hi i am sry if i have posted this thread in the wrong section but yh some one plzz help me on this question about linear programming and by the way i am doing AQA D1
Each day, a factory makes three types of hinge: basic, standard and luxury. The hinges produced need three different components: type A, type B and type C.
Basic hinges need 2 components of type A, 3 components of type B and
1 component of type C.
Standard hinges need 4 components of type A, 2 components of type B and 3 components of type C.
Luxury hinges need 3 components of type A, 4 components of type B and 5 components of type C.
Each day, there are 360 components of type A available, 270 of type B and 450 of type C.
Each day, the factory must use at least 720 components in total.
Each day, the factory must use at least 40% of the total components as type A.
Each day, the factory makes x basic hinges, y standard hinges and z luxury hinges.
In addition to x≥0 , y≥0 , z≥0 , find five inequalities, each involving x, y and z, which must be satisfied. Simplify each inequality where possible.
Now i have been able to find three ineqaulities which are 2x+4y+3z≤360, 3x+2y+4z≤ 270, x+3y+5z≤ 450
now i need help to find the other inequalities i checked the mark scheme they give these other ineqalities 6x+9y +12z ≥ 720
⇒2x+ 3y + 4z ≥ 240 (simplified), 2x+4y+3z≥ 2/5 (6x+9y+12z)
2y ≥ 2x + 9z
i don't get how they have come up with the other inequalities so plzzzz help i have exam soon
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- Thread Starter
- 17-01-2010 10:57