Let the farmer buy x kg of fertilizer F1​ and y kg of fertilizer F2​. Therefore, x≥0 and y≥0 The given information can be complied in a table as follows.F1​ consists of 10 % nitrogen and F2​ consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.∴ 10% of x + 5 % of y≥1410x​+20y​≥14 2x+y≥280 F1​ consists of 6 % phosphoric acid and F2​ consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.∴ 6% of x + 10% of y≥14 1006x​+10010y​≥14 3x+56y≥700 Total cost of fertilizers, z = 6x + 5yThe mathematical formulation of the given problem isMinimize 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 follows.It can be seen that the feasible region is unbounded.ointe are A(3700​,0),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 not be the minimum value of Z.For this, we draw a graph of the inequality, 6x + 5y It can be seen that the feasible region has no common point with 6x + 5 y Therefore, 100 kg of fertiliser F1​ and 80 kg of fertilizer F2​ should be used to minimize the cost. The minimum cost is Rs 1000 .