OCR Further Discrete 2019 June — Question 7 12 marks

Exam BoardOCR
ModuleFurther Discrete (Further Discrete)
Year2019
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear Programming
TypeConstraint derivation verification
DifficultyStandard +0.3 This is a straightforward linear programming question requiring standard constraint formulation and feasibility checking. Part (a) is trivial identification, part (b) uses the same method as the given example, part (c) is simple substitution, and part (d) requires systematic checking of integer solutions—all routine techniques for Further Maths Decision students with no novel problem-solving required.
Spec7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables

7 Sam is making pies.
There is exactly enough pastry to make 7 large pies or 20 medium pies or 36 small pies, or some mixture of large, medium and small pies. This is represented as a constraint \(180 x + 63 y + 35 z \leqslant 1260\).
  1. Write down what \(\mathrm { X } , \mathrm { Y }\) and Z represent. There is exactly enough filling to make 5 large pies or 12 medium pies or 18 small pies, or some mixture of large, medium and small pies.
  2. Express this as a constraint of the form \(a x + b y + c z \leqslant d\), where \(a , b , c\) and \(d\) are integers. The number of small pies must equal the total number of large and medium pies.
  3. Show that making exactly 9 small pies is inconsistent with the constraints.
  4. Determine the maximum number of large pies that can be made.

Question 7:
AnswerMarks Guidance
7(a) x = number of large pies made
y = number of medium pies made
AnswerMarks
z = number of small pies madeB1
B13.3
1.1Number of (pies made)
x = large, y = medium, z = small
AnswerMarks Guidance
(b)36x + 15y + 10z < 180 M1
A13.3
1.1Coefficients in ratio
1 1 1
36, 15, 10, 180 or any positive
5:12:18
AnswerMarks
integer multiple of this setFollow through their
definitions of x, y, z if
appropriate
AnswerMarks
(c)z = 9 ⇒ 180x + 63y < 945 ⇒ 20x + 7y < 105
36x + 15y < 90 ⇒ 12x + 5y < 30
and x + y = 9
AnswerMarks
⇒ 5x + 5y = 45 so 12x + 5y cannot be < 30E1
E13.5c
2.1May be argued in words,
algebraically or graphically
AnswerMarks
Filling constraint cannot be satisfiedEliminate one variable to get
two inequalities and total
Or using one inequality to
show that a variable is
negative, or equivalent
contradiction
AnswerMarks
(d)There is enough filling for 5 large pies, so x < 5
180x + 63y + 35z < 1260
36x + 15y + 10z < 180
x + y = z
Eliminating z:
215x + 98y < 1260 and 46x + 25y < 180
180 ÷ 46 = 3.913, so x < 3
x = 3 ⇒ 25y < 42 ⇒ y = 0, z = 3 or y = 1, z = 4
Verify that their solution satisfies the constraints
180x+63y+35z 36x+15y+10z
x=3, y=0, z=3 645 138
x=3, y=1, z=4 743 163
Upper limit 1260 180
AnswerMarks
The maximum number of large pies is 3B1
M1
A1
M1
A1
AnswerMarks
B13.5c
3.3
2.4
3.4
2.2a
AnswerMarks
3.4Upper limit 5
Refining the model to take the new
constraint into account
Or branch-and-bound
Or simplex with obj (max x) and at
least 3 constraints
Eliminating a variable
(-117y+215z<1260, -21y+46z<180)
(117x+98z<1260, 21x+25z<180
Or solve simplex with 4 correct
constraints
Either of these integer solutions
Shown to be consistent with all
constraints (or implied from algebra)
AnswerMarks
Maximum is 3May be argued in words,
algebraically or graphically
Or ‘make no medium so y =
0 and x = z‘
Or checking at least 3
feasible solutions
Feasible cases are (x, y, z) =
(0, k, k) for k = 0, 1, …, 6
(1, k, k+1) for k = 0, 1, …, 5
(2, k, k+2) for k = 0, 1, 2, 3
(3, k, k+3) for k = 0, 1
[12]
AnswerMarks Guidance
180x+63y+35z36x+15y+10z
x=3, y=0, z=3645 138
x=3, y=1, z=4743 163
Upper limit1260 180
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
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 (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
Question 7:
7 | (a) | x = number of large pies made
y = number of medium pies made
z = number of small pies made | B1
B1 | 3.3
1.1 | Number of (pies made)
x = large, y = medium, z = small
(b) | 36x + 15y + 10z < 180 | M1
A1 | 3.3
1.1 | Coefficients in ratio
1 1 1
36, 15, 10, 180 or any positive
5:12:18
integer multiple of this set | Follow through their
definitions of x, y, z if
appropriate
(c) | z = 9 ⇒ 180x + 63y < 945 ⇒ 20x + 7y < 105
36x + 15y < 90 ⇒ 12x + 5y < 30
and x + y = 9
⇒ 5x + 5y = 45 so 12x + 5y cannot be < 30 | E1
E1 | 3.5c
2.1 | May be argued in words,
algebraically or graphically
Filling constraint cannot be satisfied | Eliminate one variable to get
two inequalities and total
Or using one inequality to
show that a variable is
negative, or equivalent
contradiction
(d) | There is enough filling for 5 large pies, so x < 5
180x + 63y + 35z < 1260
36x + 15y + 10z < 180
x + y = z
Eliminating z:
215x + 98y < 1260 and 46x + 25y < 180
180 ÷ 46 = 3.913, so x < 3
x = 3 ⇒ 25y < 42 ⇒ y = 0, z = 3 or y = 1, z = 4
Verify that their solution satisfies the constraints
180x+63y+35z 36x+15y+10z
x=3, y=0, z=3 645 138
x=3, y=1, z=4 743 163
Upper limit 1260 180
The maximum number of large pies is 3 | B1
M1
A1
M1
A1
B1 | 3.5c
3.3
2.4
3.4
2.2a
3.4 | Upper limit 5
Refining the model to take the new
constraint into account
Or branch-and-bound
Or simplex with obj (max x) and at
least 3 constraints
Eliminating a variable
(-117y+215z<1260, -21y+46z<180)
(117x+98z<1260, 21x+25z<180
Or solve simplex with 4 correct
constraints
Either of these integer solutions
Shown to be consistent with all
constraints (or implied from algebra)
Maximum is 3 | May be argued in words,
algebraically or graphically
Or ‘make no medium so y =
0 and x = z‘
Or checking at least 3
feasible solutions
Feasible cases are (x, y, z) =
(0, k, k) for k = 0, 1, …, 6
(1, k, k+1) for k = 0, 1, …, 5
(2, k, k+2) for k = 0, 1, 2, 3
(3, k, k+3) for k = 0, 1
[12]
180x+63y+35z | 36x+15y+10z
x=3, y=0, z=3 | 645 | 138
x=3, y=1, z=4 | 743 | 163
Upper limit | 1260 | 180
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
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 (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
7 Sam is making pies.\\
There is exactly enough pastry to make 7 large pies or 20 medium pies or 36 small pies, or some mixture of large, medium and small pies.

This is represented as a constraint $180 x + 63 y + 35 z \leqslant 1260$.
\begin{enumerate}[label=(\alph*)]
\item Write down what $\mathrm { X } , \mathrm { Y }$ and Z represent.

There is exactly enough filling to make 5 large pies or 12 medium pies or 18 small pies, or some mixture of large, medium and small pies.
\item Express this as a constraint of the form $a x + b y + c z \leqslant d$, where $a , b , c$ and $d$ are integers.

The number of small pies must equal the total number of large and medium pies.
\item Show that making exactly 9 small pies is inconsistent with the constraints.
\item Determine the maximum number of large pies that can be made.

\begin{itemize}
  \item Your reasoning should be in the form of words, calculations or algebra.
  \item You must check that your solution is feasible.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR Further Discrete 2019 Q7 [12]}}