Let x be the weight of fertilizer F1 and y be the weight for fertilizer F2. We can construct the following table with the information given in the problem:
Clearly, x, y ≥ 0. Let us construct the following table from the given data

F1 (x) kg 
F2 (y) kg 
Requirements 
Nitrogen 
10% 
5% 
At least 14kg 
Phosphoric Acid 
6% 
10% 
At least 14kg 
Cost Rs. 
6 
5 

We have the following constraints:
10% x + 5% y ≥ 14 $\to$10x + 5y ≥ 1400 $\to$2x + y ≥ 280
6%x + 10% y ≥ 14 $\to$6x + 10y ≥ 1400 $\to$3x + 5y ≥ 700
We need to minimize the cost while the nutrient requirements are met, i.e minimize Z = 6x + 5y.
$\textbf{Plotting the constraints}$:
Plot the straight lines 2x + y = 280 and 3x + 5y = 700
First draw the graph of the line 2x + y = 280.
If x = 0, y = 280 and if y = 0, x = 140. So, this is a straight line between (0,280) and (140,0). At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 0.
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 0. So the area associated with this inequality is unbounded and away from the origin.
Similarly, draw the graph of the line 3x + 5y = 700
If x = 0, y = 700/5 = 140 and if y =0, x = 700/3. So, this is a straight line between (0,140) and (700/3,0).
At (0,0), in the inequality, we have 0 + 0 = 0 which is not ≥ 0. So the area associated with this inequality is unbounded and away from the origin.
$\textbf{Finding the feasible region}$:
We can see that the feasible region is bounded and in the first quadrant.
On solving the equations 2x + y = 280 and 3x + 5y = 700, we get,
3x + 5(2802x) = 700 $\to$3x + 1400 – 10x = 700 $\to$7x = 700 $\to$x = 100.
If x = 100, y = 280200 = 80.
$\Rightarrow x = 100, y = 80 $
Therefore the feasible region has the corner points (0,280), (100,80) and (700/3, 0) as shown in the figure.
$\textbf{Solving the objective function using the corner point method}$
The values of Z at the corner points are calculated as follows:
Corner Point 
Z = 6x+5y 
(0, 280) 
1400 
(100, 80) 
1000 (Minimum Value) 
(700/3, 0) 
1400 
However, we still don’t know if 1000 is indeed the minimum value for Z. We have to check through the inequality Z $\leq$ 1000, i.e, 6x + 5y $\leq$ 1000
If we graph this, we can see that there is no common point w/ the feasible region.
Therefore the minimum cost has to be 1000.
$\textbf{The minimum feasible value for the mixture is 1000.}$