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```NCEA Answers – Linear Programming
2003 Sample
Q2b
2bVertex
(2,6)
(3,6)
(4,8)
(3,12)
(2,13)
R = 54 000m + 36 000c
324 000
378 000
504 000
594 000
576 000
maximum return, R = \$594 000
when m = 3 and c = 12.
Q7a
x = number of Xvi bought
y = number of Yco bought
The constraints are:
x + y ≤ 60
1
60x + 120y ≤ 4800
x and y non-negative
Objective function:
R = 10x + 12y
Vertex
R = 10x + 12y
(0,0)
(60,0)
(40,20) 640
(0,40)
0
600
480
maximum return, R = \$640
when Xvi = 40 and Yco = 20.
Q7b
New objective function:
R = 10x + 20y
Vertex
(0, 0)
(60, 0)
(40, 20)
(0, 40)
R = 10x + 20y
0
600
800
800
maximum return, R = \$800
when you choose any integer (x, y) with 0 ≤ x ≤ 40 and
x + 2y = 80.
Multiple solutions because the objective function and a constraint have the same slope.
2004
Q3
Must calculate return at the vertices (0,320), (600,120), (720,0) or use the
“parallel line test” to identify the optimal vertex.
(i)
s = 600 and b = 120.
(ii)
Maximum return R = \$2400
2
Q4
(a)
The constraints are:
100a + 100b ≤ 60 000
50a ≤ 25 000
50a + 100b ≤ 45 000
a and b non-negative.
Objective function: P = 2.45a + 1.95b
Calculate R at the vertices (0,450), (300,300) and (500,100), or use the “parallel line test” to identify the optimal
vertex.
Maximum return R = \$1420 when a = 500 and b = 100.
(b) Gradient of the objective function is now the same as the first constraint and is valid between 300 ≤ a ≤ 500 and
a + b = 600.
2005
Q3 (a)
Vertex P = 9x + 12y
(20,60) 900
(40,50) 960
(56,40) 984
(b)
Maximum profit, P = \$984. ([Accept P = \$975 for (55,40) or P = \$993 for (57,40).]
Q7
x = number of medium frames made per week.
y = number of large frames made per week.
The constraints are:
4 x + 6 y ≤ 480
6 000 x + 12 000 y ≤ 900 000
20 ≤ x ≤ 60
20 ≤ y ≤ 60
Objective function: P = 6 x + 10 y
3
(a)
Vertex P = 6 x + 10 y
(60,40) 760
(30,60) 780
maximum profit
when x = 30 and y = 60.
(b)
Moving the line 4x + 6y = 480 upwards, until it reaches the edge of the region bounded by the other
constraints, leads to it crossing at the point where x = 60.
⇒ x = 60 and y = 45.
⇒ P = 810 >780, so this is the new maximum profit.
⇒ minimum hours = 8.5 h (8 hours & 30 minutes, 510min)
OR
Removing the Time constraint introduces the point (60, 45) into the feasible region.
⇒ P = 810 >780, so this is the new maximum profit.
⇒ minimum hours = 8.5 h (8 hours and 30 minutes, 510min)
2006
Q3 (b)
(b)
Identify vertices and value of I
(15,12) \$35.55
(15,40) \$74.75
(30,30) \$79.50
(45,12) \$73.05 maximum income when 30 of each soap is produced.
OR
Use parallel line method to identify the point (30,30) as the optimal solution
along with statement of number of each soap needed.
[Evidence needed – maybe on the graph.]
4
Q6
(a)
(i)
The constraints are:
30x + 40y ≤ 1800
2x + 5y ≤ 190
x ≥ 10
y ≥ 15
(ii)
Objective function is P = 8x + 12y
EITHER
calculate P at the vertices
Vertex P = 8x + 12y
(10,15) 260
(10,34) 488
(20,30) 520
(40,15) 500
OR
Use the parallel line test to identify the optimal vertex.
Therefore maximum profit is when Vili produces 20 sun shelters and 30 tents.
(b) There will now be multiple solutions as the objective function is now parallel to the constraint 30x + 40y = 1800.
Therefore all solutions will be integer points on the line 30x + 40y = 1800 between x = 20 and x = 40.
Need
• more than 2 solutions
• where the solutions are
• boundaries.
Desirable
• Whole number solutions.
2007
Q3
(a)
(b)
Vertex P = 9x + 16y
(0,0) 0
(0,45) 720
5
(20,30) 660
(40,0) 360
OR
Parallel line method to identify point (0,45) as optimal solution. [evidence needed – maybe on graph]
AND
So need to make 0 gnomes and 45 frogs .
[Accept x = 0 and y = 45 as x and y defined in question].
[A clear logical argument as to why “as many frogs as possible” may gain A].
2007
Q6(a)
(i)
x = no. of standard butterflies
y = no. of monarch butterflies
Constraints:
40x + 90y ≤ 2160
8x + 12y ≤ 336
x≥9
y≥9
(ii)
Objective Function
P = 18x + 60y
EITHER
Vertex P
(9,9) 702
(9,20) 1362
(18,16) 1284
(28,9) 1044
[or (28.5,9) giving P = \$1053
or (27,10) giving P = \$1086]
Max. profit when 9 standard and 20 monarch produced each day.
OR
Shown by parallel line method.
(b)
Vertex (9,20) no longer in feasible region.
Maximum profit of \$1218 now at point (21,14)
Ie: 21 standard and 14 monarch .
May use parallel line method.
2008
Q7
6
(a)
(b)
Line passing through (0,90) and (90,0) evident
and feasible region shown
Points for optimisation:
(30,45) I = \$1 425 000
(60,30) I = \$1 350 000
(80,10) I = \$1 050 000
OR
Parallel line method
The grower should plant 30 hectares of
tomatoes and 45 hectares of artichokes
Q8
x = sheep, y = cows
0.3 x + 0.4 y ≤ 120
180 x + 150 y ≤ 54 000
y ≤ 250
Intersection points for feasible region are
Points of particular interest are at (0, 250),
æ 2
ö æ
ö
1
ç 66 , 250 ÷ , ç133 , 200 ÷ , (300, 0)
è 3
ø è
ø
3
(66,250) = \$258 700 income
(67,249) = \$258 550 income
(133,200) = \$263 100 income
(132,201) = \$263 250 income
The farmer should run 132 sheep and 201 cows
(accept also 133 sheep and 200 cows).
(b)
Optimisation equation has changed to 560x + 850y .
New income is \$249 730 (from 68 sheep and 249 cows)
OR
New income is \$249 460 (from 66 sheep and 250 cows).
2009
Q3 Only need to focus on two key points (others have to be smaller by coordinates).(20, 255/11)⇒ I = \$101 363
7
(accept 25.5, 25.45 ⇒ I =\$101 500, \$101 350)
(13, 30) ⇒ I = \$107
187.50 (accept \$107 187.50, \$107 188)
He should grow 13 . tonnes of corn and 30 tonnes of tomatoes. (Must have 13. Or 13.75.)
(b)
New constraint: 9c + 8t ≤ 360 (or equivalent)
New vertices are (13⅓, 30) ⇒ I = \$131 666 and (20, 22 ) ⇒ I = \$118 750
Maximises income using13⅓ tonnes of corn and 30 tonnes of tomatoes .
(c)
For multiple solutions to occur, the function to be optimised is parallel to one of the constraints.
(Do not accept the constraints are all in the same place.)
E.G.
2010
Q2
(a)
C = 4.5a + 8b
(b)
Vertex Cost in \$
(60,60) 750
(20,60) 570
(40,20) 340
(60,10) 350
OR
Parallel line method to identify point (40,20) as optimal solution.
[Evidence needed – maybe on graph.]
So need to make 40 trips in Van A and 20 Trips in Van B.
(b)
New objective function C = 5a + 10b has the same gradient as the constraint 6a + 12b ≥ 480.
Therefore any integer (whole) coordinates along the line 6a + 12b = 480, for 40 ≤ a ≤ 60, will minimise the fuel cost.
If answer is correct except no mention of integer (whole) number solutions.
2011
8
(a)
Vertices of feasible region are:
Point Vertex (m,a)
A (5,10) 40 minutes
B (8,7) 37 minutes
C (8,10) 46 minutes
So therefore the assessment should have 7 algebra Q and 8 measurement questions .
(b)
Constraints are:
a≥m
a + m ≤ 20
a + m ≥ 11
a≥ 8
Vertices of feasible region are:
Point Vertex (a,m) Object fn = 4a + 6m
A (10,10) 100 minutes
B (20,0) 80 minutes
C (11,0) 44 minutes
D (8,3) 50 minutes
E (8,8) 80 minutes
therefore minimum time is 44 minutes.
(c)
Test would now have between 6 and 20 algebra questions and between 0 and 10 measurement questions
OR similar
9
```