Question

A housewife wishes to mix together two kinds of food, X and F in such a way that the mixtures contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C.
The vitamin contents of one kg of food are given below :

	Vitamin A	Vitamin B	Vitamin C
Food X	1	2	3
Food Y	2	2	1

One kg of food X costs Rs 6 and one kg of food F cost Rs 10. Find the least cost of the mixture which will produce the diet.

Answer 1

Let x kg of food X and Y kg of food F are mixed together to make the mixture. 

Since one kg of food X contains one unit of Vitamin A and one kg of food Y contains 2 units of Vitamin A. 

Therefore x kg of food X and y kg of food Y will contain x + 2y units of Vitamin A. 

But the mixture should contain at least 10 units of Vitamins A. 

Therefore, x + 2y ≥ 10 

Similarly, x kg of food X and y kg of food F will produce 2x + 2y units of Vitamin B and 3x + y units of Vitamin C. 

But minimum requirements of Vitamin B and Vitamin C are respectively of 12 and 8 units. 

Therefore, 2x + 2y ≥ 12 and 3x + y ≥ 8 

Since, the quantity of food X and food y cannot be negative. 

∴ x ≥ 0, y ≥ 0 

It is given that one kg of food X costs Rs 6 and one kg of food Y costs Rs 10. 

So, x kg of food X and y kg of food F will cost Rs (6x + 10y). 

Thus, the given linear programming problem is 

Minimize Z = 6x +10y 

Subject to the constraints, 

x + 2y ≥ 10 

2x + 2y ≥ 12 

3x + y ≥ 8 

and x ≥ 0, y ≥ 0 

Converting the in equation into equations 

x + 2y – 10 

2x + 2y = 12 

3x + y = 8

Region represented by x + 2y ≥ 10 : 

The line x + 2y = 10, meets the coordinate axis at ,A(10, 0) and B(0, 5). x + 2y = 10

x	10	0
y	0	5

A (10, 0); 5(0,5) Join the points A and B to obtain the line. We find that, the point (0, 0) does not satisfy the in equation x + 2y ≥ 10. 

So, the region opposite to the origin represents the solution set of the in equation. 

Region represented by 2x + 2y ≥ 12 : 

The line 2x + 2y = 12, meets the coordinate axis at C(6, 0) and D(0, 6). 2x + 2y = 12

x	6	0
y	0	6

C(6, 0); D(0, 6) Joint the points C and D to obtain the line. We find that, the point (0, 0) does not satisfy the in equation 2x + 2y ≥ 10. 

So, the region opposite to the origin, represents the solution set of the in equation. 

Region represented by 3x + y ≥ 8 : 

The line 3x + y = 8 meets the coordinate axis at E (8/3 ,0);F(0,8) 

3x + y = 8

x	8/3	0
y	0	8

E (8/3 ,0); F(0,8) Join the points E and F to obtain the line. 

We find that, the point (0,0) does not satisfy the inequation 3x + y ≥ 8. 

So, the region opposite to the origin repesents the solution set of the in equation. 

Region represented by x ≥ 0, y ≥ 0: 

Since every point in the first quadrant satisfies the in equations. 

So, the first quadrant is the region represented by the in equation x ≥ 0 and y ≥ 0. 

The shaded region AGHB represents the region of the above in equations. This region the feasible region of the given LLP.

The coordinates of the comer points of the shaded feasible region are .4(10, 0), G(2, 4), H(1, 5) and B(0, 8). 

Where the points G and H have been obtained by solving the equations of the corresponding intersecting lines, simu-ltaneously.

The value of the objective function at these points are given in the following table :

Point	x-coordinates	y-coordinates	Objective function Z = 6x + 10y
A	10	0	ZA = 6(10) + 10(0) = 60
G	2	4	ZG = 6(2) + 10(4) = 52
H	1	5	ZE = 6(1) + 10(5) = 56
B	0	8	ZB = 6(0)+ 10(8) = 80

 Clearly, Z is minimum at x = 2 and y = 4. 

The minimum value of Z is 52. 

We observe that the open half plane represented by 6x + 10y < 52 does not have points in common with the feasible region.

So, Z has minimum value equal to 52.