Question 6 - A-Level Maths

OCR MEI D1 2009 January — Question 6 16 marks

Exam Board	OCR MEI
Module	D1 (Decision Mathematics 1)
Year	2009
Session	January
Marks	16
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear Programming
Type	Three-variable constraint reduction
Difficulty	Standard +0.8 This is a multi-part linear programming question requiring formulation, algebraic manipulation to reduce from 3 to 2 variables, and graphical solution. Part (ii) demands genuine insight into constraint substitution and objective function transformation—not routine application. The variable reduction step is non-standard for D1 level, making this moderately challenging despite being a textbook LP context.
Spec	7.06a LP formulation: variables, constraints, objective function 7.06d Graphical solution: feasible region, two variables

6 A company is planning its production of "MPowder" for the next three months.

Over the next 3 months 20 tonnes must be produced.
Production quantities must not be decreasing. The amount produced in month 2 cannot be less than the amount produced in month 1 , and the amount produced in month 3 cannot be less than the amount produced in month 2.
No more than 12 tonnes can be produced in total in months 1 and 2.
Production costs are $\pounds 2000$ per tonne in month $1 , \pounds 2200$ per tonne in month 2 and $\pounds 2500$ per tonne in month 3.

The company planner starts to formulate an LP to find a production plan which minimises the cost of production: $$\begin{array} { l l } \text { Minimise } & 2000 x _ { 1 } + 2200 x _ { 2 } + 2500 x _ { 3 } \\ \text { subject to } & x _ { 1 } \geq 0 x _ { 2 } \geq 0 x _ { 3 } \geq 0 \\ & x _ { 1 } + x _ { 2 } + x _ { 3 } = 20 \\ & x _ { 1 } \leq x _ { 2 } \\ & \bullet \cdot \cdot \end{array}$$

Explain what the variables $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ represent, and write down two more constraints to complete the formulation.
Explain how the LP can be reformulated to: $$\begin{array} { l l } \text { Maximise } & 500 x _ { 1 } + 300 x _ { 2 } \\ \text { subject to } & x _ { 1 } \geq 0 x _ { 2 } \geq 0 \\ & x _ { 1 } \leq x _ { 2 } \\ & x _ { 1 } + 2 x _ { 2 } \leq 20 \\ & x _ { 1 } + x _ { 2 } \leq 12 \end{array}$$
Use a graphical approach to solve the LP in part (ii). Interpret your solution in terms of the company's production plan, and give the minimum cost.

Show mark scheme Show mark scheme source

Question 6:

Part (i)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$x_i$ represents the number of tonnes produced in month $i$	M1	quantities
$x_2 \leq x_3$	A1	tonnes
$x_1 + x_2 \leq 12$	B1
B1

Part (ii)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Substitute $x_3 = 20 - x_1 - x_2$	M1
$x_2 \leq x_3 \Rightarrow x_1 + 2x_2 \leq 20$	A1
$\text{Min } 2000x_1 + 2200x_2 + 2500x_3 \to \text{Max } 500x_1 + 300x_2$	A1

Part (iii)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Correct graph drawn (scaled correctly)	M1	sca
Three constraint lines correct	A3	lines
Correct shading of feasible region	A1	shading
More than one vertex evaluated or profit line drawn	M1	$>1$ evaluated point or profit line
Optimal point $(6, 6)$ giving value $4800$ identified	A1	$(6,6)$ or $4800$
Production plan: 6 tonnes in month 1, 6 tonnes in month 2, 8 tonnes in month 3	M1	$\checkmark$ all 3
Cost = £45200	A1	cao

# Question 6:

## Part (i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_i$ represents the number of tonnes produced in month $i$ | M1 | quantities |
| $x_2 \leq x_3$ | A1 | tonnes |
| $x_1 + x_2 \leq 12$ | B1 | |
| | B1 | |

## Part (ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $x_3 = 20 - x_1 - x_2$ | M1 | |
| $x_2 \leq x_3 \Rightarrow x_1 + 2x_2 \leq 20$ | A1 | |
| $\text{Min } 2000x_1 + 2200x_2 + 2500x_3 \to \text{Max } 500x_1 + 300x_2$ | A1 | |

## Part (iii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct graph drawn (scaled correctly) | M1 | sca |
| Three constraint lines correct | A3 | lines |
| Correct shading of feasible region | A1 | shading |
| More than one vertex evaluated or profit line drawn | M1 | $>1$ evaluated point or profit line |
| Optimal point $(6, 6)$ giving value $4800$ identified | A1 | $(6,6)$ or $4800$ |
| Production plan: 6 tonnes in month 1, 6 tonnes in month 2, 8 tonnes in month 3 | M1 | $\checkmark$ all 3 |
| Cost = £45200 | A1 | cao |

Show LaTeX source

6 A company is planning its production of "MPowder" for the next three months.

\begin{itemize}
  \item Over the next 3 months 20 tonnes must be produced.
  \item Production quantities must not be decreasing. The amount produced in month 2 cannot be less than the amount produced in month 1 , and the amount produced in month 3 cannot be less than the amount produced in month 2.
  \item No more than 12 tonnes can be produced in total in months 1 and 2.
  \item Production costs are $\pounds 2000$ per tonne in month $1 , \pounds 2200$ per tonne in month 2 and $\pounds 2500$ per tonne in month 3.
\end{itemize}

The company planner starts to formulate an LP to find a production plan which minimises the cost of production:

$$\begin{array} { l l } 
\text { Minimise } & 2000 x _ { 1 } + 2200 x _ { 2 } + 2500 x _ { 3 } \\
\text { subject to } & x _ { 1 } \geq 0 x _ { 2 } \geq 0 x _ { 3 } \geq 0 \\
& x _ { 1 } + x _ { 2 } + x _ { 3 } = 20 \\
& x _ { 1 } \leq x _ { 2 } \\
& \bullet \cdot \cdot
\end{array}$$

(i) Explain what the variables $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ represent, and write down two more constraints to complete the formulation.\\
(ii) Explain how the LP can be reformulated to:

$$\begin{array} { l l } 
\text { Maximise } & 500 x _ { 1 } + 300 x _ { 2 } \\
\text { subject to } & x _ { 1 } \geq 0 x _ { 2 } \geq 0 \\
& x _ { 1 } \leq x _ { 2 } \\
& x _ { 1 } + 2 x _ { 2 } \leq 20 \\
& x _ { 1 } + x _ { 2 } \leq 12
\end{array}$$

(iii) Use a graphical approach to solve the LP in part (ii). Interpret your solution in terms of the company's production plan, and give the minimum cost.

\hfill \mbox{\textit{OCR MEI D1 2009 Q6 [16]}}

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