AS June 2019 Q5
5. Ben 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.
Ben decides to order the minimum number of roses that satisfy his constraints.
| Scheme | Marks | AO |
|---|---|---|
| Minimise \((P =)\, x + 5y + 4z\) | B1 | 3.3 |
| Subject to \(x \geqslant \dfrac{3}{5}(x + y + z)\ (\Rightarrow 2x \geqslant 3y + 3z)\) | B1 | 3.3 |
| \(3y \geqslant 2z\) | B1 | 3.3 |
| \(x + y + z = 1000\) | B1 | 3.3 |
| \(z = 1000 - x - y\) substituted into objective and constraints gives | M1 | 3.1a |
| Minimise \((P =)\, y - 3x\,(+\ 4000)\) subject to \(x \geqslant 600\) and \(2x + 5y \geqslant 2000\) | A1 A1 | 1.1b 1.1b |
| (7) |
Notes
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
B1: \(x \geqslant \dfrac{3}{5}(x + y + z)\) oe – need not be simplified for this mark, accept \(x \geqslant \dfrac{3}{5}(1000)\)
B1: \(3y \geqslant 2z\) or 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 (\(x \geqslant 600\) and \(2x + 5y \geqslant 2000\) - must be integer coefficients for this mark)
| Scheme | Marks | AO |
|---|---|---|
| (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
(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 constraint \(x + y + z = 1000\) (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 constraint \(2y \geqslant 3z\) 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)