There are two types of fertilizers $F_1$ 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 farmer finds that she needs at least 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?

1000

Let the farmer buy x kg of fertilizer $F_1$​ and y kg of fertilizer $F_2$​. Therefore, $x≥0$ and $y≥0$

The given information can be compiled in a table as follows.

$F_1$​ consists of 10% nitrogen and $F_2$​ consists of 5%  nitrogen. However, the farmer requires at least 14kg of nitrogen.

$∴ 10\% \,of\, x + 5\% \,of\, y≥14$

$\frac{x}{10}​+\frac{y}{20}​≥14$

$2x+y≥280$

$F_1$​ consists of 6% phosphoric acid and $F_2$​ consists of 10% phosphoric acid. However, the farmer requires at least 14kg of phosphoric acid

$∴ 6\%\, of\, x + 10\% \,of\, y≥14$

$\frac{6x}{100}​+\frac{10y}{100}​≥14$

$3x+56y≥700$

Total cost of fettilizers, $Z=6x+5y$

The mathematical formulation of the given problem is

Minimise $Z=6x+5y$.....(1)

subject to the constraints

$2x+y≥280$......(2)

$3x+5y≥700$.....(3)

$x,y≥0$......(4)

The feasible region determined by the system of constraints is as shown

It can be seen that the feasible region is unbounded.

The corner points are A($\frac{700​}{3}$),B(100,80) and C(0,280)

The values of Z at these points are as follows

As the feasible region is unbounded, therefore, 1000 may or may ot be the minimum value of Z

For this, we draw a graph of the inequality $6x+5y<1000$ and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with

$6x+5y<1000$

Therefore, 100kg of fertilizer $F_1$​ and 80 kg of fertilizer $F_2$​ should be used to minimize the cost. The minimum cost is Rs.1000.