1. Linear Programming
Model Answers

3. Type A x + 4y + 3z
≤ 360
Type B x + 2y + 4z
≤ 270
Type C x + 3y + 5z
≤ 450
Altogether x + 9y + 12z
≥ 720 (the minimum you can make it)
Simplified x + 3y + 4z
≥ 240 (the minimum you can make it)
2x + 4y + 3z (Type A) ≥ 0.4(40%) x (6x + 9y + 12z)
10x + 20y + 15z ≥ 2(6x + 9y + 12z) (multiplying by 5)
10x + 20y + 15z ≥ 12x + 18y + 24z (Expand brackets)
You must leave ONE value on the left – as y would be the only value to leave a positive value, bring the y’s over from the right
2y ≥ 2x + 9z

5. Type A x + 4y + 3z
≤ 180 (3 hours is 180 minutes)
Type B x + 8y + 10z
≤ 240 (4 hours is 240 minutes)
Type C x + 12y + 18z
≤ 540 (9 hours is 540 minutes)
Simplified becomes
Type A x + 4y + 3z
≤ 180 (No Common Factors)
Type B x + 4y + 5z
≤ 120 (Divide by 2)
Type C x + 2y + 3z
≤ 90 (Divide by 6)

6. x > y
y > z
x ≥ 0.4 (x + y + z)
5x ≥ 2 (x + y + z)
5x ≥ 2x + 2y + 2z)
3x ≥ 2y + 2z

8. Type A x + 3y + 4z
≤ 360
Type B x + y + 5z
≤ 300
Type C x + 3y + 2z
≤ 400
Type A > Type B x + 3y + 4z >
3x + y + 5z
You must leave ONE value on the left – as y would be the only value to leave a positive value, bring the y’s over from the right
2y > x + z

9. 5x + 4y + 9z ≥ 4x + 3y + 2z
So x + y + 7z ≥ 0
4x + 3y + 2z ≥ 0.4 (9x + 7y + 11z)
20x + 15y + 10z ≥ 2 (9x + 7y + 11z)
20x + 15y + 10z ≥ 18x + 14y + 22z
You must leave at least ONE value on the left – as x and y would leave positive values bring the x and y’s over from the right
2x + y ≥ 12z

11. Type A x + 4y + 2z
≤ 240
Type B x + 3y + 9z
≤ 300
Type C x + 18y + 6z
≤ 900
Simplified becomes
Type A x + 2y + z
≤ 120 (Divide by 2)
Type B x + y + 3z
≤ 100 (Divide by 3)
Type C x + 3y + z
≤ 150 (Divide by 6)
Type C ≥ 2 x Type B x + 18y + 6z ≥ 2(6x + 3y + 9z)
12x + 18y + 6z ≥ 12x + 6y + 18z
12y ≥ 12z
y ≥ z

12. So luxury (z) = basic (x) z = x
Type A x + 2y + z ≤ 120 becomes
Type A x + 2y ≤ 120 (as z = x)
Type A x + y ≤ 60 (Simplified)
Type B x + y + 3z ≤ 100 becomes
Type B x + y ≤ 100 (as z = x)
Type C x + 3y + z ≤ 150 becomes
Type C x + 3y ≤ 150 (as z = x)
Type C x + y ≤ 50 (Simplified)
Lastly if y ≥ z, then y ≥ x (as z = x)

14. Slow x ≥ 190
Medium y ≥ 50
Fast z ≥ 50
Total is x + y + z ≥ 300
Cost x + 2y + 2z ≤ 1000
Cost x + 4y + 4z ≤ 2000
As well x ≥ 0.6(x + y + z)
5x ≥ 3(x + y + z)
5x ≥ 3x + 3y + 3z
2x ≥ 3y + 3z

15. Medium balls (y) = Fast balls (z) so z = y
Slow x ≥ 190 (unaffected)
Fast y ≥ 50 (as z = y)
Total is x + y + z ≥ 300 becomes x + 2y ≥ 300 (as z = y)
Cost x + 4y + 4z ≤ 2000 becomes 5x + 8y ≤ 2000 (as z = y)
As well x ≥ 0.6(x + y + z) = 2x ≥ 3y + 3z
becomes 2x ≥ 6y (as z = y)
becomes x ≥ 3y
Re-arranged to y ≤ ⅓x