Question

A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods A and B, are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?

Answer 1

Let the sick person takes x units and y units of food I and II respectively that were taken in the diet.
Since, per unit of food I costs Rs 4 and that of food II costs Rs 3.
Therefore, x units of food I costs Rs 4x and y units of food II costs Rs 3y.
Total cost = Rs (4x + 3y)
Let Z denote the total cost
Then, Z = 4x + 3y
If one unit of A contains 200 units of vitamin and one unit of food B contains 100 units of vitamin.
Thus, x units of food I and y units of food II contains 200x + 100y units of vitamin.
But a diet for a sick person must contain at least 4000 units of vitamins.

∴  \[200x + 100y \geq 4000\]
If one unit of A contains 1 unit of mineral and one unit of food B contains 2 units of mineral.
Thus, x units of food I and y units of food II contains x + 2y units of mineral.

But a diet for a sick person must contain at least 50 units of vitamins.

∴  \[x + 2y \geq 50\] If one unit of A contains 40 calories and one unit of food B contains 40 calories.
Thus, x units of food I and y units of food II contains 40x + 40y units of calories.

But a diet for a sick person must contain at least 1400 calories.

\[\therefore 40x + 40y \geq 1400\]

Finally, the quantities of food I and food II are non negative values.
So

\[x, y \geq 0\]

Hence, the required LPP is as follows:
Min Z = 4x + 3y
subject to 

\[200x + 100y \geq 4000\]
\[x + 2y \geq 50\]
\[40x + 40y \geq 1400\]
\[x, y \geq 0\]

First, we will convert the given inequations into equations, we obtain the following equations:
200x + 100y = 4000, x +2y = 50, 40x + 40y =1400, x = 0 and y = 0

Region represented by 200x + 100y ≥ 4000:
The line 200x + 100y = 4000 meets the coordinate axes at A₁(20, 0) and B₁(0,40) respectively. By joining these points we obtain the line
200x + 100y = 4000.Clearly (0,0) does not satisfies the inequation 200x + 100y ≥ 4000. So,the region in xy plane which does not contain the origin represents the solution set of the inequation 200x + 100y ≥ 4000.

Region represented by x +2y ≥ 50:
The line x +2y = 50 meets the coordinate axes at C₁(50, 0) and D₁(0, 25) respectively. By joining these points we obtain the line
x +2y = 50.Clearly (0,0) does not satisfies the x +2y ≥ 50. So,the region which does not contains the origin represents the solution set of the inequation x +2y ≥ 50.

Region represented by 40x + 40y ≥ 1400:
The line 40x + 40y = 1400 meets the coordinate axes at E₁(35, 0) and F₁(0, 35) respectively. By joining these points we obtain the line
40x + 40y = 1400.Clearly (0,0) does not satisfies the inequation 40x + 40y ≥ 1400. So,the region which does not contains the origin represents the solution set of the inequation 40x + 40y ≥ 1400.

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 ≥ 4000,x +2y ≥ 50, 40x + 40y ≥ 1400, x ≥ 0, and y ≥ 0 are as follows.

The corner points of the feasible region are B₁(0, 40), G₁(5, 30), H₁(20, 15) and C₁(50, 0)
 
The value of the objective function at these points are given by the following table

Points	Value of Z
B1	4(0)+3(40) = 120
G1	4(5)+3(30) = 110
H1	4(20) + 3(15) = 125
C1	4(50)+3(0) = 200

The minimum cost is Rs 110 which is at G₁(5, 30).

Hence, the required combination of food is 5 units of food A and 30 units of food B.