A gardener has supply of fertilizer A which consists of $10\%$ nitrogen and $6\%$ phosphoric acid and fertilizer B which consists of $5\%$ nitrogen and $10\%$ phosphoric acid. After testing the soil conditions, he finds that he needs at least $14\text{ kg}$ of nitrogen and $14\text{ kg}$ of phosphoric acid for his crop. If fertilizer A costs ₹ $10.60$ per kg and fertilizer B costs ₹ $8.40$ per kg, what is the minimum cost at which the farmer can meet the nutrient requirement by using a combination of both types of fertilizers?

### Concept & Strategy This is a classic Linear Programming problem. The goal is to minimize a cost function subject to two inequality constraints ("at least" a certain amount of nutrients). The minimum cost in such scenarios almost always occurs at the exact intersection of the two constraint boundary lines. ### Step-by-Step Solution * **Given:** * Let $x$ be the amount (in kg) of fertilizer A used. * Let $y$ be the amount (in kg) of fertilizer B used. * Cost function to minimize: $Z = 10.60x + 8.40y$ * Nitrogen constraint: $10\%$ of A + $5\%$ of B $\ge 14\text{ kg}$. * Phosphoric acid constraint: $6\%$ of A + $10\%$ of B $\ge 14\text{ kg}$. * **Calculation / Deduction:** * Formulate the nitrogen equation: $0.10x + 0.05y = 14$. * Multiply by $100$ to simplify: $10x + 5y = 1400$. * Divide by $5$: $2x + y = 280$ (Equation 1). * Formulate the phosphoric acid equation: $0.06x + 0.10y = 14$. * Multiply by $100$: $6x + 10y = 1400$. * Divide by $2$: $3x + 5y = 700$ (Equation 2). * Find the intersection by substituting $y$ from Eq 1 into Eq 2. * From Eq 1: $y = 280 - 2x$. * Substitute into Eq 2: $3x + 5(280 - 2x) = 700$. * $3x + 1400 - 10x = 700$ * $-7x = -700 \implies x = 100$. * Now, substitute $x$ back into $y = 280 - 2x$: * $y = 280 - 2(100) = 280 - 200 = 80$. * The optimal combination is $100\text{ kg}$ of A and $80\text{ kg}$ of B. * Calculate the total cost: * Cost = $(100 \times 10.60) + (80 \times 8.40) = 1060 + 672 = 1732$. ### Exam Strategy & Shortcut In competitive exams, solving the linear programming graph is too slow. Because both requirements are exactly equal ($14\text{ kg}$ each), the optimal point is simply the intersection of the two constraint lines treated as strict equalities. Quickly form the equations $2x + y = 280$ and $3x + 5y = 700$, solve for $(x,y)$, and plug into the cost function. ### Common Pitfall A common error is trying to guess ratios or use only one type of fertilizer. For instance, using only Fertilizer A to meet both needs would require $233.3\text{ kg}$ (to meet the phosphoric acid need), costing over ₹$2400$, which is far from optimal. Always find the intersection point. ### Final Answer **Therefore, the correct answer is ₹ 1732.**