A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg.) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg. of potash and at most 310 kg. of chlorine.
If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
|
kg. per bag |
||
|
Brand P |
Brand Q |
|
|
Nitrogen |
3 |
3.5 |
|
Phosphoric acid |
1 |
2 |
|
Potash |
3 |
1.5 |
|
Chlorine |
1.5 |
2 |
If the grower wants to maximise the amount of nitrogen added in the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
Table
We are to minimise
subject to the constraints
or
or
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 2y = 240.
For x = 0, 2 y = 240 or y = 120
For y = 0, x = 240
∴ line meets OX in A(240, 0) and OY in L(0, 120).
Now we draw the graph of
2x + y = 180.
For x = 0, y = 180
For y = 0, 2x = 180 or x = 90
∴ line meets OX in B(90, 0) and OY in M(0, 180).
Again we draw the graph of 3x + 4y = 620.
For x = 0, 4 y = 620 or y = 155
For y = 0, 3 x = 620 or x =
line meets OX in 
Since feasible region satisfies all the constraints.
∴ DEF is the feasible region.
The corner points are
D(140, 50), E(20, 140), F(40, 100).
∴ minimum amount of nitrogen is 470 kg. when 40 bags of brand P and 100 bags of brand Q are mixed.
Also maximum value = 595 at (140, 50).
∴ maximum amount of nitrogen is 595 kg. when 140 bags of brand P and 50 bags of brand Q are mixed.
