| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Parametric objective analysis |
| Difficulty | Challenging +1.2 This question requires reading constraints from a graph (routine D1 skill), then using the given conditions about minimum/maximum values to determine a range for parameter a. While it involves parametric analysis and gradient comparison at vertices, it follows a standard D1 approach: comparing objective function gradients to constraint gradients. The multi-step nature and need to interpret conditions systematically makes it moderately above average, but it's still a recognizable D1 question type without requiring novel insight. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (a) \(4y \leq 7x + 8\); \(4y \geq x + 8\); \(3x + 4y \leq 24\) | B1 B1 (2) | a1B1: One correct inequality (allow strict inequality). a2B1: All three inequalities correct (allow any equivalent forms). |
| (b) Min value of \(P\) is \(8 \Rightarrow 2b = 8 \therefore b = 4\) | B1 (1) | b1B1: \(b = 4\) (only). |
| Solve \(4y = 7x + 8\) and \(3x + 4y = 24\) simultaneously (to give \(B\left(\frac{8}{5}, \frac{24}{5}\right)\)) | M1 | b1M1: Solve correct pair of simultaneous equations to find B – this mark can be implied by correct coordinates of B stated. |
| \(C(4,3) \Rightarrow P = 4a + 12, B\left(\frac{8}{5}, \frac{24}{5}\right) \Rightarrow P = \frac{8}{5}a + \frac{96}{5}\) | M1 | b2M1: Either linear expression in terms of \(a\) only (using their value of b) for either the correct C or their B (their B must be correct or a method for solving the correct simultaneous equations to find B must be seen). |
| \(4a + 12 > \frac{8}{5}a + \frac{96}{5} \Rightarrow a > …\) | M1 | b3M1: Their linear expression in \(a\) only for C compared to their linear expression in \(a\) only for B (allow any inequality or equals) and attempting to solve for \(a\) – this mark is dependent on one correct expression in \(a\). |
| \(a > 3\) | A1 (5) 7 marks | b1A1: CAO (\(a > 3\) only). |
| Alternative approach for (b): | ||
| b1B1: \(b = 4\) (only) | ||
| b1M1: Finding the gradient of \(3x + 4y = 24\) e.g. \(y = -\frac{3}{4}x + 6 \Rightarrow m = -\frac{3}{4}\) the gradient must either be stated explicitly or used later | ||
| b2M1: Gradient of the objective function stated as \(-\frac{a}{b}\) (or used later) in terms of \(a\) only (so must have substituted their value of b) | ||
| b3M1: Their gradient of \(3x + 4y = 24\) compared to their gradient of the objective function (in terms of \(a\) only) – allow any inequality or equals and attempting to solve for \(a\) – this mark is dependent on one correct gradient (if correct then should be \(-\frac{a}{4} < -\frac{3}{4}\)) | ||
| b1A1: CAO (\(a > 3\) only) |
| Answer | Marks | Guidance |
|--------|-------|----------|
| (a) $4y \leq 7x + 8$; $4y \geq x + 8$; $3x + 4y \leq 24$ | B1 B1 (2) | a1B1: One correct inequality (allow strict inequality). a2B1: All three inequalities correct (allow any equivalent forms). |
| (b) Min value of $P$ is $8 \Rightarrow 2b = 8 \therefore b = 4$ | B1 (1) | b1B1: $b = 4$ (only). |
| | Solve $4y = 7x + 8$ and $3x + 4y = 24$ simultaneously (to give $B\left(\frac{8}{5}, \frac{24}{5}\right)$) | M1 | b1M1: Solve correct pair of simultaneous equations to find B – this mark can be implied by correct coordinates of B stated. |
| | $C(4,3) \Rightarrow P = 4a + 12, B\left(\frac{8}{5}, \frac{24}{5}\right) \Rightarrow P = \frac{8}{5}a + \frac{96}{5}$ | M1 | b2M1: Either linear expression in terms of $a$ only (using their value of b) for either the correct C or their B (their B must be correct or a method for solving the correct simultaneous equations to find B must be seen). |
| | $4a + 12 > \frac{8}{5}a + \frac{96}{5} \Rightarrow a > …$ | M1 | b3M1: Their linear expression in $a$ only for C compared to their linear expression in $a$ only for B (allow any inequality or equals) and attempting to solve for $a$ – this mark is dependent on one correct expression in $a$. |
| | $a > 3$ | A1 (5) 7 marks | b1A1: CAO ($a > 3$ only). |
| | **Alternative approach for (b):** | | |
| | b1B1: $b = 4$ (only) | | |
| | b1M1: Finding the gradient of $3x + 4y = 24$ e.g. $y = -\frac{3}{4}x + 6 \Rightarrow m = -\frac{3}{4}$ the gradient must either be stated explicitly or used later | | |
| | b2M1: Gradient of the objective function stated as $-\frac{a}{b}$ (or used later) in terms of $a$ only (so must have substituted their value of b) | | |
| | b3M1: Their gradient of $3x + 4y = 24$ compared to their gradient of the objective function (in terms of $a$ only) – allow any inequality or equals and attempting to solve for $a$ – this mark is dependent on one correct gradient (if correct then should be $-\frac{a}{4} < -\frac{3}{4}$) | | |
| | b1A1: CAO ($a > 3$ only) | | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3aa30e8f-7d55-4c3b-8b2c-55c3e822c8a0-07_1296_1586_230_301}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The graph in Figure 2 is being used to solve a linear programming problem in $x$ and $y$. The three constraints have been drawn on the graph and the rejected regions have been shaded out. The three vertices of the feasible region $R$ are labelled $\mathrm { A } , \mathrm { B }$ and C .
\begin{enumerate}[label=(\alph*)]
\item Determine the inequalities that define $R$.\\
(2)
The objective function, $P$, is given by
$$P = a x + b y$$
where $a$ and $b$ are positive constants.\\
The minimum value of $P$ is 8 and the maximum value of $P$ occurs at C .
\item Find the range of possible values of $a$. You must make your method clear.\\
(5)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2020 Q6 [7]}}