| Exam Board | AQA |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Dual objective optimization |
| Difficulty | Moderate -0.8 This is a standard linear programming question requiring graphing of constraints and evaluating objective functions at vertices. The mechanics are routine for D1: plot lines, identify feasible region, test corner points. While it has multiple parts, each follows the same algorithmic procedure with no conceptual challenges or novel problem-solving required. |
| Spec | 7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 1\) drawn correctly | B1 | Vertical line at \(x=1\) |
| \(y = 3\) drawn correctly | B1 | Horizontal line at \(y=3\) |
| \(x + y = 5\) drawn correctly | B1 | Line through \((2,3)\) and \((5,0)\) etc |
| \(x + y = 12\) drawn correctly | B1 | Line through \((4,8)\), \((9,3)\) etc |
| \(3x + 8y = 64\) drawn correctly | B1 | Line through \((8,5)\), \((0,8)\) etc |
| Feasible region correctly identified/shaded | B1 | All lines correct and region indicated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (i) Maximum of \(3x + y\) | ||
| Correct vertex identified, e.g. \((9, 3)\) | M1 | Must be a vertex of feasible region |
| Maximum value = \(30\), at \((9,3)\) | A1 | |
| (ii) Maximum of \(2x + 3y\) | ||
| Correct vertex identified | M1 | Must be a vertex of feasible region |
| Maximum value = \(32\), at \((8,\frac{16}{3})\) or \((1,10\frac{2}{3})\) accept \((1,11)\) approximately | A1 | |
| (iii) Minimum of \(-2x + y\) | ||
| Correct vertex identified | M1 | Must be a vertex of feasible region |
| Minimum value = \(-15\), at \((9,3)\) | A1 |
# Question 5:
## Part (a) - Drawing the feasible region [5 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 1$ drawn correctly | B1 | Vertical line at $x=1$ |
| $y = 3$ drawn correctly | B1 | Horizontal line at $y=3$ |
| $x + y = 5$ drawn correctly | B1 | Line through $(2,3)$ and $(5,0)$ etc |
| $x + y = 12$ drawn correctly | B1 | Line through $(4,8)$, $(9,3)$ etc |
| $3x + 8y = 64$ drawn correctly | B1 | Line through $(8,5)$, $(0,8)$ etc |
| Feasible region correctly identified/shaded | B1 | All lines correct and region indicated |
## Part (b) - Optimisation [6 marks]
| Answer | Mark | Guidance |
|--------|------|----------|
| **(i)** Maximum of $3x + y$ | | |
| Correct vertex identified, e.g. $(9, 3)$ | M1 | Must be a vertex of feasible region |
| Maximum value = $30$, at $(9,3)$ | A1 | |
| **(ii)** Maximum of $2x + 3y$ | | |
| Correct vertex identified | M1 | Must be a vertex of feasible region |
| Maximum value = $32$, at $(8,\frac{16}{3})$ or $(1,10\frac{2}{3})$ accept $(1,11)$ approximately | A1 | |
| **(iii)** Minimum of $-2x + y$ | | |
| Correct vertex identified | M1 | Must be a vertex of feasible region |
| Minimum value = $-15$, at $(9,3)$ | A1 | |
---5 The feasible region of a linear programming problem is determined by the following:
$$\begin{aligned}
x & \geqslant 1 \\
y & \geqslant 3 \\
x + y & \geqslant 5 \\
x + y & \leqslant 12 \\
3 x + 8 y & \leqslant 64
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item On the grid below, draw a suitable diagram to represent the inequalities and indicate the feasible region.
\item Use your diagram to find, on the feasible region:
\begin{enumerate}[label=(\roman*)]
\item the maximum value of $3 x + y$;
\item the maximum value of $2 x + 3 y$;
\item the minimum value of $- 2 x + y$.
In each case, state the coordinates of the point corresponding to your answer.\\[0pt]
[6 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D1 2014 Q5 [11]}}