| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2024 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Reverse engineering objective from solution |
| Difficulty | Standard +0.3 This is a straightforward reverse-engineering problem requiring students to find the objective function from one vertex (where P=540 at maximum), then evaluate it at the minimum vertex. The method is standard: substitute coordinates into P=ax+by, solve for the ratio a:b, then test remaining vertices. While it requires systematic working, it involves only basic algebra and no novel insight beyond standard linear programming procedures taught in D1. |
| Spec | 7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Objective function is maximised at \(C(9, 22)\) and minimised at \(A(6, 8)\) | B1 | Recognises max at \(C\) and min at \(A\); award if correct coordinates \((9,22)\) and \((6,8)\) used in calculations. Sight of \(9a+22b=540\) and \(6a+8b=k\) earns this mark |
| Let \(P = \lambda(4x + 4.5y)\) | M1 | Setting up objective function of form \(\lambda(4x+4.5y)\) or \(\lambda(4.5x+4y)\); allow consideration of \(4x+4.5y\) or any multiple; considers gradient of objective function \(-\frac{8}{9}\); considers ratio between \(a\) and \(b\) e.g. \(4b=4.5a\) |
| \(540 = \lambda(4(9) + 4.5(22)) \Rightarrow \lambda = \ldots\) | M1 | Correct approach to find the objective function. \((P=)\ 16x+18y\) earns this mark (if no incorrect working seen). May be implied by consideration of e.g. \(540/135\). Candidates may adopt alternative algebraic approaches using \(9a+22b\) and forming simultaneous equations |
| \(P_{\min} = \lambda'(4(6) + 4.5(8))\) | dM1 | Dependent on previous M mark – using point \(A\) and their objective function of the form \(\lambda(4x+4.5y)\) where \(\lambda \neq 1\) to get a value for \(P_{\min}\) |
| \(P_{\min} = 240\) | A1 | CAO (\(k=240\)) |
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Objective function is maximised at $C(9, 22)$ and minimised at $A(6, 8)$ | B1 | Recognises max at $C$ and min at $A$; award if correct coordinates $(9,22)$ and $(6,8)$ used in calculations. Sight of $9a+22b=540$ and $6a+8b=k$ earns this mark |
| Let $P = \lambda(4x + 4.5y)$ | M1 | Setting up objective function of form $\lambda(4x+4.5y)$ or $\lambda(4.5x+4y)$; allow consideration of $4x+4.5y$ or any multiple; considers gradient of objective function $-\frac{8}{9}$; considers ratio between $a$ and $b$ e.g. $4b=4.5a$ |
| $540 = \lambda(4(9) + 4.5(22)) \Rightarrow \lambda = \ldots$ | M1 | Correct approach to find the objective function. $(P=)\ 16x+18y$ earns this mark (if no incorrect working seen). May be implied by consideration of e.g. $540/135$. Candidates may adopt alternative algebraic approaches using $9a+22b$ and forming simultaneous equations |
| $P_{\min} = \lambda'(4(6) + 4.5(8))$ | dM1 | Dependent on previous M mark – using point $A$ and their objective function of the form $\lambda(4x+4.5y)$ where $\lambda \neq 1$ to get a value for $P_{\min}$ |
| $P_{\min} = 240$ | A1 | CAO ($k=240$) |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4814ebd7-f48a-49cf-8ca2-045d84abd63c-6_883_986_219_552}
\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, $R$.
The four vertices of $R$ are $A ( 6,8 ) , B ( 13,12 ) , C ( 9,22 )$ and $D ( 5,18 )$.\\
An objective line has been drawn and labelled on the graph.\\
When the objective function, $P$, is maximised, the value of $P$ is 540\\
When the objective function, $P$, is minimised, the value of $P$ is $k$\\
Determine the value of $k$. You must make your method and working clear.\\
(You may assume that the objective function, $P$, takes the form $a x + b y$ where $a$ and $b$ are constants.)\\
\hfill \mbox{\textit{Edexcel D1 2024 Q5 [5]}}