, in terms of $x$ and $y$ only.\\[0pt]
[2 marks]
\item \begin{enumerate}[label=(\roman*)]
\item By drawing an objective line on your graph, or otherwise, find the values of $x$ and $y$ which give the maximum profit.\\[0pt]
[2 marks]
\item State the maximum profit and the amount of each product that must be made.\\[0pt]
[2 marks]
\section*{Answer space for question 9}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f5890e58-38c3-413c-8762-6f80ce6dcec7-21_1349_1728_310_148}
\end{center}
QUESTION\\
PART\\
REFERENCE\\
\includegraphics[max width=\textwidth, alt={}, center]{f5890e58-38c3-413c-8762-6f80ce6dcec7-24_2488_1728_219_141}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D1 2015 Q9 [17]}}