| Exam Board | Edexcel |
|---|---|
| Module | FD1 AS (Further Decision 1 AS) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Three-variable constraint reduction |
| Difficulty | Standard +0.3 This is a straightforward linear programming formulation question from Further Decision 1. Part (a) requires translating word constraints into inequalities and eliminating one variable using the equality constraint—standard textbook procedure. Part (b) involves simple substitution and arithmetic once the minimum roses constraint is applied. The question is methodical rather than conceptually challenging, making it slightly easier than average for A-level standard. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06d Graphical solution: feasible region, two variables |
| Answer | Marks |
|---|---|
| 5(a) | Minimise ( P=) x+5y+4z |
| Answer | Marks |
|---|---|
| x≥600 and 2x+5y≥2000 | B1 |
| Answer | Marks |
|---|---|
| A1 | 3.3 |
| Answer | Marks |
|---|---|
| (b) | (i) Using least value of x to find y and z |
| Answer | Marks |
|---|---|
| (ii) £2360 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.4 |
Question 5:
--- 5(a) ---
5(a) | Minimise ( P=) x+5y+4z
3
Subject to x≥ ( x+ y+z )(⇒2x≥3y+3z )
5
3y≥2z
x+ y+z =1000
z =1000−x− y substituted into objective and constraints gives
Minimise ( P=) y−3x (+ 4000 ) subject to
x≥600 and 2x+5y≥2000 | B1
B1
B1
B1
M1
A1
A1 | 3.3
3.3
3.3
3.3
3.1a
1.1b
1.1b
(7)
(b) | (i) Using least value of x to find y and z
600 roses, 160 hydrangeas and 240 peonies
(ii) £2360 | M1
A1
A1 | 3.4
3.2a
1.1b
(3)
(10 marks)
Notes
(a)
B1: CAO (for objective) – must contain ‘minimise’ or ‘min’ only (so not ‘minimum’) either when
stated in terms of x, y and z or x and y only
3 3
B1: x≥ ( x+ y+z ) oe – need not be simplified for this mark, accept x≥ ( 1000 )
5 5
B1: 3y≥2zor any equivalent form (need not be simplified nor integer coefficients for this mark)
B1: x+ y+z =1000(could be implied by earlier/later working)
M1: Eliminating z from either the objective or both constraints using the constraint x+ y+z =1000
A1: Correct objective in terms of x and y only – condone lack of ‘minimise’
A1: Both constraints correct (x≥600 and 2x+5y≥2000- must be integer coefficients for this
mark)
(b)(i)
M1: Using their least value of x to find both y and z (with both y and z being positive integers) –
note that all values must satisfy the constraintx+ y+z =1000(and must all be integers)
A1: All three types of flowers correct (in context – so not just in terms of x, y and z) – must come
from correct constraints in (a)
(ii)
A1: CAO for cost (condone lack of units but not 2360p) – must come from correct constraints in (a)
SC for (b) – for those candidates with the constraint2y≥3z in (a) leading to 600 roses, 240
hydrangeas and 160 peonies (so not just in terms of x, y and z) together with (£)2440 award SC
M1A1A0 in (b)
PMT
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United KingdomBen is a wedding planner. He needs to order flowers for the weddings that are taking place next month. The three types of flower he needs to order are roses, hydrangeas and peonies.
Based on his experience, Ben forms the following constraints on the number of each type of flower he will need to order.
• At least three-fifths of all the flowers must be roses.
• For every 2 hydrangeas there must be at most 3 peonies.
• The total number of flowers must be exactly 1000
The cost of each rose is £1, the cost of each hydrangea is £5 and the cost of each peony is £4
Ben wants to minimise the cost of the flowers.
Let $x$ represent the number of roses, let $y$ represent the number of hydrangeas and let $z$ represent the number of peonies that he will order.
\begin{enumerate}[label=(\alph*)]
\item Formulate this as a linear programming problem in $x$ and $y$ only, stating the objective function and listing the constraints as simplified inequalities with integer coefficients. [7]
\end{enumerate}
Ben decides to order the minimum number of roses that satisfy his constraints.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the number of each type of flower that he will order to minimise the cost of the flowers.
\item Calculate the corresponding total cost of this order. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD1 AS 2019 Q5 [10]}}