D1 June 2009 Q7
7. Rose makes hanging baskets which she sells at her local market. She makes two types, large and small.
Rose makes \(x\) large baskets and \(y\) small baskets.
Each large basket costs £7 to make and each small basket costs £5 to make. Rose has £350 she can spend on making the baskets.
Two further constraints are
\[\begin{aligned} &y \leqslant 20 \text{ and}\\ &y \leqslant 4x.\end{aligned}\]Rose makes a profit of £2 on each large basket and £3 on each small basket. Rose wishes to maximise her profit, £\(P\).
| Scheme | Marks |
|---|---|
| \(7x + 5y \leqslant 350\) | M1 A1 |
| (2) |
Notes
(a) 1M1: Coefficients correct (condone swapped \(x\) and \(y\) coefficients) need 350 and any inequality
1A1: cso.
| Scheme | Marks |
|---|---|
| \(y \leqslant 20\) e.g. make at most 20 small baskets | B1 |
| \(y \leqslant 4x\) e.g. the number of small (\(y\)) baskets is at most 4 times the number of large baskets (\(x\)). {E.g if \(y = 40\), \(x = 10, 11, 12\) etc. or if \(x = 10\), \(y = 40, 39, 38\)} | B1 |
| (2) |
Notes
(b) 1B1: cao
2B1: cao, test their statement, need both = and < aspects.
| Scheme | Marks |
|---|---|
| (see graph) Draw three lines correctly Label R ![]() | B3, 2, 1, 0 B1 |
| (4) |
Notes
(c) 1B1: One line drawn correctly
2B1: Two lines drawn correctly
3B1: Three lines drawn correctly. Check (10, 40) (0, 0) and axes
4B1: R correct, but allow if one line is slightly out (1 small square).
| Scheme | Marks |
|---|---|
| \((P =)\ 2x + 3y\) | B1 |
| (1) |
Notes
(d) 1B1: cao accept an expression.
| Scheme | Marks |
|---|---|
| Profit line or point testing. | M1 A1 |
| \(x = 35.7\) \(y = 20\) precise point found. | B1 |
| Need integers so optimal point in R is (35, 20); Profit (£)130 | B1;B1 |
| (5) | |
| (14 marks) |
Notes
(e) 1M1: Attempt at profit line or attempt to test at least two vertices in their feasible region.
1A1: Correct profit line or correct testing of at least three vertices.
Point testing: (0,0) P= 0; (5,20) P = 70; (50,0) P = 100
\[\left(35\tfrac{5}{7}, 20\right) = \left(\dfrac{250}{7}, 20\right) \quad P = 131\tfrac{3}{7} = \dfrac{920}{7}\]also (35, 20) P = 130. Accept (36,20) P = 132 for M but not A.
Objective line: Accept gradient of 1/m for M mark or line close to correct gradient.
1B1: cao – accept \(x\) co-ordinates which round to 35.7
2B1: cao
3B1: cao
