| Exam Board | OCR |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Constraint derivation verification |
| Difficulty | Moderate -0.5 This is a standard linear programming question requiring routine constraint formulation from word problems, graph plotting, and optimization. The constraint derivation is straightforward arithmetic (fabric cost: 8x×8 + 3y×8 ≤ 600 simplifies to 4x+3y≤20), and the remaining parts follow textbook LP procedures with no novel problem-solving required. Slightly easier than average due to simple numbers and standard D1 format. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| Each dress takes 20 hours, each outfit 15 hours; total hours \(\leq 100\): \(20x + 15y \leq 100 \Rightarrow 4x + 3y \leq 20\) | B1 | Derivation shown |
| Material cost: \((8\times8 + 35)x + (8\times3+80)y \leq 600 \Rightarrow 99x + 104y \leq 600\) | B1 | |
| \(x \geq 1\) | B1 | |
| \(y \geq 1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Correct lines drawn for all four constraints | M1 M1 | One mark each for correct boundary lines |
| Feasible region correctly identified and shaded | A1 A1 |
| Answer | Marks |
|---|---|
| Maximum of \(x + y\) read from graph = 6 (or as appropriate from feasible region vertex) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total cost \(= (8\times8+35+20\times15)x + (8\times3+80+15\times15)y = (64+35+300)x+(24+80+225)y\) | M1 | |
| \(= 399x + 329y\) | A1 | Simplified expression |
| Evaluate at vertices of feasible region from (iii); find minimum cost solution | A1 | With interpretation |
# Question 6:
**(i)**
| Each dress takes 20 hours, each outfit 15 hours; total hours $\leq 100$: $20x + 15y \leq 100 \Rightarrow 4x + 3y \leq 20$ | B1 | Derivation shown |
| Material cost: $(8\times8 + 35)x + (8\times3+80)y \leq 600 \Rightarrow 99x + 104y \leq 600$ | B1 | |
| $x \geq 1$ | B1 | |
| $y \geq 1$ | B1 | |
**(ii)**
| Correct lines drawn for all four constraints | M1 M1 | One mark each for correct boundary lines |
| Feasible region correctly identified and shaded | A1 A1 | |
**(iii)**
| Maximum of $x + y$ read from graph = **6** (or as appropriate from feasible region vertex) | B1 | |
**(iv)**
| Total cost $= (8\times8+35+20\times15)x + (8\times3+80+15\times15)y = (64+35+300)x+(24+80+225)y$ | M1 | |
| $= 399x + 329y$ | A1 | Simplified expression |
| Evaluate at vertices of feasible region from (iii); find minimum cost solution | A1 | With interpretation |6 William is making the bridesmaid dresses and pageboy outfits for his sister's wedding. He expects it to take 20 hours to make each bridesmaid dress and 15 hours to make each pageboy outfit. Each bridesmaid dress uses 8 metres of fabric. Each pageboy outfit uses 3 metres of fabric. The fabric costs $\pounds 8$ per metre. Additional items cost $\pounds 35$ for each bridesmaid dress and $\pounds 80$ for each pageboy outfit.
William's sister wants to have at least one bridesmaid and at least one pageboy. William has 100 hours available and must not spend more than $\pounds 600$ in total on materials.
Let $x$ denote the number of bridesmaids and $y$ denote the number of pageboys.\\
\begin{enumerate}[label=(\roman*)]
\item Show why the constraint $4 x + 3 y \leqslant 20$ is needed and write down three more constraints on the values of $x$ and $y$, other than that they must be integers.
\item Plot the feasible region where all four constraints are satisfied.
William's sister wants to maximise the total number of attendants (bridesmaids and pageboys) at her wedding.
\item Use your graph to find the maximum number of attendants.
\item William costs his time at $\pounds 15$ an hour. Find, and simplify, an expression, in terms of $x$ and $y$, for the total cost (for all materials and William's time). Hence find, and interpret, the minimum cost solution to part (iii).
\end{enumerate}
\hfill \mbox{\textit{OCR D1 2016 Q6 [12]}}