Problem 39 - Precalculus Mathematics for Calculus

Question

Agriculture A farmer has 1200 acres of land on which he grows corn, wheat, and soybeans. It costs \(\$ 45\) per acre to grow corn, \(\$ 60\) to grow wheat, and \(\$ 50\) to grow soybeans. Because of market demand, the farmer will grow twice as many acres of wheat as of com. He has allocated \(\$ 63,750\) for the cost of growing his crops. How many acres of each crop should he plant?

Step-by-Step Solution

Answer

The farmer should plant 250 acres of corn, 500 acres of wheat, and 450 acres of soybeans.

Step 1: Define Variables

Let \( x \) be the number of acres for corn, \( y \) be the number of acres for wheat, and \( z \) be the number of acres for soybeans. We need to find the values of \( x \), \( y \), and \( z \).

Step 2: Set up the Equations

We have three main equations. Based on the land:\[ x + y + z = 1200 \]Based on the cost:\[ 45x + 60y + 50z = 63750 \]Based on the market demand:\[ y = 2x \]

Step 3: Substitute Demand Equation

Substitute \( y = 2x \) in the land equation:\[ x + 2x + z = 1200 \]which simplifies to:\[ 3x + z = 1200 \]

Step 4: Substitute Demand Equation in Cost Equation

Substitute \( y = 2x \) in the cost equation:\[ 45x + 60(2x) + 50z = 63750 \]\[ 45x + 120x + 50z = 63750 \]\[ 165x + 50z = 63750 \]

Step 5: Solve the System of Two Equations

We have two equations:1. \( 3x + z = 1200 \)2. \( 165x + 50z = 63750 \)From equation 1, express \( z \) in terms of \( x \):\[ z = 1200 - 3x \]Substitute into equation 2:\[ 165x + 50(1200 - 3x) = 63750 \]\[ 165x + 60000 - 150x = 63750 \]\[ 15x = 3750 \]\[ x = 250 \]

Step 6: Find Wheat and Soybean Acres

Calculate \( y \) and \( z \):For wheat, since \( y = 2x \):\[ y = 2 \times 250 = 500 \]Substitute \( x = 250 \) into \( z = 1200 - 3x \):\[ z = 1200 - 3(250) \]\[ z = 1200 - 750 \]\[ z = 450 \]

Step 7: Verify the Solution

Verify by substituting \( x = 250 \), \( y = 500 \), and \( z = 450 \) into the original cost equation:\[ 45(250) + 60(500) + 50(450) \]\[ = 11250 + 30000 + 22500 \]\[ = 63750 \]The costs match the budget, confirming the solution.

Key Concepts

Systems of Equations

Often in complex problems, we encounter multiple conditions that need to be satisfied simultaneously. In mathematical terms, these conditions are represented using systems of equations. A system of equations consists of two or more equations that share common variables. Solving these systems allows us to find the values of the variables that satisfy all the given conditions simultaneously.

For instance, in the farmer's problem, we have three conditions: the total land usage, the cost of planting different crops, and the market demand constraint. These are captured in three equations with common variables representing different acre allocations. Systems of equations can be solved using various methods, including graphing, substitution, and elimination, each offering a unique way to approach the solution. Using these methods, you can uncover the optimal combination that meets all the criteria laid out by the problem.

Substitution Method

The substitution method is a powerful technique used to solve systems of equations, particularly when one of the equations can be easily solved for one variable. The method involves substituting the expression obtained for one variable into the other equations. This reduces the system to fewer equations, often making them easier to handle and solve.

In the farmer's scenario, we use the equation derived from market demand, where wheat acreage is twice that of corn. By expressing wheat in terms of corn (i.e., \(y = 2x\)), this expression is substituted into the other two conditions: total land used and total cost. This substitution simplifies the original three-equation system into a two-equation system, making it easier to find the specific acreages needed for corn, wheat, and soybeans.

• Start by solving one equation for one variable.
• Substitute this expression into the remaining equations.
• Solve the simplified system and backtrack to find values for all variables.

By breaking down complex dependencies into simpler terms, the substitution method efficiently determines variable values.

Agricultural Optimization

In agricultural optimization, farmers make decisions to maximize outputs or minimize costs under certain constraints like land size, budget, and market demands. This often involves determining how much of each crop to plant to achieve desired economic outcomes within these constraints.

For the farmer in our exercise, choices about crop distribution are crucial for both profitability and sustainability. By using linear equations to model the various constraints, the farmer can quantitatively analyze which planting strategies will lead to the most efficient use of resources. Successful agricultural optimization uses math to balance costs (like labor and material) and benefits (such as crop yield and market prices).

• Consider all possible constraints: land capacity, budget limits, demand forecasts.
• Use systems of equations to model these constraints mathematically.
• Apply mathematical methods, like substitution, to find optimal solutions that adhere to constraints.

These calculations help farmers make informed decisions that influence their economic success and resource sustainability.