| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Graphical optimization with objective line |
| Difficulty | Moderate -0.3 This is a standard textbook linear programming question requiring routine application of well-defined techniques: formulating inequalities, graphing constraints, and finding the optimal vertex. Part (iv) adds mild complexity by requiring evaluation of whether to purchase extra resources, but the overall question follows a predictable template with no novel insight required. Slightly easier than average due to its mechanical nature and clear structure. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Let \(x\) be the number of (10s of) litres of stew and \(y\) the number of (10s of) litres of soup that Ian makes. | B1, B1 | "number of", referring to soup & stew; identification of soup and stew variables |
| Carrots: \(0.15x + 0.1y < 100\), i.e. \(3x + 2y < 2000\); Beans: \(0.1x + 0.075y < 70\), i.e. \(4x + 3y < 2800\); Tomatoes: \(0.15x + 0.15y < 110\), i.e. \(3x + 3y < 2200\) | B1, B1, B1 | -1 each scaling or systematic error, e.g. equalities |
| (ii) Intercepts are (666.7,0) and (0,1000); (700,0) and (0,933.3); (733.3,0) and (0,733.3) | B1 | axes consistently labelled and scaled |
| [Graph showing feasible region with axes labeled, broken axis scores 0 for 6(ii)] | B1, B1, B1, B1 | line 1; line 2; line 3 all ∨ subject to negative gradients shading giving feasible quadrilateral bounded by axes ... or identified by vertices |
| (iii) Line 2 irrelevant. Comparing at (0, 733.3), (533.3+10, 200±10) and (666.7, 0) (accuracy quoted is for graphical solutions). Max profit at intersection of lines 1 and 3 (533.33,200) with profit £3466.67 (accuracy from 3375 to 3560) (cf £3333.33 and £2933.33) | M1, A1, A1 | comparing 3 vertices (not origin) or profit line with approximately correct gradient (-5/4); stew and soup (cao); profit (cao) |
| Answer | Marks | Guidance |
|---|---|---|
| (iv) Best solution now at (0, 933.3) ... profit £3733.33 (£373.33); So best new solution uses 30 kg extra tomatoes (140 kg total) | M1, A1 | 30kg (allow 140 new total) cao |
| Extra profit is £(3733.33 – 3466.67 – 30*2.5) = £191.67 | A1 | (allow £3658.33 new total) cao |
**(i)** Let $x$ be the number of (10s of) litres of stew and $y$ the number of (10s of) litres of soup that Ian makes. | B1, B1 | "number of", referring to soup & stew; identification of soup and stew variables
Carrots: $0.15x + 0.1y < 100$, i.e. $3x + 2y < 2000$; Beans: $0.1x + 0.075y < 70$, i.e. $4x + 3y < 2800$; Tomatoes: $0.15x + 0.15y < 110$, i.e. $3x + 3y < 2200$ | B1, B1, B1 | -1 each scaling or systematic error, e.g. equalities
**(ii)** Intercepts are (666.7,0) and (0,1000); (700,0) and (0,933.3); (733.3,0) and (0,733.3) | B1 | axes consistently labelled and scaled
[Graph showing feasible region with axes labeled, broken axis scores 0 for 6(ii)] | B1, B1, B1, B1 | line 1; line 2; line 3 all ∨ subject to negative gradients shading giving feasible quadrilateral bounded by axes ... or identified by vertices
**(iii)** Line 2 irrelevant. Comparing at (0, 733.3), (533.3+10, 200±10) and (666.7, 0) (accuracy quoted is for graphical solutions). Max profit at intersection of lines 1 and 3 (533.33,200) with profit £3466.67 (accuracy from 3375 to 3560) (cf £3333.33 and £2933.33) | M1, A1, A1 | comparing 3 vertices (not origin) or profit line with approximately correct gradient (-5/4); stew and soup (cao); profit (cao)
So make 533.33 litres of stew and 200 litres of soup, giving a profit of £3466.67 (3375 – 3560).
**(iv)** Best solution now at (0, 933.3) ... profit £3733.33 (£373.33); So best new solution uses 30 kg extra tomatoes (140 kg total) | M1, A1 | 30kg (allow 140 new total) cao
Extra profit is £(3733.33 – 3466.67 – 30*2.5) = £191.67 | A1 | (allow £3658.33 new total) cao6 Ian the chef is to make vegetable stew and vegetable soup for distribution to a small chain of vegetarian restaurants. The recipes for both of these require carrots, beans and tomatoes.
10 litres of stew requires 1.5 kg of carrots, 1 kg of beans and 1.5 kg of tomatoes.\\
10 litres of soup requires 1 kg of carrots, 0.75 kg of beans and 1.5 kg of tomatoes.
Ian has available 100 kg of carrots, 70 kg of beans and 110 kg of tomatoes.\\
(i) Identify appropriate variables and write down three inequalities corresponding to the availabilities of carrots, beans and tomatoes.\\
(ii) Graph your inequalities and identify the region corresponding to feasible production plans.
The profit on a litre of stew is $\pounds 5$, and the profit on a litre of soup is $\pounds 4$.\\
(iii) Find the most profitable production plan, showing your working. Give the maximum profit.
Ian can buy in extra tomatoes at $\pounds 2.50$ per kg .\\
(iv) What extra quantity of tomatoes should Ian buy? How much extra profit would be generated by the extra expenditure?
\section*{END OF QUESTION PAPER}
\section*{OCR}
\hfill \mbox{\textit{OCR MEI D1 2014 Q6 [16]}}