| Exam Board | Edexcel |
|---|---|
| Module | FD1 AS (Further Decision 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Reverse engineering objective from solution |
| Difficulty | Standard +0.8 This question requires reverse-engineering an objective function from a given optimal value and graphical constraints, involving multiple steps: reading constraints from a graph, finding intersection points algebraically, and working backwards from the optimal value to derive the objective function. While the individual techniques are standard A-level material, the reverse-engineering aspect and multi-stage reasoning elevate it above typical linear programming questions. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x + y \leqslant 14\) | M1 | One correct non-trivial inequality; condone strict inequality; must be simplified to three terms |
| \(2y - x \leqslant 12\) | A1 | Two correct non-trivial inequalities |
| \(3x - y \leqslant 15\) | A1 | All three non-trivial inequalities with integer coefficients |
| \((x \geqslant 0, y \geqslant 0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to solve two equations to find optimal vertex | M1 | Solve \(x+y=14\) and \(2y-x=12\) simultaneously |
| \(\left(\dfrac{16}{3}, \dfrac{26}{3}\right)\) | A1 | Must be exact; clearly stated as optimal vertex |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = k(4x + 10y)\) | M1 | Expression comprising constant multiple/factor of \(2x+5y\) |
| \(216 = k\left(4 \times \dfrac{16}{3} + 10 \times \dfrac{26}{3}\right)\) | ddM1 | Dependent on both previous M marks; forming equation with \(k(4x+10y)\), the 216, and optimal vertex |
| \((P =) 8x + 20y\) | A1 | cao — accept \(8x+20y\); not \(8x+20y=0\) or \(216\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 6 small (flower pots) and 8 large (flower pots) | B1 | Not for \((6,8)\) or \(x=6, y=8\) — must be in context |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x + y \leqslant 14$ | M1 | One correct non-trivial inequality; condone strict inequality; must be simplified to three terms |
| $2y - x \leqslant 12$ | A1 | Two correct non-trivial inequalities |
| $3x - y \leqslant 15$ | A1 | All three non-trivial inequalities with integer coefficients |
| $(x \geqslant 0, y \geqslant 0)$ | | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to solve two equations to find optimal vertex | M1 | Solve $x+y=14$ and $2y-x=12$ simultaneously |
| $\left(\dfrac{16}{3}, \dfrac{26}{3}\right)$ | A1 | Must be exact; clearly stated as optimal vertex |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = k(4x + 10y)$ | M1 | Expression comprising constant multiple/factor of $2x+5y$ |
| $216 = k\left(4 \times \dfrac{16}{3} + 10 \times \dfrac{26}{3}\right)$ | ddM1 | Dependent on both previous M marks; forming equation with $k(4x+10y)$, the 216, and optimal vertex |
| $(P =) 8x + 20y$ | A1 | cao — accept $8x+20y$; not $8x+20y=0$ or $216$ |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 6 small (flower pots) and 8 large (flower pots) | B1 | Not for $(6,8)$ or $x=6, y=8$ — must be in context |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c8134d3b-71cb-4b92-ac54-81a4ff8f3011-06_1504_1733_210_173}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the constraints of a maximisation linear programming problem in $x$ and $y$, where $x \geqslant 0$ and $y \geqslant 0$. The unshaded area, including its boundaries, forms the feasible region, $R$. An objective line has been drawn and labelled on the graph.
\begin{enumerate}[label=(\alph*)]
\item List the constraints as simplified inequalities with integer coefficients.
The optimal value of the objective function is 216
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the exact coordinates of the optimal vertex.
\item Hence derive the objective function.
Given that $x$ represents the number of small flower pots and $y$ represents the number of large flower pots supplied to a customer,
\end{enumerate}\item deduce the optimal solution to the problem.
TOTAL FOR DECISION MATHEMATICS 1 IS 40 MARKS END
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD1 AS 2022 Q4 [9]}}