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A farmer mixes two brands P and Q of cat...

A farmer mixes two brands `P` and `Q` of cattle feed. Brand `P` costing Rs. 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand `Q` costing Rs. 200 per bag contains 1.5 units of nutritonal element A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A,B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

Text Solution

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Let the farmer mixed `x` bags of type `P` and `y` bags of type `Q` cattle feeds. Then

Minimise `Z=250x+200y`…………..1
and constraints
`3x+1.5yge18implies2x+yge12`…………2
`2.5x+11.25yge45implies2x+9yge36`….............3
`2x+3yge24`...................4
`xge0,yge0`...................5
First draw the graph of the line `2x+y=12`

Put `(0,0)` in the inequation `2x+yge12`,
`2xx0+0ge12implies0ge12` (False)
Therefore, the half plane does not contain the origin.
Now, draw the graph of the line `2x+9y=36`.

Put `(0,0)` in the inequation `2x+9yge36`,
`2xx0+9xx0ge36implies0ge36`(False)
Therefore the half plane does not contain the origin.
Now, draw the graph of the line `2x+3y=24`

Put `(0,0)` in the inequation `2x+3yge24`,
`2xx0+3xx0ge40gt2ge24` (False)
Thus, the half plane does not contain the origin,.
Since `x,yge0` so the feasible solution will be in first quadrant. From equations `3x+1.5y=18` and `2x+3y=24` the point of intersection is `C(3,6)`.
Similarly, the point of intersection of the equations `2.5x+11.25y=45` and `2x+3y=24` is `B(9,2)`. Thus, the vertices of the feasible region are `A(18,0),B(9,2),C(3,6)` and `D(0,12)`. 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 1950. For this we draw the graph of inequation `250x+200ylt1950` or `5x+4y39`. There is no common point here. Therefore minimum value of `Z` is 1950 at point `C(3,6)`. Therefore, minimum cost is Rs. 1950 which is obtained by using 3 bags of type `P` and 6 bags of type `Q` cattle feed.
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