Let $x$ be the amount (in kg) of fertilizer $F_1$ used.
Let $y$ be the amount (in kg) of fertilizer $F_2$ used.
$F_1$ contains 10% nitrogen, so nitrogen from $F_1$ = $0.10x$ kg.
$F_2$ contains 5% nitrogen, so nitrogen from $F_2$ = $0.05y$ kg.
Total nitrogen requirement is at least 14 kg.
So, constraint for nitrogen: $0.10x + 0.05y \geq 14$
Similarly, $F_1$ contains 6% phosphoric acid → $0.06x$ kg.
$F_2$ contains 10% phosphoric acid → $0.10y$ kg.
Total phosphoric acid requirement is at least 14 kg.
So, constraint for phosphoric acid: $0.06x + 0.10y \geq 14$
The cost of $F_1$ is ₹ 6 per kg, so cost from $x$ kg = ₹ $6x$.
The cost of $F_2$ is ₹ 5 per kg, so cost from $y$ kg = ₹ $5y$.
Total cost to minimize: $Z = 6x + 5y$
Also, $x \geq 0$, $y \geq 0$ (can’t use negative quantity of fertilizer).
Hence, the linear programming problem is:
Minimize $Z = 6x + 5y$
Subject to:
$0.10x + 0.05y \geq 14$
$0.06x + 0.10y \geq 14$
$x \geq 0$, $y \geq 0$