| Exam Board | OCR |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Reverse engineering constraints from graph |
| Difficulty | Moderate -0.8 This is a routine linear programming question requiring standard techniques: reading constraints from a graph, finding vertices by solving simultaneous equations, and evaluating an objective function at vertices. Part (v) adds mild algebraic manipulation but no conceptual challenge. Easier than average A-level as it's mostly procedural with clear visual guidance. |
| Spec | 7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x \leq 2\) | B1 | Strict inequalities used, penalise first time only |
| \(y \geq 1\) | B1 | |
| \(y \leq 2x\) | B1 | All inequalities reversed, penalise first time only |
| \(x + y \leq 4\) | B1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2, 1),\ (2, 2)\) | B1 | Both of these |
| \(\left(\frac{1}{2}, 1\right)\) | B1 | This vertex in any exact form |
| \(\left(1\tfrac{1}{3},\ 2\tfrac{2}{3}\right)\) | B1 [3] | This vertex in any exact form or correct to 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Check \(P = x + 2y\) at vertices: \((2,1)\to4\); \((2,2)\to6\); \(\left(\frac{1}{2},1\right)\to2\tfrac{1}{2}\); \(\left(1\tfrac{1}{3},2\tfrac{2}{3}\right)\to6\tfrac{2}{3}\) | M1 | Evidence of checking value at any vertex or using a sliding profit line |
| \(x = 1\tfrac{1}{3},\ y = 2\tfrac{2}{3}\) (may be given in coordinate form) | A1 | Their \(x\) and \(y\) values at maximum in any exact form or correct to 3 sf |
| \(P = 6\tfrac{2}{3}\) | A1 [3] | Their maximum \(P\) value in any exact form or correct to 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Check \(Q = 2x - y\) at vertices: \((2,1)\to3\); \((2,2)\to2\); \(\left(\frac{1}{2},1\right)\to0\); \(\left(1\tfrac{1}{3},2\tfrac{2}{3}\right)\to0\) | M1 | Evidence of checking value at any vertex or using a sliding profit line |
| \(Q = 0\) | A1 | 0 (cao) |
| \((x, y)\) can be any point on the line segment joining \(\left(\frac{1}{2}, 1\right)\) and \(\left(1\tfrac{1}{3},\ 2\tfrac{2}{3}\right)\) | A1 [3] | The edge of the feasible region where \(y = 2x\); No follow through |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = Q \Rightarrow 2x - y = x + 2y \Rightarrow x = 3y\) | M1, A1 | For considering \(P = Q\), or equivalent; For this line, or any equivalent reasoning |
| \(y = \tfrac{1}{3}x\) lies entirely in the shaded region | A1 [3] | For explanation of why there are no solutions |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x \leq 2$ | B1 | Strict inequalities used, penalise first time only |
| $y \geq 1$ | B1 | |
| $y \leq 2x$ | B1 | All inequalities reversed, penalise first time only |
| $x + y \leq 4$ | B1 [4] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2, 1),\ (2, 2)$ | B1 | Both of these |
| $\left(\frac{1}{2}, 1\right)$ | B1 | This vertex in any exact form |
| $\left(1\tfrac{1}{3},\ 2\tfrac{2}{3}\right)$ | B1 [3] | This vertex in any exact form or correct to 3 sf |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Check $P = x + 2y$ at vertices: $(2,1)\to4$; $(2,2)\to6$; $\left(\frac{1}{2},1\right)\to2\tfrac{1}{2}$; $\left(1\tfrac{1}{3},2\tfrac{2}{3}\right)\to6\tfrac{2}{3}$ | M1 | Evidence of checking value at any vertex or using a sliding profit line |
| $x = 1\tfrac{1}{3},\ y = 2\tfrac{2}{3}$ (may be given in coordinate form) | A1 | Their $x$ and $y$ values at maximum in any exact form or correct to 3 sf |
| $P = 6\tfrac{2}{3}$ | A1 [3] | Their maximum $P$ value in any exact form or correct to 3 sf |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Check $Q = 2x - y$ at vertices: $(2,1)\to3$; $(2,2)\to2$; $\left(\frac{1}{2},1\right)\to0$; $\left(1\tfrac{1}{3},2\tfrac{2}{3}\right)\to0$ | M1 | Evidence of checking value at any vertex or using a sliding profit line |
| $Q = 0$ | A1 | 0 (cao) |
| $(x, y)$ can be any point on the **line segment** joining $\left(\frac{1}{2}, 1\right)$ and $\left(1\tfrac{1}{3},\ 2\tfrac{2}{3}\right)$ | A1 [3] | The edge of the feasible region where $y = 2x$; No follow through |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = Q \Rightarrow 2x - y = x + 2y \Rightarrow x = 3y$ | M1, A1 | For considering $P = Q$, or equivalent; For this line, or any equivalent reasoning |
| $y = \tfrac{1}{3}x$ lies entirely in the shaded region | A1 [3] | For explanation of why there are no solutions |
---4 The constraints of a linear programming problem are represented by the graph below. The feasible region is the unshaded region, including its boundaries.\\
\includegraphics[max width=\textwidth, alt={}, center]{f2b85dfb-49df-4ea5-b118-9b95f0b27bad-03_1025_826_374_657}\\
(i) Write down inequalities that define the feasible region.\\
(ii) Find the coordinates of the four vertices of the feasible region.
The objective is to maximise $P$, where $P = x + 2 y$.\\
(iii) Find the values of $x$ and $y$ that maximise $P$, and the corresponding maximum value of $P$.
The objective is changed to minimise $Q$, where $Q = 2 x - y$.\\
(iv) Find the minimum value of $Q$ and describe the set of feasible points for which $Q$ takes this value.\\
(v) Show that there are no points in the feasible region for which the value of $P$ is the same as the value of $Q$.
\hfill \mbox{\textit{OCR D1 2006 Q4 [16]}}