The cost of one bag of rice is ₹1800 and one bag of wheat is ₹1200. The total money available is ₹15000, leading to the cost constraint 1800x + 1200y ≤ 15000. The storage capacity is 10 bags, so the total number of bags cannot exceed 10, giving the constraint x + y ≤ 10. Also, the number of bags cannot be negative, so x ≥ 0 and y ≥ 0.

The cost of one bag of rice is ₹1800 and one bag of wheat is ₹1200. The total money available is ₹15000, leading to the cost constraint 1800x + 1200y ≤ 15000. The storage capacity is 10 bags, so the total number of bags cannot exceed 10, giving the constraint x + y ≤ 10. Also, the number of bags cannot be negative, so x ≥ 0 and y ≥ 0.

A man has ₹ 15,000 for purchasing rice and wheat. A bag of rice and a bag of wheat cost ₹ 1,800 and ₹ 1,200 respectively. He has a storage capacity of 10 bags. He earns a profit of ₹ 100 and ₹ 90 per bag of rice and wheat respectively. Assuming that he can sell all the items that he can buy, he purchases x bags of rice and y bags of wheat. Based on the above information and by formulation of Linear Programming Problem (LPP), answer the following questions : (i) Write the objective function which represents the total profit from the sale of total bags of both types. (ii) Write the constraints that relate the total cost of both types of bags. (iii) (a) How many bags of each type should the man buy to get maximum profit ? OR (iii) (b) Find the profit that the man can earn by selling all the bags.

Answer: (i) Objective function: $Z = 100x + 90y$ . (ii) Constraints: $1800x + 1200y \le 15000$ (or $3x + 2y \le 25$ ), $x + y \le 10$ , $x \ge 0$ , $y \ge 0$ . (iii) (a) The man should buy 5 bags of rice and 5 bags of wheat. (iii) (b) The maximum profit the man can earn is ₹950.

Step-by-step solution

Step 1: Define Variables and Objective Function

Let $x$ be the number of bags of rice and $y$ be the number of bags of wheat. The profit from one bag of rice is ₹100 and from one bag of wheat is ₹90. Therefore, the objective function, which represents the total profit $Z$ , is the sum of the profits from $x$ bags of rice and $y$ bags of wheat.

Step 2: Formulate Constraints

The cost of one bag of rice is ₹1800 and one bag of wheat is ₹1200. The total money available is ₹15000, leading to the cost constraint $1800x + 1200y \le 15000$ . The storage capacity is 10 bags, so the total number of bags cannot exceed 10, giving the constraint $x + y \le 10$ . Also, the number of bags cannot be negative, so $x \ge 0$ and $y \ge 0$ .

Step 3: Simplify Constraints and Identify Corner Points

The cost constraint $1800x + 1200y \le 15000$ can be simplified by dividing by 600, resulting in $3x + 2y \le 25$ . We need to find the feasible region defined by these constraints and the non-negativity conditions. The corner points of this feasible region are found by solving the equations of the boundary lines.

Step 4: Calculate Corner Points

The corner points of the feasible region are: the origin $(0{,}0)$ , the intersection of $3x+2y=25$ with the $x$ -axis $(25/3, 0)$ , the intersection of $x+y=10$ with the $y$ -axis $(0{,}10)$ , and the intersection of $3x+2y=25$ and $x+y=10$ . Solving $x+y=10 \implies y=10-x$ and substituting into $3x+2y=25$ gives $3x+2(10-x)=25 \implies 3x+20-2x=25 \implies x=5$ . Then $y=10-5=5$ . So, the intersection point is $(5{,}5)$ .

Step 5: Evaluate Objective Function at Corner Points

We evaluate the objective function $Z = 100x + 90y$ at each corner point of the feasible region. The maximum value of $Z$ will correspond to the maximum profit.

Step 6: Determine Maximum Profit

Comparing the values of $Z$ at the corner points, the maximum profit of ₹950 is achieved when $x=5$ bags of rice and $y=5$ bags of wheat are purchased.

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