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 nutritional 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?

Explanation

Let the mixture contain
` " " " ` `x` bags of brand P & `y` bags of brand Q



According to Question:

Element A
Brand P `to` 3 units
Brand Q `to` 15 units
Least Requirement `to` 18 units
`therefore`
`3x+1.5y>=18`
`2x+y>=12`

Element B
Brand P `to` 2.5 units
Brand Q `to` 11.25 units
Least Requirement `to` 45 units
`therefore`
`2.5x+11.25y>=45`
`2x+9y>=36`

Element C
Brand P `to` 2 units
Brand Q `to` 3 units
Least Requirement `to` 24 units
`therefore`
`2x+3y>=24`

Also. `x,y>=0`

We need to Minimize the cost, hence the function used here is Minimum`Z`.

Now,
` " " " `Cost of brand P `to` Rs 250
` " " " `Cost of brand Q`to` Rs 200

`therefore` Minimize `Z=250x+200y`

Combining all constraints
` " " " " ` Min `Z=250x+200y`

Subject to constraints
` " " " " ` `2x+y>=12`
` " " " " ` `2x+9y>=36`
` " " " " ` `2x+3y>=24`
` " " " " ` `x>=0,y>=0`




As the feasible region is unbounded, hence, 1950 may or may not be the minimum value of `Z`

For this we need graph inequality
`250x+200y<1950`

Since there is no common point between the feasible region of the inequality.

Hence, the cost will be minimum if

The mixture contains = 3 bags of brand P
and
The mixture contains = 6 bags of brand Q