Question -
Answer -
Let the first kind of cakes be x and second kind of cakes be y.Hence,
x ≥0 and y ≥0
The given information can be compiled in a table as shown below
| Flour (g) | Fat (g) |
| Cakes of first kind, x | 200 | 25 |
| Cakes of second kind, y | 100 | 50 |
| Availability | 5000 | 1000 |
So, 200x + 100y ≤ 5000
2x + y ≤ 50
25x + 50y ≤ 1000
x + 2y ≤ 40
Total number of cakes Z that can be made are
Z = x + y
The mathematical formulation of the given problem can be writtenas
Maximize Z = x + y (i)
Here, subject to the constraints,
2x + y ≤ 50 (ii)
x + 2y ≤ 40 (iii)
x, y ≥ 0 (iv)
The feasible region determined by the system of constraints isgiven as below
A (25, 0), B (20, 10), O (0, 0) and C (0, 20) are the cornerpoints
The values of Z at these corner points are as given below
| Corner point | Z = x + y | |
| A (25, 0) | 25 | |
| B (20, 10) | 30 | Maximum |
| C (0, 20) | 20 | |
| O (0,0) | 0 | |
Hence, the maximum numbers of cakes that can be made are 30 (20cakes of one kind and 10 cakes of other kind)