| Exam Board | AQA |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Graphical optimization with objective line |
| Difficulty | Moderate -0.8 This is a standard textbook linear programming question requiring routine graphical methods: plotting constraints, identifying the feasible region, and using objective lines to find optimal vertices. The constraints are straightforward, vertices are at integer coordinates, and the method is algorithmic with no problem-solving insight required. Easier than average A-level content. |
| Spec | 7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Line \(x + y = 60\) drawn correctly | B1 | Straight line through \((60,0)\) and \((0,60)\) |
| Line \(2x + y = 80\) drawn correctly | B1 | Straight line through \((40,0)\) and \((0,80)\) — note \((0,80)\) is off grid |
| Lines \(y = 20\), \(x = 15\), \(y = x\) all drawn correctly | B1 | All three lines correct |
| Correct region indicated/shaded | B2 | B1 for partially correct region |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Maximum at vertex \((15, 45)\) | M1 | Testing objective function at vertices of feasible region |
| \(P = 15 + 4(45) = 195\) | A1 | \(x = 15\), \(y = 45\) stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct method — testing vertices or using objective line | M1 | |
| Identifies correct vertex \((40, 20)\) | A1 | |
| \(P = 4(40) + 20 = 180\) | A1 | \(x = 40\), \(y = 20\) stated |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Line $x + y = 60$ drawn correctly | B1 | Straight line through $(60,0)$ and $(0,60)$ |
| Line $2x + y = 80$ drawn correctly | B1 | Straight line through $(40,0)$ and $(0,80)$ — note $(0,80)$ is off grid |
| Lines $y = 20$, $x = 15$, $y = x$ all drawn correctly | B1 | All three lines correct |
| Correct region indicated/shaded | B2 | B1 for partially correct region |
## Part (b)(i) $P = x + 4y$
| Answer | Marks | Guidance |
|--------|-------|----------|
| Maximum at vertex $(15, 45)$ | M1 | Testing objective function at vertices of feasible region |
| $P = 15 + 4(45) = 195$ | A1 | $x = 15$, $y = 45$ stated |
## Part (b)(ii) $P = 4x + y$
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct method — testing vertices or using objective line | M1 | |
| Identifies correct vertex $(40, 20)$ | A1 | |
| $P = 4(40) + 20 = 180$ | A1 | $x = 40$, $y = 20$ stated |
---5 The feasible region of a linear programming problem is defined by
$$\begin{aligned}
x + y & \leqslant 60 \\
2 x + y & \leqslant 80 \\
y & \geqslant 20 \\
x & \geqslant 15 \\
y & \geqslant x
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item On the grid opposite, draw a suitable diagram to represent these inequalities and indicate the feasible region.
\item In each of the following cases, use your diagram to find the maximum value of $P$ on the feasible region. In each case, state the corresponding values of $x$ and $y$.
\begin{enumerate}[label=(\roman*)]
\item $P = x + 4 y$
\item $P = 4 x + y$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D1 2013 Q5 [10]}}