OCR MEI D1 2007 June — Question 3 8 marks

Exam BoardOCR MEI
ModuleD1 (Decision Mathematics 1)
Year2007
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear Programming
TypeGraphical optimization with objective line
DifficultyModerate -0.8 This is a standard textbook linear programming question requiring routine application of the graphical method: plot constraints, identify feasible region, and test vertices to find the maximum. It involves no conceptual difficulty beyond basic D1 syllabus content and follows a completely algorithmic procedure with no problem-solving insight required.
Spec7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables

3 Use a graphical approach to solve the following LP. $$\begin{aligned} & \text { Maximise } \quad 2 x + 3 y \\ & \text { subject to } \quad x + 5 y \leqslant 14 \\ & \quad x + 2 y \leqslant 8 \\ & \quad 3 x + y \leqslant 21 \\ & \quad x \geqslant 0 \\ & y \geqslant 0 \end{aligned}$$ Section B (48 marks)

Question 3:
Plot constraints and find feasible region:
- \(x + 5y = 14\)
- \(x + 2y = 8\)
- \(3x + y = 21\)
- \(x \geq 0\), \(y \geq 0\)
Find vertices of feasible region and evaluate \(2x + 3y\):
Intersection of \(x+5y=14\) and \(x+2y=8\): subtract gives \(3y=6\), \(y=2\), \(x=4\) → objective \(= 8+6=14\)
Intersection of \(x+2y=8\) and \(3x+y=21\): from first \(x=8-2y\), sub: \(24-6y+y=21\), \(5y=3\), \(y=\frac{3}{5}\), \(x=\frac{34}{5}\) → objective \(= \frac{68}{5}+\frac{9}{5}=\frac{77}{5}=15.4\)
Intersection of \(3x+y=21\) and \(y=0\): \(x=7\), \(y=0\) → objective \(=14\)
AnswerMarks Guidance
Maximum is \(2x+3y = 15.4\) at \(x = \frac{34}{5}\), \(y = \frac{3}{5}\)B3 for correct graph/region M1 for finding vertices
# Question 3:

Plot constraints and find feasible region:
- $x + 5y = 14$
- $x + 2y = 8$
- $3x + y = 21$
- $x \geq 0$, $y \geq 0$

Find vertices of feasible region and evaluate $2x + 3y$:

Intersection of $x+5y=14$ and $x+2y=8$: subtract gives $3y=6$, $y=2$, $x=4$ → objective $= 8+6=14$

Intersection of $x+2y=8$ and $3x+y=21$: from first $x=8-2y$, sub: $24-6y+y=21$, $5y=3$, $y=\frac{3}{5}$, $x=\frac{34}{5}$ → objective $= \frac{68}{5}+\frac{9}{5}=\frac{77}{5}=15.4$

Intersection of $3x+y=21$ and $y=0$: $x=7$, $y=0$ → objective $=14$

Maximum is $2x+3y = 15.4$ at $x = \frac{34}{5}$, $y = \frac{3}{5}$ | B3 for correct graph/region | M1 for finding vertices | A1 each for correct intersections | A1 for correct maximum value | A1 for correct coordinates

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3 Use a graphical approach to solve the following LP.

$$\begin{aligned}
& \text { Maximise } \quad 2 x + 3 y \\
& \text { subject to } \quad x + 5 y \leqslant 14 \\
& \quad x + 2 y \leqslant 8 \\
& \quad 3 x + y \leqslant 21 \\
& \quad x \geqslant 0 \\
& y \geqslant 0
\end{aligned}$$

Section B (48 marks)\\

\hfill \mbox{\textit{OCR MEI D1 2007 Q3 [8]}}