One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
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.
