| Exam Board | Edexcel |
|---|---|
| Module | FD1 AS (Further Decision 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Reverse engineering objective from solution |
| Difficulty | Challenging +1.2 This question requires reading constraints from a graph (routine skill) and reverse-engineering an objective function from given max/min values at vertices. While the reverse direction is less common than standard optimization, the algebraic work is straightforward: substitute two vertex coordinates into P=ax+by and solve simultaneous equations. The fractional answers suggest non-integer coefficients but the method is mechanical once vertices are identified. |
| Spec | 7.06a LP formulation: variables, constraints, objective function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x \leqslant 8\), \(3x + 8y \geqslant 32\), \(3y \leqslant 4x + 5\), \(4x + 15y \leqslant 120\) | B2, 1, 0 | B1: Any two correct inequalities. Condone strict inequalities. B2: All four correct inequalities (not strict) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to solve correct two equations to find either optimal vertex | M1 | Considering either pair: \(3y = 4x+5, 3x+8y=32\) or \(x=8, 4x+15y=120\) |
| Coordinates of 'minimum' point is \(\left(\frac{56}{41}, \frac{143}{41}\right)\) | A1 | cao — must be seen exact at some point. Do not need to associate with 'minimum' |
| Coordinates of 'maximum' point is \(\left(8, \frac{88}{15}\right)\) | A1 | cao — must be seen exact at some point. Do not need to associate with 'maximum' |
| Setting up pair of simultaneous equations using two points and objective function of form \(ax + by\): \(56a + 143b = 883\) and \(15a + 11b = 100\) (oe) | dM1 | Must use solution of \(3y=4x+5, 3x+8y=32\) together with \(\frac{883}{41}\) and solution of \(x=8, 4x+15y=120\) together with \(\frac{160}{3}\) |
| Objective function is \((P =) 3x + 5y\) | A1 | Allow equal to any other letter but not a value. \(3x+5y=0\) is A0. Allow exact expression only, not any multiple or factor |
## Question 4:
### Part 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x \leqslant 8$, $3x + 8y \geqslant 32$, $3y \leqslant 4x + 5$, $4x + 15y \leqslant 120$ | B2, 1, 0 | B1: Any two correct inequalities. Condone strict inequalities. B2: All four correct inequalities (not strict) |
### Part 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to solve correct two equations to find either optimal vertex | M1 | Considering either pair: $3y = 4x+5, 3x+8y=32$ **or** $x=8, 4x+15y=120$ |
| Coordinates of 'minimum' point is $\left(\frac{56}{41}, \frac{143}{41}\right)$ | A1 | cao — must be seen exact at some point. Do not need to associate with 'minimum' |
| Coordinates of 'maximum' point is $\left(8, \frac{88}{15}\right)$ | A1 | cao — must be seen exact at some point. Do not need to associate with 'maximum' |
| Setting up pair of simultaneous equations using two points and objective function of form $ax + by$: $56a + 143b = 883$ and $15a + 11b = 100$ (oe) | dM1 | Must use solution of $3y=4x+5, 3x+8y=32$ together with $\frac{883}{41}$ **and** solution of $x=8, 4x+15y=120$ together with $\frac{160}{3}$ |
| Objective function is $(P =) 3x + 5y$ | A1 | Allow equal to any other letter but not a value. $3x+5y=0$ is **A0**. Allow exact expression only, not any multiple or factor |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9edb5209-4244-4916-b3ee-d77e395e8cab-05_997_1379_260_456}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the constraints of a linear programming problem in $x$ and $y$. 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 State the inequalities that define the feasible region.
The maximum value of the objective function is $\frac { 160 } { 3 }$
The minimum value of the objective function is $\frac { 883 } { 41 }$
\item Determine the objective function, showing your working clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD1 AS 2023 Q4 [7]}}