| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Optimal vertex with additional constraint |
| Difficulty | Standard +0.3 This is a standard linear programming question requiring graphing feasible regions and finding optimal vertices, plus identifying when a constraint becomes redundant. While it involves multiple steps, the techniques are routine for D1 students—plotting lines, finding intersections, and testing vertices. The redundancy part requires slightly more insight but follows directly from examining which constraints are active at the optimum, making this slightly easier than average overall. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | B3 lines, B1 shading | |
| (ii) Intersection of \(2x + 5y = 60\) and \(x + y = 18\) is at \((10, 8)\). \(10 + 2 \times 8 = 26\) | M1, A1 |
**(i)** | | B3 lines, B1 shading |
**(ii)** Intersection of $2x + 5y = 60$ and $x + y = 18$ is at $(10, 8)$. $10 + 2 \times 8 = 26$ | M1, A1 |3 Consider the following linear programming problem:\\
Maximise $\quad 3 x + 4 y$\\
subject to $\quad 2 x + 5 y \leqslant 60$\\
$x + 2 y \leqslant 25$\\
$x + y \leqslant 18$\\
(i) Graph the inequalities and hence solve the LP.\\
(ii) The right-hand side of the second inequality is increased from 25 . At what new value will this inequality become redundant?
\hfill \mbox{\textit{OCR MEI D1 2009 Q3 [8]}}