| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2021 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Formulation from word problem |
| Difficulty | Moderate -0.8 This is a straightforward linear programming formulation requiring translation of verbal constraints into inequalities. The constraints are clearly stated, the objective function is obvious (minimise 2x + 3y), and the algebraic manipulation needed (e.g., converting '80% large' to x ≥ y/4) is routine for D1 students. No problem-solving insight or multi-step reasoning required. |
| Spec | 7.06a LP formulation: variables, constraints, objective function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| Minimise \(C = 2x + 3y\) | B1 | Expression correct \((2x + 3y)\) together with 'minimise' or 'min' (but not 'minimum') – if 'simplified' e.g. \(x + 1.5y\) then must see \(2x + 3y\) at some point |
| \(x + y \geq 85\) | B1 | CAO – any equivalent form provided integer coefficients and only one term in \(x\) and one term in \(y\) e.g. \(x \geq 85 - y\) |
| \(y \geq 2x\) | M1 | \(y \square 2x\) where \(\square\) is any inequality or equals. Accept \(2y \geq x\) for this mark |
| \(y \leq \frac{4}{5}(x + y)\) | M1 | \(y \square \frac{4}{5}(x+y)\) where \(\square\) is any inequality or equals – if no bracket then correct rhs must be implied by later working. \(y \square 4x\) where \(\square\) is any inequality or equals implies this mark. Use of % symbol only is M0 unless correctly replaced by a fraction or decimal later |
| \(y \geq 2x\) and \(y \leq 4x\) | A1 | Both \(y \geq 2x\) and \(y \leq 4x\) CAO – must be single terms in \(x\) and \(y\) but allow any equivalent form provided integer coefficients e.g. \(2x - y \leq 0\), \(2y - 8x \leq 0\) etc. |
# Question 2:
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Minimise $C = 2x + 3y$ | B1 | Expression correct $(2x + 3y)$ together with 'minimise' or 'min' (but not 'minimum') – if 'simplified' e.g. $x + 1.5y$ then must see $2x + 3y$ at some point |
| $x + y \geq 85$ | B1 | CAO – any equivalent form provided integer coefficients and only one term in $x$ and one term in $y$ e.g. $x \geq 85 - y$ |
| $y \geq 2x$ | M1 | $y \square 2x$ where $\square$ is any inequality or equals. Accept $2y \geq x$ for this mark |
| $y \leq \frac{4}{5}(x + y)$ | M1 | $y \square \frac{4}{5}(x+y)$ where $\square$ is any inequality or equals – if no bracket then correct rhs must be implied by later working. $y \square 4x$ where $\square$ is any inequality or equals implies this mark. Use of % symbol only is M0 unless correctly replaced by a fraction or decimal later |
| $y \geq 2x$ **and** $y \leq 4x$ | A1 | Both $y \geq 2x$ **and** $y \leq 4x$ CAO – must be single terms in $x$ and $y$ but allow any equivalent form provided integer coefficients e.g. $2x - y \leq 0$, $2y - 8x \leq 0$ etc. |
**Total: 5 marks**
---2. A restaurant sells two sizes of pizza, small and large. The restaurant owner knows that, each evening, she needs to make
\begin{itemize}
\item at least 85 pizzas in total
\item at least twice as many large pizzas as small pizzas
\end{itemize}
In addition, at most $80 \%$ of the pizzas must be large.\\
Each small pizza costs $\pounds 2$ to make and each large pizza costs $\pounds 3$ to make.\\
The restaurant owner wants to minimise her costs.
Let $x$ represent the number of small pizzas made each evening and let $y$ represent the number of large pizzas made each evening.
Formulate the information above as a linear programming problem. State the objective and list the constraints as simplified inequalities with integer coefficients. You should not attempt to solve the problem.\\
\hfill \mbox{\textit{Edexcel D1 2021 Q2 [5]}}