Problem 64 - Algebra and Trigonometry Real Mathematics, Real People

Question

Consider a company that specializes in potting soil. Each bag of potting soil for seedlings requires 2 units of sand, 1 unit of loam, and 1 unit of peat moss. Each bag of potting soil for general potting requires 1 unit of sand, 2 units of loam, and 1 unit of peat moss. Each bag of potting soil for hardwood plants requires 2 units of sand, 2 units of loam, and 2 units of peat moss. Find the numbers of bags of the three types of potting soil that the company can produce with the given amounts of raw materials. 500 units of sand 750 units of loam 450 units of peat moss

Step-by-Step Solution

Answer

The company can create 50 bags of potting soil for seedlings, 150 bags for general potting, and 125 bags for hardwood plants.

Step 1: Set up the equations

First, let's set up the system of equations based on the given problem. Let \(x\), \(y\), and \(z\) represent the number of bags of potting soil for seedlings, general potting, and hardwood plants, respectively. From the problem, we can set up the following equations: \nFor Sand: \(2x + y + 2z = 500\)\nFor Loam: \(x + 2y + 2z = 750\)\nFor Peat moss: \(x + y + 2z = 450\)

Step 2: Solve the equations

Now, we can use a method to solve these three equations simultaneously. One efficient method is substitution or elimination. First, we can subtract the third equation from the first to get: \(x = 500 - 450 = 50\). Then we substitute \(x = 50\) into the second and third equation, we get \(2y + 2z = 700\), \(y + 2z = 400\), which simplifies to \(y = 150\), \(z = 125\)

Step 3: Interpret the solution

So, the given amounts of materials can produce 50 bags of potting soil for seedlings, 150 bags for general potting, and 125 bags for hardwood plants.

Key Concepts

Linear Algebra

Linear algebra is a key element in many areas of mathematics and is primarily concerned with vector spaces and linear mappings between these spaces. It provides a way to solve for unknown variables in systems of linear equations, like the ones presented when determining the possible production of different types of potting soil given certain resources.

In the given exercise, we are presented with three linear equations, each representing a constraint based on the available raw materials. This is a classic application of linear algebra, where we aim to find a solution set that satisfies all equations simultaneously, representing the number of bags of each potting soil that can be produced. These sets of equations can be represented in matrix form, which is a compact and structured way to handle multiple equations. However, since our focus here is on methods that do not require matrices, such as substitution and elimination, we will not delve further into matrix operations in this explanation.

Understanding the fundamental concepts of linear algebra and how to manipulate and solve linear equations is crucial for students as it forms the basis for more complex mathematical modeling and problem-solving.

Substitution Method

The substitution method is a powerful technique in solving systems of equations, especially when the system involves equations which can be easily isolated for one variable. In our potting soil example, the system of equations is ideally suited for substitution, as we simplified one equation to isolate the variable \(x\).

Here's how the substitution method works in general terms:

• First, solve one of the equations for one variable in terms of the others.
• Next, substitute the expression obtained into the other equations.
• This substitution provides equations with one fewer variable.
• Continue the process until you end up with an equation involving only one variable, solve for it, and backtrack to find the other variables.

This method is particularly useful when equations are not conducive to the elimination method, or when dealing with non-linear systems that can still be reduced to linear equations through substitution.

Elimination Method

The elimination method is another strategic approach to solving systems of equations that works by cancelling out one or more variables to reduce the system to a simpler one. In the context of the potting soil problem, although the step-by-step solution primarily used substitution, elimination could also have been applied.

For a general understanding, the elimination method involves the following steps:

• First, if necessary, multiply the equations by factors so that the coefficients of one of the variables are opposites in two of the equations.
• Add or subtract the equations to eliminate one variable, forming a new system with one fewer equation and one fewer variable.
• Repeat the process as necessary to reduce the system to a single linear equation, which can be solved straightforwardly.
• Substitute the obtained value(s) into the other equations to find the remaining unknowns.

The method is particularly useful when the coefficients are already set up to make elimination straightforward. However, it can become cumbersome if a lot of multiplication or division of the entire equations is required to match coefficients. Understanding when to apply substitution versus elimination can be a subtle skill, and solving various types of systems will help build this intuition.