OCR Further Discrete AS 2023 June — Question 7 12 marks

Exam BoardOCR
ModuleFurther Discrete AS (Further Discrete AS)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear Programming
TypeParametric objective analysis
DifficultyChallenging +1.2 Part (a) is a standard graphical linear programming problem requiring plotting constraints and finding the optimal vertex. Part (b) introduces a parametric constraint y ≥ kx and asks when the optimal solution changes, requiring analysis of how this line intersects the feasible region and affects the objective function—this adds conceptual depth beyond routine exercises but remains within expected Further Maths problem-solving without requiring exceptional insight.
Spec7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients

7 A linear programming problem is
Maximise \(P = 4 x + y\) subject to $$\begin{aligned} 3 x - y & \leqslant 30 \\ x + y & \leqslant 15 \\ x - 3 y & \leqslant 6 \end{aligned}$$ and \(x \geqslant 0 , y \geqslant 0\)
  1. Use a graphical method to find the optimal value of \(P\), and the corresponding values of \(x\) and \(y\). An additional constraint is introduced.
    This constraint means that the value of \(y\) must be at least \(k\) times the value of \(x\), where \(k\) is a positive constant.
    1. Determine the set of values of \(k\) for which the optimal value of \(P\) found in part (a) is unchanged.
    2. Determine, in terms of \(k\), the values of \(x , y\) and \(P\) in the cases when the optimal solution is different from that found in part (a).

Question 7:
7
AnswerMarks Guidance
7(a) x y P = 4x + y
6 0 24
10.5 1.5 43.5
11.25 3.75 48.75
0 15 15
P = 48.75
AnswerMarks
x = 11.25, y = 3.75B1
B1
B1
M1
A1
A1
AnswerMarks
[6]1.1
1.1
1.1
1.1
1.1
AnswerMarks
1.1Boundary line y = 3x –30, at least from (10, 0) to (12, 6)
Boundary line y = 15 – x, at least from (6, 9) to (12, 3)
1
Boundary line y = x – 2, at least from (6, 0) to (12, 2)
3
Ignore any extra lines
Checking any vertex (other than the origin)
or using a sliding profit line with gradient –4 (approx.) seen,
or implied from answer
cao
cao
AnswerMarks Guidance
xy P = 4x + y
7(b) (i)
1
3.75  11.25 =
3
1
Solution is unchanged for k 
AnswerMarks
3M1
A1ft
AnswerMarks
[2]3.1a
1.11
or (their y)  (their x)
3
1
 (their)
3
1
Accept with a lower bound of 0, 0 < k  (their) , provided their
3
upper bound is positive, but not with a lower bound other than 0
AnswerMarks Guidance
7(b) (ii)
For k > , optimal solution is at intersection
3
of y = kx and x + y = 15  x + kx = 15
4+𝑘
P = 15( )
1+𝑘
1
x = 15( )
1+𝑘
𝑘
y = 15( )
AnswerMarks
1+𝑘M1
A1
A1
A1
AnswerMarks
[4]3.1a
1.1
1.1
AnswerMarks
1.1Evidence of finding point where y = kx meets x + y = 15
May be implied from any of x, y, P correct
cao (or equivalent)
cao (or equivalent) without wrong working
cao (or equivalent) without wrong working
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Question 7:
7
7 | (a) | x y P = 4x + y
6 0 24
10.5 1.5 43.5
11.25 3.75 48.75
0 15 15
P = 48.75
x = 11.25, y = 3.75 | B1
B1
B1
M1
A1
A1
[6] | 1.1
1.1
1.1
1.1
1.1
1.1 | Boundary line y = 3x –30, at least from (10, 0) to (12, 6)
Boundary line y = 15 – x, at least from (6, 9) to (12, 3)
1
Boundary line y = x – 2, at least from (6, 0) to (12, 2)
3
Ignore any extra lines
Checking any vertex (other than the origin)
or using a sliding profit line with gradient –4 (approx.) seen,
or implied from answer
cao
cao
x | y | P = 4x + y
7 | (b) | (i) | Additional constraint y  kx
1
3.75  11.25 =
3
1
Solution is unchanged for k 
3 | M1
A1ft
[2] | 3.1a
1.1 | 1
or (their y)  (their x)
3
1
 (their)
3
1
Accept with a lower bound of 0, 0 < k  (their) , provided their
3
upper bound is positive, but not with a lower bound other than 0
7 | (b) | (ii) | 1
For k > , optimal solution is at intersection
3
of y = kx and x + y = 15  x + kx = 15
4+𝑘
P = 15( )
1+𝑘
1
x = 15( )
1+𝑘
𝑘
y = 15( )
1+𝑘 | M1
A1
A1
A1
[4] | 3.1a
1.1
1.1
1.1 | Evidence of finding point where y = kx meets x + y = 15
May be implied from any of x, y, P correct
cao (or equivalent)
cao (or equivalent) without wrong working
cao (or equivalent) without wrong working
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
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01223 553998
Alternatively, you can email us on
support@ocr.org.uk
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ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
7 A linear programming problem is\\
Maximise $P = 4 x + y$\\
subject to

$$\begin{aligned}
3 x - y & \leqslant 30 \\
x + y & \leqslant 15 \\
x - 3 y & \leqslant 6
\end{aligned}$$

and $x \geqslant 0 , y \geqslant 0$
\begin{enumerate}[label=(\alph*)]
\item Use a graphical method to find the optimal value of $P$, and the corresponding values of $x$ and $y$.

An additional constraint is introduced.\\
This constraint means that the value of $y$ must be at least $k$ times the value of $x$, where $k$ is a positive constant.
\item \begin{enumerate}[label=(\roman*)]
\item Determine the set of values of $k$ for which the optimal value of $P$ found in part (a) is unchanged.
\item Determine, in terms of $k$, the values of $x , y$ and $P$ in the cases when the optimal solution is different from that found in part (a).
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Discrete AS 2023 Q7 [12]}}