Let's break down and formulate the problem step-by-step: --- # Step 1: Define Decision Variables Let: - \( x_1 \) = Acres allocated to **tomatoes** - \( x_2 \) = Acres allocated to **lettuce** - \( x_3 \) = Acres allocated to **radishes** --- # Step 2: Objective Function (Profit Maximization) ### **Revenue:** - Tomatoes: \( 2,000 \) kg/acre × ₹1.00/kg × \( x_1 \) acres = ₹2,000\( x_1 \) - Lettuce: \( 3,000 \) heaps/acre × ₹.75/heap × \( x_2 \) acres = ₹2,250\( x_2 \) - Radishes: \( 1,000 \) kg/acre × ₹2.00/kg × \( x_3 \) acres = ₹2,000\( x_3 \) ### **Fertilizer Cost:** - Tomatoes: 100 kg/acre × ₹.50/kg × \( x_1 \) = ₹50\( x_1 \) - Lettuce: 100 kg/acre × ₹.50/kg × \( x_2 \) = ₹50\( x_2 \) - Radishes: 50 kg/acre × ₹.50/kg × \( x_3 \) = ₹25\( x_3 \) ### **Labour Cost:** - Tomatoes: 5 man-days/acre × ₹20/man-day × \( x_1 \) = ₹100\( x_1 \) - Lettuce: 6 man-days/acre × ₹20/man-day × \( x_2 \) = ₹120\( x_2 \) - Radishes: 5 man-days/acre × ₹20/man-day × \( x_3 \) = ₹100\( x_3 \) ### **Net Profit:** \[ \begin{align*} \text{Tomatoes:} & \quad ₹2,000x_1 - ₹50x_1 - ₹100x_1 = ₹1,850x_1 \\ \text{Lettuce:} & \quad ₹2,250x_2 - ₹50x_2 - ₹120x_2 = ₹2,080x_2 \\ \text{Radishes:} & \quad ₹2,000x_3 - ₹25x_3 - ₹100x_3 = ₹1,875x_3 \\ \end{align*} \] **Total profit to maximize:** \[ \text{Maximize:} \quad Z = 185x_1 + 208x_2 + 1875x_3 \] --- # Step 3: Constraints ### **1. Land Constraint:** \[ x_1 + x_2 + x_3 \leq 100 \] ### **2. Labour Constraint:** \[ 5x_1 + 6x_2 + 5x_3 \leq 400 \] ### **3. Non-negativity:** \[ x_1 \geq ,\quad x_2 \geq ,\quad x_3 \geq \] --- # Step 4: Complete Linear Programming Model \[ \begin{align*} \text{Maximize:} \quad & Z = 185x_1 + 208x_2 + 1875x_3 \\ \text{Subject to:} \quad & x_1 + x_2 + x_3 \leq 100 \\ & 5x_1 + 6x_2 + 5x_3 \leq 400 \\ & x_1 \geq ,\ x_2 \geq ,\ x_3 \geq \end{align*} \] --- # **Conclusion** This linear programming model will help the farmer determine the optimal allocation of his 100-acre farm among tomatoes, lettuce, and radishes to **maximize the total profit**, considering land, fertilizer, and labor constraints. If you want to solve for the actual values of \( x_1, x_2, x_3 \), you can use the simplex method or any LP solver.

# Step-by-Step Solution for Linear Programming Problem ## Step 1: Define Decision Variables We begin by defining our decision variables, which represent the acres allocated to each crop: - **\( x_1 \)**: Acres allocated to **tomatoes** - **\( x_2 \)**: Acres allocated to **lettuce** - **\( x_3 \)**: Acres allocated to **radishes** This sets the foundation for the model, allowing us to quantify the areas dedicated to each crop. --- ## Step 2: Objective Function (Profit Maximization) ### Revenue Calculation We calculate the revenue generated from each crop based on the yield per acre and the selling price: - **Tomatoes**: \[ \text{Revenue} = 2,000 \, \text{kg/acre} \times ₹1.00/\text{kg} \times x_1 = ₹2,000x_1 \] - **Lettuce**: \[ \text{Revenue} = 3,000 \, \text{heaps/acre} \times ₹0.75/\text{heap} \times x_2 = ₹2,250x_2 \] - **Radishes**: \[ \text{Revenue} = 1,000 \, \text{kg/acre} \times ₹2.00/\text{kg} \times x_3 = ₹2,000x_3 \] ### Cost Calculation Next, we calculate the costs associated with fertilizers and labor for each crop: - **Fertilizer Costs**: - **Tomatoes**: \[ 100 \, \text{kg/acre} \times ₹0.50/\text{kg} \times x_1 = ₹50x_1 \] - **Lettuce**: \[ 100 \, \text{kg/acre} \times ₹0.50/\text{kg} \times x_2 = ₹50x_2 \] - **Radishes**: \[ 50 \, \text{kg/acre} \times ₹0.50/\text{kg} \times x_3 = ₹25x_3 \] - **Labour Costs**: - **Tomatoes**: \[ 5 \, \text{man-days/acre} \times ₹20/\text{man-day} \times x_1 = ₹100x_1 \] - **Lettuce**: \[ 6 \, \text{man-days/acre} \times ₹20/\text{man-day} \times x_2 = ₹120x_2 \] - **Radishes**: \[ 5 \, \text{man-days/acre} \times ₹20/\text{man-day} \times x_3 = ₹100x_3 \] ### Net Profit Calculation We can now determine the net profit for each crop: - **Tomatoes**: \[ ₹2,000x_1 - ₹50x_1 - ₹100x_1 = ₹1,850x_1 \] - **Lettuce**: \[ ₹2,250x_2 - ₹50x_2 - ₹120x_2 = ₹2,080x_2 \] - **Radishes**: \[ ₹2,000x_3 - ₹25x_3 - ₹100x_3 = ₹1,875x_3 \] ### Total Profit Function The objective function to maximize total profit becomes: \[ \text{Maximize: } Z = 1850x_1 + 2080x_2 + 1875x_3 \] --- ## Step 3: Constraints ### 1. Land Constraint The total acres used cannot exceed the available land: \[ x_1 + x_2 + x_3 \leq 100 \] ### 2. Labour Constraint The total labour required should not exceed the available man-days: \[ 5x_1 + 6x_2 + 5x_3 \leq 400 \] ### 3. Non-negativity Constraints The acres allocated to each crop must be non-negative: \[ x_1 \geq 0, \quad x_2 \geq 0, \quad x_3 \geq 0 \] --- ## Step 4: Complete Linear Programming Model We can summarize the complete linear programming model as follows: \[ \begin{align*} \text{Maximize:} \quad & Z = 1850x_1 + 2080x_2 + 1875x_3 \\ \text{Subject to:} \quad & x_1 + x_2 + x_3 \leq 100 \\ & 5x_1 + 6x_2 + 5x_3 \leq 400 \\ & x_1 \geq 0, \quad x_2 \geq 0, \quad x_3 \geq 0 \end{align*} \] --- ## Conclusion This linear programming model provides a structured approach for the farmer to determine the optimal allocation of his 100-acre farm among tomatoes, lettuce, and radishes. The objective is to maximize total profit while adhering to land and labor constraints. To find the specific values of \( x_1, x_2, \) and \( x_3 \), one can apply the simplex method or use any linear programming solver.

# Step-by-Step Solution for Linear Programming Problem ## Step 1: Define Decision Variables We begin by defining our decision variables, which represent the acres allocated to each crop: - **\( x_1 \)**: Acres allocated to **tomatoes** - **\( x_2 \)**: Acres allocated to **lettuce** - **\( x_3 \)**: Acres allocated to **radishes** ### Explanation Defining decision variables is crucial as they represent the unknowns we aim to solve for in the optimization problem. These variables will be used to form our objective function and constraints. --- ## Step 2: Objective Function (Profit Maximization) ### Revenue Calculation We calculate the revenue generated from each crop based on the yield per acre and the selling price: - **Tomatoes**: \[ \text{Revenue} = 2,000 \, \text{kg/acre} \times ₹1.00/\text{kg} \times x_1 = ₹2,000x_1 \] - **Lettuce**: \[ \text{Revenue} = 3,000 \, \text{heaps/acre} \times ₹0.75/\text{heap} \times x_2 = ₹2,250x_2 \] - **Radishes**: \[ \text{Revenue} = 1,000 \, \text{kg/acre} \times ₹2.00/\text{kg} \times x_3 = ₹2,000x_3 \] ### Cost Calculation Next, we calculate the costs associated with fertilizers and labor for each crop: - **Fertilizer Costs**: - **Tomatoes**: \[ 100 \, \text{kg/acre} \times ₹0.50/\text{kg} \times x_1 = ₹50x_1 \] - **Lettuce**: \[ 100 \, \text{kg/acre} \times ₹0.50/\text{kg} \times x_2 = ₹50x_2 \] - **Radishes**: \[ 50 \, \text{kg/acre} \times ₹0.50/\text{kg} \times x_3 = ₹25x_3 \] - **Labour Costs**: - **Tomatoes**: \[ 5 \, \text{man-days/acre} \times ₹20/\text{man-day} \times x_1 = ₹100x_1 \] - **Lettuce**: \[ 6 \, \text{man-days/acre} \times ₹20/\text{man-day} \times x_2 = ₹120x_2 \] - **Radishes**: \[ 5 \, \text{man-days/acre} \times ₹20/\text{man-day} \times x_3 = ₹100x_3 \] ### Net Profit Calculation We can now determine the net profit for each crop: - **Tomatoes**: \[ ₹2,000x_1 - ₹50x_1 - ₹100x_1 = ₹1,850x_1 \] - **Lettuce**: \[ ₹2,250x_2 - ₹50x_2 - ₹120x_2 = ₹2,080x_2 \] - **Radishes**: \[ ₹2,000x_3 - ₹25x_3 - ₹100x_3 = ₹1,875x_3 \] ### Total Profit Function The objective function to maximize total profit becomes: \[ \text{Maximize: } Z = 1850x_1 + 2080x_2 + 1875x_3 \] ### Explanation The objective function encapsulates our goal of maximizing profit while accounting for revenues and costs associated with each crop. This function will guide the optimization process. --- ## Step 3: Constraints ### 1. Land Constraint The total acres used cannot exceed the available land: \[ x_1 + x_2 + x_3 \leq 100 \] ### 2. Labour Constraint The total labour required should not exceed the available man-days: \[ 5x_1 + 6x_2 + 5x_3 \leq 400 \] ### 3. Non-negativity Constraints The acres allocated to each crop must be non-negative: \[ x_1 \geq 0, \quad x_2 \geq 0, \quad x_3 \geq 0 \] ### Explanation Constraints are essential to ensure that the solution is feasible. They represent limitations in resources like land and labor, ensuring that the farmer operates within realistic boundaries. --- ## Step 4: Complete Linear Programming Model We can summarize the complete linear programming model as follows: \[ \begin{align*} \text{Maximize:} \quad & Z = 1850x_1 + 2080x_2 + 1875x_3 \\ \text{Subject to:} \quad & x_1 + x_2 + x_3 \leq 100 \\ & 5x_1 + 6x_2 + 5x_3 \leq 400 \\ & x_1 \geq 0, \quad x_2 \geq 0, \quad x_3 \geq 0 \end{align*} \] ### Explanation This formulation brings together the objective function and constraints, creating a complete linear programming model. It provides a structured framework for optimization. --- ## Conclusion This linear programming model provides a structured approach for the farmer to determine the optimal allocation of his 100-acre farm among tomatoes, lettuce, and radishes. The objective is to maximize total profit while adhering to land and labor constraints. To find the specific values of \( x_1, x_2, \) and \( x_3 \), one can apply the simplex method or use any linear programming solver.