Problem 41 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Jackson Farms has allotted a certain amount of land for cultivating soybeans, corn, and wheat. Cultivating 1 acre of soybeans requires 2 labor-hours, and cultivating 1 acre of corn or wheat requires 6 labor-hours. The cost of seeds for 1 acre of soybeans is \(\$ 12\), for 1 acre of corn is \(\$ 20\), and for 1 acre of wheat is \(\$ 8\). If all resources are to be used, how many acres of each crop should be cultivated if the following hold? a. 1000 acres of land are allotted, 4400 labor-hours are available, and \(\$ 13,200\) is available for seeds. b. 1200 acres of land are allotted, 5200 labor-hours are available, and \(\$ 16,400\) is available for seeds.

Step-by-Step Solution

Answer

In scenario a, 200 acres of soybeans, 280 acres of corn, and 520 acres of wheat should be cultivated. In scenario b, 300 acres of soybeans, 280 acres of corn, and 620 acres of wheat should be cultivated.

Step 1: Scenario a: Setup equations

Given the constraints in scenario a: 1. Total land allotted: 1000 acres 2. Total labor-hours available: 4400 hours 3. Total money available for seeds: $13,200 Let's denote the number of acres of soybeans, corn, and wheat by s, c, and w, respectively. We can create a system of 3 linear equations based on these constraints: (1) Land usage constraint: \(s + c + w = 1000\) (2) Labor-hour constraint: \(2s + 6c + 6w = 4400\) (3) Seed cost constraint: \(12s + 20c + 8w = 13200\)

Step 2: Scenario a: Solve for variables

From equation (1), we can express w as a function of s and c: \(w = 1000 - s - c\) Now we can substitute this expression for w in equations (2) and (3) to eliminate w: (2'): \(2s + 6c + 6(1000 - s - c) = 4400\) (3'): \(12s + 20c + 8(1000 - s - c) = 13200\) Let's solve for c in terms of s in equation (2') and substitute it in equation (3'): (2''): \(c = \frac{2200 - 4s}{5}\) Now substitute this c value into equation (3'): (3''): \(12s + 20(\frac{2200 - 4s}{5}) + 8(1000 - s - \frac{2200 - 4s}{5}) = 13200\) After solving the equation (3'') for s, one would get: \(s = 200\) Now, plug the value of s in equation (2'') to find the value of c: \(c = \frac{2200 - 4(200)}{5} = 280\) Finally, use s and c values in the equation (1) to find w: \(w = 1000 - 200 - 280 = 520\)

Step 3: Scenario a: Conclusion

In scenario a, 200 acres of soybeans, 280 acres of corn, and 520 acres of wheat should be cultivated.

Step 4: Scenario b: Setup equations

Given the constraints in scenario b: 1. Total land allotted: 1200 acres 2. Total labor-hours available: 5200 hours 3. Total money available for seeds: $16,400 We can create a new system of 3 linear equations based on these constraints: (1) Land usage constraint: \(s + c + w = 1200\) (2) Labor-hour constraint: \(2s + 6c + 6w = 5200\) (3) Seed cost constraint: \(12s + 20c + 8w = 16400\)

Step 5: Scenario b: Solve for variables

From equation (1), we can express w as a function of s and c: \(w = 1200 - s - c\) Now we can substitute this expression for w in equations (2) and (3) to eliminate w: (2'): \(2s + 6c + 6(1200 - s - c) = 5200\) (3'): \(12s + 20c + 8(1200 - s - c) = 16400\) Let's solve for c in terms of s in equation (2') and substitute it in equation (3'): (2''): \(c = \frac{2600 - 4s}{5}\) Now substitute this c value into equation (3'): (3''): \(12s + 20(\frac{2600 - 4s}{5}) + 8(1200 - s - \frac{2600 - 4s}{5}) = 16400\) After solving the equation (3'') for s, one would get: \(s = 300\) Now, plug the value of s into equation (2'') to find the value of c: \(c = \frac{2600 - 4(300)}{5} = 280\) Finally, use s and c values in the equation (1) to find w: \(w = 1200 - 300 - 280 = 620\)

Step 6: Scenario b: Conclusion

In scenario b, 300 acres of soybeans, 280 acres of corn, and 620 acres of wheat should be cultivated.

Key Concepts

Systems of Linear Equations

In many real-world situations, solving a problem requires organizing and computing several interrelated variables. This is where systems of linear equations come into play. For instance, consider Jackson Farms' problem where they need to determine the number of acres dedicated to different crops like soybeans, corn, and wheat.
Each equation represents a specific constraint, such as land, labor, and seed costs. These constraints form a linear system:

• Land usage: the sum of all acres must equal the total land available.
• Labor-hour: multiplying the acres by labor required per acre matches the total labor hours.
• Seed cost: the cost should not exceed the budget for seeds.

The solution to the system reveals the exact allocation of resources across different crops, adhering to all constraints. By solving these equations, farmers can optimize their production according to available resources.

Resource Allocation

Resource allocation is a critical concept within linear programming as it involves distributing limited resources effectively among competing needs. In the farm scenario, Jackson Farms must decide how to allocate their limited land, labor hours, and seed budget among the three types of crops.
Proper resource allocation ensures that all resources are used efficiently. In our exercise, this involves covering as much land as possible with the available hours and within the budget. The use of algebraic equations helps identify the best possible mix of soybean, corn, and wheat acres.
Resource allocation helps answer key questions, such as:

• How much of each crop should be planted?
• Which resource constraints are binding or most critical?

Using linear programming ensures optimal decisions that make the best use of available resources.

Optimization

Optimization is the process of making a system as effective as possible. In mathematical terms, it's about maximizing or minimizing an objective function given a set of constraints. In the case of Jackson Farms, optimization involves determining the best number of acres for each crop to maximize land use while staying within labor and budget limits.
This process is crucial since resources are inherently limited, and optimal solutions provide the best use of these resources. The calculation typically involves:

• Setting up an objective function, such as profit or yield, to be optimized.
• Solving the linear equations to find the values that lead to the best possible outcome.

In our scenario, the optimal solution means fully using the allotted land, labor hours, and budget, resulting in maximum possible cultivation.

Constraints in Mathematics

Constraints are the conditions that any solution to a problem must meet. In the context of Jackson Farms, constraints include total available land, maximum labor hours, and seed budget. These constraints frame the problem and define the feasible region in linear programming.
The challenge lies in finding solutions that do not violate these constraints. Constraints are represented mathematically as linear inequalities or equations:

• Total land constraint: acres of soybean, corn, and wheat should not exceed the total land allotted.
• Labor constraint: the total labor required must be within available hours.
• Budget constraint: cost calculation for seeds must stay within financial limits.

Understanding and complying with these constraints are essential, as they ensure the feasibility and legality of any solution reached.