Linear Programming
Let x and y be the number of cakes of first and second type that can be made. Clearly x ≥ 0, y ≥ 0.
Let Z be the number of cakes.
Table
Kind |
Number of Cakes |
Flour ret|uired (in gms). |
Fat required (in gms) |
I |
x |
200 x |
25 x |
II |
y |
100 y |
50 y |
Total |
x + y |
200 x + 100 y |
25 x + 50 y |
Mathematical formulation of the problem is as follows:
Maximise Z = x + y subject to the constraints:
200x + 100y ≤ 5000 i.e. 2x + y ≤ 50
25x + 50y ≤ 1000 i.e. x + 2y ≤ 40
x, y ≥ 0
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 2x + y = 50
For x = 0, y = 50
For y = 0, 2 x = 50 or x = 25
∴ line meets OX in A(25, 0) and OY in L(0, 50)
Again, we draw the graph of x + 2y = 40
For x = 0, 2y = 40 or y = 20
For y = 0, x = 40
∴ line meets OX in B(40, 0) and OY in M(0, 20).
Since feasible region satisfies all the constraints.
∴ OACM is the feasible region.
The corner points arc O(0, 0), A(25, 0), C(20, 10), M(0, 20).
At O(0, 0), Z = 0 + 0 = 0
At A(25, 0), Z = 25 + 0 = 25
At C(20, 10), Z = 20 + 10 = 30
At M(0, 20), Z = 0 + 20 = 20
∴ maximum value = 30 at (20, 10)
∴ maximum number of cakes is 20 of one kind and 10 of second kind.
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Solve the following problem graphically:
Minimise and Maximise Z = 3x + 9y
subject to the constraints:
x + 3y ≤ 60
x + y ≥ 10
x ≤ y
x ≥ 0, y ≥ 0
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