There are two types of fertilisers `F_1`and `F_2`. `F_1`, consists of 10% nitrogen and 6% phosphoric acid and `F_2`consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a fanner finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If `F_1` cost Rs.6/kg and `F_2` costs Rs.5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Explanation

Let the fertilizer `F_1` requirement be `x` kg
and the fertilizer `F_2` requirement be `y` kg



According to Question:

Nitrogen
Quantity of Nitrogen in `F_1=10%`
Quantity of Nitrogen in `F_2=5%`
Minimum Available = 14 kg
`therefore`
`10%` of `x+5% `of `y >=14`
`(10)/(100)x+(5)/(100)y>=14`
`2x+y>=280`

Phosphoric Acid
Quantity of Phosphoric Acid in `F_1=6%`
Quantity of Phosphoric Acid in `F_2=10%`
Minimum Available = 14 kg
`therefore`
`6%` of `x+10% `of `y >=14`
`(6)/(100)x+(10)/(100)y>=14`
`3x+5y>=700`
Also,
`x>=0,y>=0`

As we need to minimize the cost, hence the function used here is Minimize `Z`

Cost of fertilizer `F_1` = Rs 6/kg
Cost of fertilizer `F_2` = Rs 5/kg

Minimize `Z=6x+5y`

Combining all constraints:
Minimize `Z=6x+5y`
`2x+y>=280`
`3x+5y>=700`
`x,y>=0`







As the feasible area is unbounded
Hence 1000 may or may not be the minimum value of Z.

For this we need graph inequality





Since, there is no common point between feasible region and `6x+5y<1000`

Hence the cost will be minimum, if
` " " " `Fertilizer `F_1` used = 100 kg
` " " " `Fertilizer `F_2` used = 80 kg

Minimum cost = Rs 1000