Problem 69 - Precalculus: Mathematics for Calculus

Question

A farmer has 500 acres of arable land on which he wants to plant potatoes and corn. The farmer has \(\$ 40,000\) available for planting and \(\$ 30,000\) for fertilizer. Planting 1 acre of potatoes costs \(\$ 90,\) and planting 1 acre of corn costs \(\$ 50 .\) Fertilizer costs \(\$ 30\) for 1 acre of potatoes and \(\$ 80\) for 1 acre of com. (a) Find a system of inequalities that describes the number of acres of each crop that the farmer can plant with the available resources. Graph the feasible region. (b) Can the farmer plant 300 acres of potatoes and 180 acres of corn? (c) Can the farmer plant 150 acres of potatoes and 325 acres of corn?

Step-by-Step Solution

Answer

(b) Yes, (c) No.

Step 1: Define Variables

Let's introduce two variables: let \(x\) represent the acres of potatoes and \(y\) represent the acres of corn that the farmer plants.

Step 2: Formulate Cost Constraints

According to the problem, planting 1 acre of potatoes costs \\(90 and planting 1 acre of corn costs \\)50. The farmer has \$40,000 for planting. Therefore, the cost constraint is: \[90x + 50y \leq 40,000\]

Step 3: Formulate Fertilizer Constraints

For fertilizer, it costs \\(30 per acre for potatoes and \\)80 per acre for corn with a total budget of \$30,000. Thus, the fertilizer constraint is: \[30x + 80y \leq 30,000\]

Step 4: Land Area Constraint

The total land available is 500 acres, so the constraint for the total land is: \[x + y \leq 500\]

Step 5: Non-Negativity Constraints

Since negative values of planting do not make sense, we have the following constraints: \[x \geq 0\] and \[y \geq 0\]

Step 6: Graph the Feasible Region

Graph the inequalities on a coordinate plane to identify the feasible region: - Plot lines for each equation: 1. \(90x + 50y = 40,000\) 2. \(30x + 80y = 30,000\) 3. \(x + y = 500\)- The feasible region is the area where all inequalities are satisfied and is bounded by constraints.

Step 7: Evaluate Situation (b)

Check if \(x = 300\) and \(y = 180\) satisfy all constraints: 1. \(90(300) + 50(180) = 27,000 + 9,000 = 36,000 \leq 40,000\) 2. \(30(300) + 80(180) = 9,000 + 14,400 = 23,400 \leq 30,000\)3. \(300 + 180 = 480 \leq 500\)All constraints are satisfied.

Step 8: Evaluate Situation (c)

Check if \(x = 150\) and \(y = 325\) satisfy all constraints: 1. \(90(150) + 50(325) = 13,500 + 16,250 = 29,750 \leq 40,000\)2. \(30(150) + 80(325) = 4,500 + 26,000 = 30,500 > 30,000\)The fertilizer constraint is not satisfied.

Key Concepts

Linear Programming

Linear programming is a mathematical technique used to find the best outcome in a situation with various constraints. In the context of farming decisions, it's about maximizing land use while managing limited resources like money or materials. This specific problem involves deciding how many acres to plant for two crops: potatoes and corn.

The goal is to create a system of inequalities that represents the constraints like costs for planting and fertilizing, along with the total available land. By solving these inequalities, we can determine the best way to utilize the available resources and reach an optimal solution, or determine the best planting strategy that won't exceed resource limits.

Resource Allocation

Resource allocation refers to how resources such as money, land, and materials are distributed to different options or tasks. In this scenario, the farmer needs to decide how to allocate funds across the acreage for potatoes and corn.

To make the best decisions, costs for each crop for planting and fertilizing are considered within the total available budget. For potatoes, each acre planted costs \\(90, and for corn, each acre requires \\)50. Fertilizer costs are \\(30 per acre for potatoes and \\)80 per acre for corn.

This information allows the farmer to strategically allocate resources to stay within budget and cover as much land as possible for crop growth.

Feasible Region

The feasible region is a visual representation where all constraints are satisfied. Essentially, any combination of planting potatoes and corn inside this region will meet all restrictions that a farmer faces. These include land acreage, money for planting, and fertilizer costs.

The region is defined by plotting the lines of each constraint inequality:

• The planting cost constraint: \(90x + 50y \leq 40,000\)
• The fertilizer cost constraint: \(30x + 80y \leq 30,000\)
• The total land constraint: \(x + y \leq 500\)

This visual helps quickly identify which planting strategies are viable by seeing what combinations fall within the allowed boundaries.

Cost Constraints

Cost constraints are specific restrictions based on the budget available for planting crops. The farmer in this example has \\(40,000 available for planting costs and \\)30,000 for fertilizers. Each acre of potatoes costs \\(90 to plant and \\)30 to fertilize, while corn costs \\(50 per acre to plant and \\)80 to fertilize.

These costs turn into mathematical inequalities:

• Planting cost: \(90x + 50y \leq 40,000\)
• Fertilizing cost: \(30x + 80y \leq 30,000\)

The farmer must plant within these limits to ensure he does not exceed his budget, guiding the decision-making process for how many acres of each crop to plant.

Fertilizer Constraints

Fertilizer constraints outline the limits on the amount of fertilizer money that can be spent. The budget is constrained at \\(30,000. Potatoes require \\)30 of fertilizer per acre, while corn demands a more substantial \$80 per acre.

For the farmer, maintaining this budget means setting up constraints like: \[30x + 80y \leq 30,000\]This inequality needs to be satisfied in line with the other constraints related to land and planting costs, to find the balance that allows productive planting without overspending on fertilizers.