Problem 10 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A farmer plans to plant two crops, A and B. The cost of cultivating crop \(\mathrm{A}\) is $$\$ 40 $$ acre whereas that of crop \(\mathrm{B}\) is $$\$ 60$$ acre. The farmer has a maximum of $$\$ 7400$$ available for land cultivation. Each acre of crop A requires 20 labor-hours, and each acre of crop \(\mathrm{B}\) requires 25 labor-hours. The farmer has a maximum of 3300 labor- hours available. If she expects to make a profit of $$\$ 150$$ acre on crop \(\mathrm{A}\) and $$\$ 200$$ acre on crop \(\mathrm{B}\), how many acres of each crop should she plant in order to maximize her profit?

Step-by-Step Solution

Answer

The farmer should plant 165 acres of crop A and 0 acres of crop B to maximize her profit at $$\$ 24,750$$.

Step 1: Formulate the inequalities and objective function based on the given information

The inequalities representing the constraints and the objective function are: 1. Cost constraint: \(40x + 60y \leq 7400\) 2. Labor-hour constraint: \(20x + 25y \leq 3300\) 3. Non-negativity constraint: \(x \geq 0, y \geq 0\) Objective function: Maximize profit P = 150x + 200y

Step 2: Identify the feasible region for the inequalities

To find the feasible region, let's solve and graph the inequality constraints: 1. Cost constraint: \(y \leq \frac{7400 - 40x}{60}\) (This is a downward sloping line) 2. Labor-hour constraint: \(y \leq \frac{3300 - 20x}{25}\) (This is another downward sloping line) 3. Non-negativity constraint: The graph should be in the first quadrant where x and y are both non-negative. Plot all these inequalities on a graph to form the feasible region.

Step 3: Find the corner points of the feasible region

The feasible region in this case is the intersection of all constraints. To find the corner points, check the intersections of the lines formed by the constraints: 1. Intersection of the cost constraint and y-axis (x = 0): \(y \leq \frac{7400 - 0}{60}\), y ≤ 123.33 2. Intersection of the labor-hour constraint and y-axis (x = 0): \(y \leq \frac{3300 - 0}{25}\), y ≤ 132 3. Intersection of the cost constraint and x-axis (y = 0): \(x \leq \frac{7400 - 0}{40}\), x ≤ 185 4. Intersection of the labor-hour constraint and x-axis (y = 0): \(x \leq \frac{3300 - 0}{20}\), x ≤ 165 5. Intersection of the cost constraint and labor-hour constraint: Solve the equations simultaneously, \(40x + 60y = 7400\) and \(20x + 25y = 3300\).

Step 4: Evaluate the objective function at each corner point

Calculate the profit at each corner point using the objective function P = 150x + 200y: 1. (x = 0, y = 0): P = 150(0) + 200(0) = 0 (Minimum profit when no crops are planted) 2. (x = 0, y = 123.33): P = 150(0) + 200(123.33) ≈ 24666 3. (x = 165, y = 0): P = 150(165) + 200(0) = 24750 4. (x = 185, y = 0): This point is not in the feasible region because it is not a part of the labor-hour constraint. 5. (x, y) found by solving the two equations simultaneously: Substitute x and y values into P = 150x + 200y to find the profit.

Step 5: Determine the maximum profit and the corresponding acres for each crop

Compare the profit at each corner point and choose the one with the maximum profit. The point with the maximum profit will be the number of acres to plant for each crop (x = crop A acres, y = crop B acres).

Key Concepts

Formulating Linear Inequalities

When dealing with linear programming in agricultural planning, formulating linear inequalities is a crucial initial step. These inequalities represent the constraints under which the farmer operates. For example, in the given exercise, there are cost and labor-hour constraints that dictate how many acres of crops A and B the farmer can plant.

The cost and labor constraints are presented as mathematical inequalities involving the variables x (acres of crop A) and y (acres of crop B). To formulate them, we use the given information about costs, labor hours, and available resources:

• The cost per acre for each crop is multiplied by the number of acres planted for that crop.
• The labor hours required per acre for each crop are multiplied by the number of acres planted for that crop.

The non-negativity constraints, in this case, simply mean that the number of acres planted cannot be negative, a practical requirement in real-world scenarios.

Objective Function in Optimization

The objective function is the heart of an optimization problem in linear programming. It's the formula we want to maximize or minimize, depending on the goal. For a farmer aiming to maximize profit, the objective function will equate to the total expected profit from all crops.

In our scenario, the objective function is defined as:
\[ P = 150x + 200y \]
where P represents the total profit, 'x' corresponds to the number of acres of crop A, and 'y' corresponds to the number of acres of crop B. The coefficients (150 and 200) represent the profit per acre for crops A and B, respectively. The objective function puts a clear numerical value on the outcomes of different planting decisions, guiding us towards the most profitable option.

Feasible Region in Linear Programming

The feasible region is the set of all possible points (x, y) that satisfy all the constraints of a linear programming problem. It's where all the conditions of the problem come together to create a 'playground' for the possible solutions.

To visualize it, each constraint is graphed on the same set of axes. The area where these graphs overlap is the feasible region, a polygon shaped by the intersection of constraints. Our feasible region is in the first quadrant of the coordinate plane, due to the non-negativity constraints which state that we cannot plant a negative number of acres. The feasible region is very important because the best solution to a linear programming problem lies on its boundary, specifically at one of its corner points.

Corner Point Method

The corner point method is a powerful technique used in linear programming to identify the optimal solution. It is based on the fundamental theorem of linear programming, which states that if there is an optimal solution, it will be at a corner point of the feasible region.

To apply the corner point method, one must first graph the constraints and find the feasible region. Then, identify the corner points of this region by looking for intersections of the constraint lines. Finally, calculate the value of the objective function at each of these points. The corner points are the only candidates for the optimal solution; the optimal values of x and y are the coordinates of the corner point yielding the highest (or lowest if minimizing) objective function value.

Maximizing Profit with Constraints

In agricultural planning, maximizing profit with constraints involves finding the most profitable combination of crops within the limits of resources such as budget, land, and labor. The constraints are expressed as linear inequalities and can include numerous factors, but ultimately, they reflect the limitations faced by the farmer.

The objective function, on the other hand, quantifies profit based on the decision variables, which are the acres of each type of crop to plant in this case. The farmer's task is to choose the number of acres for each crop that maximizes profit while staying within the feasible region defined by the constraints. Through linear programming and specifically the corner point method, the farmer can determine the precise mix of crops that will yield the maximum profit given the available resources.