| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Parametric objective analysis |
| Difficulty | Standard +0.8 This question requires identifying inequalities from a graph, then solving a parametric optimization problem where the objective function coefficient is unknown. Part (b) demands testing multiple vertices algebraically to find which gives P=38, requiring systematic case analysis and solving simultaneous equations with a parameter—significantly more sophisticated than routine linear programming questions. |
| Spec | 7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x + y \leq 8\) | B1 | oe |
| \(5y \geq x + k\) | B1 | oe |
| \(y = -\dfrac{8}{4}x + 8\) | M1 | Any inequality symbol replacing equals |
| \(2x + y \geq 8\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = 5x + ky\); if \((0,8)\) is optimal vertex then \(k = \dfrac{19}{4}\ (= 4.75)\) | B1 | |
| Other possible optimal vertex is intersection of \(x+y=8\) and \(5y=x+k\); attempt to solve simultaneously or express \(x\) and \(k\) in terms of \(y\) | M1 | |
| \(\left(\dfrac{40-k}{6}, \dfrac{8+k}{6}\right)\) or stating \(k = 40 - 6x\) or \(k = 6y - 8\) | A1 | |
| \(5\!\left(\dfrac{40-k}{6}\right) + k\!\left(\dfrac{8+k}{6}\right) = 38\) or \(5x + (40-6x)(8-x) = 38\) or \(5(8-y)+(6y-8)y=38\) | dM1 | |
| \(k^2 + 3k - 28 = 0 \Rightarrow (k-4)(k+7) = 0\) or \(6x^2 - 83x + 282=0\) or \(6y^2-13y+2=0\) | ddM1 | |
| \(k = 4\) | A1 | |
| If \(k=4\): \((6,2) \to P=38\) and \((0,8)\to P=32\); if \(k=\dfrac{19}{4}\): \((0,8)\to P=38\) and \(\left(\dfrac{47}{8},\dfrac{17}{8}\right)\to P=\dfrac{1263}{32}\ (=39.468\ldots)\); so \(k=4\) only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x + y \leq 8\) | a1B1 | Any equivalent inequality (not strict inequality) |
| \(5y \geq x + k\) | a2B1 | Any equivalent inequality (not strict inequality) |
| Correct equation of line through \((0, 8)\) and \((4, 0)\) | a1M1 | Correct equation or with any inequality symbol |
| Any equivalent form with three terms only | a1A1 | Condone non-integer coefficients; any equivalent inequality (not strict) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k = \dfrac{19}{4}\) | b1B1 | Exact value (or equivalent) seen |
| Correct method solving \(x + y = 8\) and \(5y = x + k\) simultaneously to find \(x\), \(y\) in terms of \(k\) | b1M1 | Or express \(x\) and \(k\) in terms of \(y\) only, or \(y\) and \(k\) in terms of \(x\) only |
| e.g. \(x = \dfrac{40-k}{6}\), \(y = 8 - \dfrac{40-k}{6}\) or \(\begin{cases}y = 8-x \\ k = 40-6x\end{cases}\) or \(\begin{cases}x = 8-y \\ k = 6y-8\end{cases}\) | b1A1 | CAO for coordinates of optimal vertex in terms of \(k\); allow unsimplified; implied by correct equation in \(k\), \(x\) or \(y\) only |
| Setting up equation in \(k\), \(x\) or \(y\) only using intersection of \(x+y=8\) and \(5y=x+k\), together with \(5x + ky = 38\) | b2dM1 | Dependent on previous M mark |
| Solving three-term quadratic in \(k\), \(x\) or \(y\); finding at least one positive value of \(k\) | b3ddM1 | Dependent on both previous M marks; if quadratic formula used must show correct formula with values; if factorising, expanding brackets must give two terms of 3-term quadratic. Note: \(5\!\left(\dfrac{40-k}{6}\right)+k\!\left(\dfrac{8+k}{6}\right)=38 \Rightarrow k=4\) can imply this and next A mark |
| \(k = 4\) | b2A1 | Ignore mention of other values e.g. \(k=-7\); must come from correct working |
| Clear rejection of \(k = \dfrac{19}{4}\) (showing \(P > 38\) for this value) and evidence of second root e.g. \(k = -7\) or \((k+7)(k-4)=0\) seen (also rejected), and stating \(k=4\) only | b3A1 | Dependent on all previous marks in (b); e.g. \((k+7)(k-4)=0 \Rightarrow k=4\) sufficient; must not give more than one value of \(k\) |
# Question 7:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + y \leq 8$ | B1 | oe |
| $5y \geq x + k$ | B1 | oe |
| $y = -\dfrac{8}{4}x + 8$ | M1 | Any inequality symbol replacing equals |
| $2x + y \geq 8$ | A1 | oe |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = 5x + ky$; if $(0,8)$ is optimal vertex then $k = \dfrac{19}{4}\ (= 4.75)$ | B1 | |
| Other possible optimal vertex is intersection of $x+y=8$ and $5y=x+k$; attempt to solve simultaneously or express $x$ and $k$ in terms of $y$ | M1 | |
| $\left(\dfrac{40-k}{6}, \dfrac{8+k}{6}\right)$ or stating $k = 40 - 6x$ or $k = 6y - 8$ | A1 | |
| $5\!\left(\dfrac{40-k}{6}\right) + k\!\left(\dfrac{8+k}{6}\right) = 38$ or $5x + (40-6x)(8-x) = 38$ or $5(8-y)+(6y-8)y=38$ | dM1 | |
| $k^2 + 3k - 28 = 0 \Rightarrow (k-4)(k+7) = 0$ or $6x^2 - 83x + 282=0$ or $6y^2-13y+2=0$ | ddM1 | |
| $k = 4$ | A1 | |
| If $k=4$: $(6,2) \to P=38$ and $(0,8)\to P=32$; if $k=\dfrac{19}{4}$: $(0,8)\to P=38$ and $\left(\dfrac{47}{8},\dfrac{17}{8}\right)\to P=\dfrac{1263}{32}\ (=39.468\ldots)$; so $k=4$ only | A1 | |
## Question 7:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x + y \leq 8$ | **a1B1** | Any equivalent inequality (not strict inequality) |
| $5y \geq x + k$ | **a2B1** | Any equivalent inequality (not strict inequality) |
| Correct equation of line through $(0, 8)$ and $(4, 0)$ | **a1M1** | Correct equation or with any inequality symbol |
| Any equivalent form with three terms only | **a1A1** | Condone non-integer coefficients; any equivalent inequality (not strict) |
---
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = \dfrac{19}{4}$ | **b1B1** | Exact value (or equivalent) seen |
| Correct method solving $x + y = 8$ and $5y = x + k$ simultaneously to find $x$, $y$ in terms of $k$ | **b1M1** | Or express $x$ and $k$ in terms of $y$ only, or $y$ and $k$ in terms of $x$ only |
| e.g. $x = \dfrac{40-k}{6}$, $y = 8 - \dfrac{40-k}{6}$ **or** $\begin{cases}y = 8-x \\ k = 40-6x\end{cases}$ **or** $\begin{cases}x = 8-y \\ k = 6y-8\end{cases}$ | **b1A1** | CAO for coordinates of optimal vertex in terms of $k$; allow unsimplified; implied by correct equation in $k$, $x$ or $y$ only |
| Setting up equation in $k$, $x$ or $y$ only using intersection of $x+y=8$ and $5y=x+k$, together with $5x + ky = 38$ | **b2dM1** | Dependent on previous M mark |
| Solving three-term quadratic in $k$, $x$ or $y$; finding at least one positive value of $k$ | **b3ddM1** | Dependent on both previous M marks; if quadratic formula used must show correct formula with values; if factorising, expanding brackets must give two terms of 3-term quadratic. Note: $5\!\left(\dfrac{40-k}{6}\right)+k\!\left(\dfrac{8+k}{6}\right)=38 \Rightarrow k=4$ can imply this and next A mark |
| $k = 4$ | **b2A1** | Ignore mention of other values e.g. $k=-7$; must come from correct working |
| Clear rejection of $k = \dfrac{19}{4}$ (showing $P > 38$ for this value) **and** evidence of second root e.g. $k = -7$ or $(k+7)(k-4)=0$ seen (also rejected), **and** stating $k=4$ only | **b3A1** | Dependent on all previous marks in (b); e.g. $(k+7)(k-4)=0 \Rightarrow k=4$ sufficient; must not give more than one value of $k$ |
> **Note:** Correct value of $k$ with no working scores no marks; correct value with minimal working — send to review.7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{48e785c0-7de5-450f-862c-4dd4d169adf9-10_993_1268_221_402}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the constraints of a linear programming problem in $x$ and $y$, where $R$ is the feasible region. The equations of two of the lines have been shown in Figure 3.
Given that $k$ is a positive constant,
\begin{enumerate}[label=(\alph*)]
\item determine, in terms of $k$ where necessary, the inequalities that define $R$.
The objective is to maximise $P = 5 x + k y$\\
Given that the value of $P$ is 38 at the optimal vertex of $R$,
\item determine the possible value(s) of $k$. You must show algebraic working and make your method clear.\\
(Total 11 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2021 Q7 [11]}}