Q3Advanced Engineering Mathematics

To determine the optimal strategy for the poultry farm, we must first formulate this scenario as a Linear Programming Problem (LPP). The goal is to maximize the weekly profit subject to financial and spatial constraints.

Let the decision variables be: $x_1=\text{Number of old hens purchased}$ $x_2=\text{Number of young hens purchased}$

1. Defining the Constraints

The problem outlines two primary constraints that the farmer must adhere to:

• Budget Constraint: The farmer has a maximum of Rs. 80 to spend. Old hens cost Rs. 2 each, and young hens cost Rs. 5 each. Therefore, the total investment cannot exceed 80: $2x_1+5x_2\le 80$
• Accommodation Constraint: The total capacity of the poultry farm is limited to 20 hens. Therefore, the total number of birds cannot exceed 20: $x_1+x_2\le 20$
• Non-negativity Constraint: The number of hens cannot be negative: $x_1\ge 0,x_2\ge 0$

2. Formulating the Objective Function

We must calculate the net weekly profit generated by each type of hen. The eggs are sold at Rs. 0.30 each, and the weekly feed cost is Rs. 1.00 per hen.

For an Old Hen: Revenue per week = 3 eggs * Rs. 0.30/egg = Rs. 0.90 Cost per week = Rs. 1.00 Net Profit = Revenue - Cost = $0.90-1.00=-\text{Rs. }0.10$ (This is a net loss of 10 paise per week).

For a Young Hen: Revenue per week = 5 eggs * Rs. 0.30/egg = Rs. 1.50 Cost per week = Rs. 1.00 Net Profit = Revenue - Cost = $1.50-1.00=+\text{Rs. }0.50$ (This is a net profit of 50 paise per week).

Therefore, the objective function to maximize total profit $Z$ is: $\text{Maximize }Z=-0.10x_1+0.50x_2$

3. Solving the LPP

By observing the objective function, it is immediately apparent that purchasing old hens directly reduces the total profit because their coefficient is negative ($-0.10$). To maximize profit, the logical step is to set the number of old hens to the absolute minimum allowed by the non-negativity constraint. Therefore, we set $x_1=0$.

Now we maximize $Z=0.50x_2$ subject to the remaining constraints: $5x_2\le 80⟹x_2\le 16$ $x_2\le 20$

The most restrictive constraint here is $x_2\le 16$. Therefore, the maximum allowable number of young hens we can purchase with the available budget is 16.

Final Optimal Solution: The farmer should purchase 0 old hens ($x_1=0$) and 16 young hens ($x_2=16$). The maximum expected weekly profit will be $Z=0.50\times 16=\text{Rs. }8.00$.