D2 June 2005 Q8
8. Polly has a bird food stall at the local market. Each week she makes and sells three types of packs \(A\), \(B\) and \(C\).
Pack \(A\) contains 4 kg of bird seed, 2 suet blocks and 1 kg of peanuts.
Pack \(B\) contains 5 kg of bird seed, 1 suet block and 2 kg of peanuts.
Pack \(C\) contains 10 kg of bird seed, 4 suet blocks and 3 kg of peanuts.
Each week Polly has 140 kg of bird seed, 60 suet blocks and 60 kg of peanuts available for the packs.
The profit made on each pack of \(A\), \(B\) and \(C\) sold is £3.50, £3.50 and £6.50 respectively. Polly sells every pack on her stall and wishes to maximise her profit, \(P\) pence.
Let \(x\), \(y\) and \(z\) be the numbers of packs \(A\), \(B\) and \(C\) sold each week.
An initial Simplex tableau for the above situation is
| Basic variable | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | Value |
|---|---|---|---|---|---|---|---|
| \(r\) | \(4\) | \(5\) | \(10\) | \(1\) | \(0\) | \(0\) | \(140\) |
| \(s\) | \(2\) | \(1\) | \(4\) | \(0\) | \(1\) | \(0\) | \(60\) |
| \(t\) | \(1\) | \(2\) | \(3\) | \(0\) | \(0\) | \(1\) | \(60\) |
| \(P\) | \(-350\) | \(-350\) | \(-650\) | \(0\) | \(0\) | \(0\) | \(0\) |
Taking the most negative number in the profit row to indicate the pivotal column,
| Scheme | Marks |
|---|---|
| \(r\), \(s\) and \(t\) are unused amounts of bird seed (in kg), suet blocks and peanuts (in kg) that Polly has at the end of each week after she has made up and sold her packs | B2, 1, 0 |
| (2) |
Notes
B2 Ref to “unused” “bird seed, suet blocks & peanuts”
B1 Ref to “unused” or bird seed etc or muddled explanation.
“bad” gets B1 must engage with context
| Scheme | Marks | |||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| M1 A1 M1 A2ft, 1ft, 0 | |||||||||||||||||||||||||||||||||||||||||||||
| (5) |
Notes
The boxed entry is ringed in the mark scheme (the next pivot).
M1 correct pivot
A1 pivot row correct c.a.o. incl. b.v
M1ft correct row operations used (all 3) – at least 1 non zero or 1 term correct in each row. Where row not ft \(\Rightarrow\) M0
A2ft non-pivoted rows correct; –1 each error ft on error in pivot choice only. Penalise b.v once only
(b) Notes
1. Wrong pivot chosen in col 2 (–usually 4) M0 then for M1A2ft
(a)
| b.v. | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | value | |
|---|---|---|---|---|---|---|---|---|
| \(r\) | \(-1\) | \(2\tfrac{1}{2}\) | \(0\) | \(1\) | \(-2\tfrac{1}{2}\) | \(0\) | \(-10\) | \(\text{R}_1 - 10\text{R}_2\) |
| \(z\) | \(\tfrac{1}{2}\) | \(\tfrac{1}{4}\) | \(1\) | \(0\) | \(\tfrac{1}{4}\) | \(0\) | \(15\) | \(\text{R}_2 \div 4\) |
| \(t\) | \(-\tfrac{1}{2}\) | \(\boxed{1\tfrac{1}{4}}\) | \(0\) | \(0\) | \(-\tfrac{3}{4}\) | \(1\) | \(15\) | \(\text{R}_3 - 3\text{R}_2\) |
| \(p\) | \(-25\) | \(-187\tfrac{1}{2}\) | \(0\) | \(0\) | \(162\tfrac{1}{2}\) | \(0\) | \(9750\) | \(\text{R}_4 + 650\text{R}_2\) |
(b)
| b.v. | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | value | |
|---|---|---|---|---|---|---|---|---|
| \(r\) | \(\tfrac{2}{3}\) | \(-1\tfrac{2}{3}\) | \(0\) | \(1\) | \(0\) | \(\tfrac{-10}{3}\) | \(-60\) | \(\text{R}_1 - 10\text{R}_3\) |
| \(s\) | \(\tfrac{2}{3}\) | \(-1\tfrac{2}{3}\) | \(0\) | \(0\) | \(1\) | \(\tfrac{-4}{3}\) | \(-20\) | \(\text{R}_2 - 4\text{R}_3\) |
| \(z\) | \(\boxed{\tfrac{1}{3}}\) | \(\tfrac{2}{3}\) | \(1\) | \(0\) | \(0\) | \(\tfrac{1}{3}\) | \(20\) | \(\text{R}_3 \div 3\) |
| \(p\) | \(-133\tfrac{1}{3}\) | \(83\tfrac{1}{3}\) | \(0\) | \(0\) | \(0\) | \(216\tfrac{2}{3}\) | \(13000\) | \(\text{R}_4 + 650\text{R}_3\) |
2. MISREADS – use col \(x\) or col \(y\) (–2 A marks if earned)
(a)
| b.v. | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | value | |
|---|---|---|---|---|---|---|---|---|
| \(r\) | \(0\) | \(\boxed{3}\) | \(2\) | \(1\) | \(-2\) | \(0\) | \(20\) | \(\text{R}_1 - 4\text{R}_2\) |
| \(x\) | \(1\) | \(\tfrac{1}{2}\) | \(2\) | \(0\) | \(\tfrac{1}{2}\) | \(0\) | \(30\) | \(\text{R}_2 \div 2\) |
| \(t\) | \(0\) | \(1\tfrac{1}{2}\) | \(1\) | \(0\) | \(-\tfrac{1}{2}\) | \(1\) | \(30\) | \(\text{R}_3 - \text{R}_2\) |
| \(p\) | \(0\) | \(-175\) | \(50\) | \(0\) | \(175\) | \(0\) | \(10500\) | \(\text{R}_4 + 350\text{R}_2\) |
(b)
| b.v. | \(x\) | \(y\) | \(z\) | \(r\) | \(s\) | \(t\) | value | |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(\tfrac{4}{5}\) | \(1\) | \(2\) | \(\tfrac{1}{5}\) | \(0\) | \(0\) | \(28\) | \(\text{R}_1 - 5\) |
| \(s\) | \(\boxed{1\tfrac{1}{5}}\) | \(0\) | \(2\) | \(-\tfrac{1}{5}\) | \(1\) | \(0\) | \(32\) | \(\text{R}_2 - \text{R}_1\) |
| \(t\) | \(-\tfrac{3}{5}\) | \(0\) | \(-1\) | \(-\tfrac{2}{5}\) | \(0\) | \(1\) | \(4\) | \(\text{R}_3 - 2\text{R}_1\) |
| \(p\) | \(-70\) | \(0\) | \(50\) | \(70\) | \(0\) | \(0\) | \(9800\) | \(\text{R}_4 + 350\text{R}_2\) |
| Scheme | Marks |
|---|---|
| \(x = 0\quad y = 0\quad z = 14\quad r = 0\quad s = 4\quad t = 18\quad p =\) £91 | M1 A2ft, 1ft, 0 |
| (3) |
Notes
M1 3 variables stated – must have completed b.v. + value columns on tableau. Any negatives M0
A1ft all 7 c.a.o. Need £91 ft but accept 9100
A1ft at least 4 c.a.o. (condone \(P = 9100\)ft)
| Scheme | Marks |
|---|---|
| \(p - 90x - 25y + 65r = 9100\) (o.e.) | M1 A1ft |
| (2) |
Notes
M1ft \(P\), \((-)90x\), \((-)25y\), \(65r\) and 9100 (or 91) all present and one = sign
A1ft c.a.o. (o.e.)
(Corrected from the printed mark scheme: the \(y\) term is printed garbled as “\(-2\sqrt{y}\)”; it is \(-25y\), as in the notes and part (e).)
| Scheme | Marks |
|---|---|
| \(p = 9100 + 90x + 25y - 65r\) So increasing \(x\) or \(y\) would increase the profit | B1ft |
| (1) |
Notes
B1ft stating that increasing \(x\) or \(y\) would increase profit, probably re-arranging profit equation. Generous.
(The printed mark scheme shows a part total of 3 here; the part is worth 1 mark.)
| Scheme | Marks |
|---|---|
| The \(\dfrac{2}{5}\) in the \(x\) column and 2nd (\(s\)) row. | B2ft, 1ft, 0 |
| (2) | |
| (15 marks) |
Notes
B2ft \(\dfrac{2}{5}\) identified, \(x\) column and 2nd (\(s\)) row.
Accept ringed in last tableau
B1ft “bad” gets B1, if ft their “optional” tableau B1.