A diet is to contain at leat 80 units of vitamin A and 100 units of minerals. Two foods `F_(1)` and `F_(2)` are available. Food `F_(1)` cost Rs. 4 per unit and `F_(2)` costs Rs. 6 per unit. One unit of food `F_(1)` contains 3 units of vitamin A and 4 units of minerals. One unit of food `F_(2)` contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

Explanation

Let `x` units of `F_(1)` and `y` units of `F_(2)` be in the food. Then,

Maximise `Z=4x+6y`…………….1
Constraints `3x+6yge80` ………………….2
`4x+3yge100`……………….3
`xge0,yge0`…………………..4
First, draw the graph of the line `3x+6y=80`,.

Put `(0,0)` in the inequation `3x+6yge80`,
`3xx0+6xx0ge80implies0ge80` (False)
Thus, the half plane does not contain the origin.
Since `x,yge0` so the feasible solution is in the first quadrant.
Now, draw the graph of the line `4x+3y=100`

Put `(0,0)` in the inequation `4x+3yge100`,
`4xx0+3xx0ge100`
`implies0ge100` (False)
Thus, the half plane does not contain the origin.
From equations `3x+6y=80` and `4x+3y=100`. The point of intersection is `B(24, 4/3)`
Clearly, the feasible region is unbounded.
The vertices the feasible region are `A(80/3,0), B(24,4/3)` and `C(0,100/3)`. We find the value of `Z` at these vertices.

Since the feasible region is unbounded, so the minimum value of `z` may or may not be 104. For this we draw the graph of inequations `4x+6t lt 104` or `2x+3ylt52`. There is no common point here, so the minimum cost of mixture is Rs. 104.