Question

One kind of cake requires 200 g of flour and 25 g of fat, and another kind of cake requires 100 g of flour and 50 g 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.

Answer 1

Let there be x cakes of first kind and y cakes of second kind.
Number of cakes cannot be negative.
Therefore, x ≥ 0 and y ≥ 0

The given information can be compiled in the form of a table as:

	Flour(gm)	Fat(gm)
Cakes of first kind	200	25
Cakes of second kind	100	50
Availability	5000	1000

Therefore, constraints are:

$$\begin{gathered} 200x+100y\le 5000 \\ 25x+50y\le 1000 \end{gathered}$$

Total number of cakes that can be made = Z = x + y

Thus, the given problem can be mathematically formulated as:

Maximize Z = x + y

subject to constraints,

$$\begin{gathered} 200x+100y\le 5000 \\ 25x+50y\le 1000 \end{gathered}$$
x, y ≥ 0

First we will convert inequations into equations as follows:
200x + 100y = 5000, 25x + 50y = 1000, x = 0 and y = 0

Region represented by 200x + 100y ≤ 5000:
The line 200x + 100y = 5000 meets the coordinate axes at A(25, 0) and B(0, 50) respectively. By joining these points we obtain the line
200x + 100y = 5000 . Clearly (0,0) satisfies the 200x + 100y = 5000. So, the region which contains the origin represents the solution set of the inequation 200x + 100y ≤ 5000.

Region represented by 25x + 50y ≤ 1000:
The line 25x + 50y = 1000 meets the coordinate axes at C(40, 0) and D(0, 20) respectively. By joining these points we obtain the line
25x + 50y = 1000. Clearly (0,0) satisfies the inequation 25x + 50y ≤ 1000. So,the region which contains the origin represents the solution set of the inequation 25x + 50y ≤ 1000.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 200x + 100y ≤ 5000, 25x + 50y ≤ 1000, x ≥ 0 and y ≥ 0 are as follows.

The corner points are O(0, 0), A(25, 0), E(20, 10) and D(0, 20).

The values of Z at these corner points are:

Corner points	Z = x + y
O(0, 0)	0
A(25, 0)	25
E(20, 10)	30
D(0, 20)	20

Thus, maximum numbers of cakes that can be made from 20 of the first kind and 10 of the other kind.