There are two types of fertilisers ${F_1}$ and ${F_2} \cdot {F_1}$ 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 atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If ${F_1}$ costs Rs.6/kg and ${F_2}$ costs Rs. 5/kg, then determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost ?

.: Let x kg of fertiliser ${F_1}$ and y kg of fertiliser ${F_2}$ be required.

Clearly $x \ge 0,y \ge 0$ . We make the following table from the given data.

Since the fertiliser contains atleast 14 kg of nitrogen and 14 kg of phosphoric acid,

we have the constraints :
$\cfrac{{10}}{{100}}x + \cfrac{5}{{100}}y \ge 14;\cfrac{6}{{100}}x + \cfrac{{10}}{{100}}y \ge 14$

Total cost Z of purchasing x kg of fertiliser F, and y kg of fertiliser ${F_2}$ is $Z = 6x + 5y$

Hence, the mathematical formulation of the problem is

Minimize $Z = 6x + 5y$ ...(1)
subject to the constraints
$2x + y \ge 280$ ...(2)

$3x + 5y \ge 700$ ...(3)

$x,y \ge 0$ ...(4)

Let ${l_1}:2x + y = 280,{l_2}:3x + 5y = 700$

Let us graph the inequalities (2) to (4).

The feasible region determined by ,the system is shown in the graph.

Here again, observe that the feasible region is unbounded.

Let us evaluate Z at the comer points $C\left( {\cfrac{{700}}{3},0} \right),E(100,80)$ and B(0, 280).

In the table, we find that smallest value of Z is 1000 at the point E(100, 80).

Now, that the feasible region is unbounded,

we have to draw the graph of the inequality $6x + 5y < 1000$ to check whether

the resulting open half plane has any point common with the feasible region.

From the graph, we see that it has no points in common.

Thus, the minimum value of Z is 1000 attained at (100, 80).

Hence, 100 kg of fertiliser ${F_1}$ and 80 kg of fertiliser ${F_2}$

should be used so that the nutrient requirements are met at minimum