Problem 27 - Precalculus: Functions and Graphs

Question

A supplier of lawn products has three types of grass fertilizer, \(\mathrm{G}_{1}, \mathrm{G}_{2},\) and \(\mathrm{G}_{3},\) having nitrogen contents of \(30 \%, 20 \%,\) and \(15 \%,\) respectively. The supplier plans to mix them, obtaining 600 pounds of fertilizer with a \(25 \%\) nitrogen content. The mixture is to contain 100 pounds more of type \(\mathrm{G}_{3}\) than of type \(\mathrm{G}_{2}\). How much of each type should be used?

Step-by-Step Solution

Answer

380 pounds of \( G_1 \), 60 pounds of \( G_2 \), and 160 pounds of \( G_3 \).

Step 1: Define Variables

Let \( x \) be the pounds of fertilizer \( G_1 \), \( y \) be the pounds of fertilizer \( G_2 \), and \( z \) be the pounds of fertilizer \( G_3 \). We need to find the values of \( x, y, \text{ and } z \).

Step 2: Set Up Equations for Total Weight

The total weight of the mixture should be 600 pounds: \( x + y + z = 600 \).

Step 3: Set Up Equation for Nitrogen Content

The nitrogen content should be 25% of the total 600 pounds: \( 0.30x + 0.20y + 0.15z = 0.25 \times 600 \). Simplifying, we get: \( 0.30x + 0.20y + 0.15z = 150 \).

Step 4: Relation Between G2 and G3

The mixture contains 100 pounds more of \( G_3 \) than of \( G_2 \), hence: \( z = y + 100 \).

Step 5: Substitute and Simplify Equations

Substitute \( z = y + 100 \) into the equation from Step 2: \( x + y + (y + 100) = 600 \). This simplifies to: \( x + 2y + 100 = 600 \). Further simplifying gives: \( x + 2y = 500 \).

Step 6: Substitute and Simplify Nitrogen Equation

Substitute \( z = y + 100 \) in the nitrogen equation from Step 3: \( 0.30x + 0.20y + 0.15(y + 100) = 150 \). Simplify this to: \( 0.30x + 0.20y + 0.15y + 15 = 150 \). This further simplifies to: \( 0.30x + 0.35y = 135 \).

Step 7: Solve System of Equations

Now solve the system of equations:1. \( x + 2y = 500 \) 2. \( 0.30x + 0.35y = 135 \).Multiply the second equation by 10 for easier manual calculation: \( 3x + 3.5y = 1350 \).Now, solve for \( x \) using the first equation: \( x = 500 - 2y \). Substitute \( x \) into the second equation:\( 3(500 - 2y) + 3.5y = 1350 \).Solving gives: \( 1500 - 6y + 3.5y = 1350 \).Combine like terms: \( -2.5y = -150 \).Solve for \( y \): \( y = 60 \).

Step 8: Find Values of Other Variables

Substitute \( y = 60 \) back into the equation \( x = 500 - 2y \):\( x = 500 - 2(60) = 380 \).Now, find \( z \) using \( z = y + 100 \):\( z = 60 + 100 = 160 \).

Step 9: Verify Solution

Verify the solution by plugging back into the original equations:1. Total weight: \( 380 + 60 + 160 = 600 \).2. Nitrogen content: \( 0.30 \times 380 + 0.20 \times 60 + 0.15 \times 160 = 114 + 12 + 24 = 150 \).Both checks confirm the solution is correct.

Key Concepts

Nitrogen Content

The nitrogen content of a fertilizer is an essential indicator of its strength in providing nutrients to plants. Nitrogen is a critical element in plant growth, as it is a major component of chlorophyll, which plants use in photosynthesis.
In this specific exercise, the nitrogen content is expressed as a percentage for each type of grass fertilizer: - Fertilizer \( G_1 \) has a nitrogen content of 30%.- Fertilizer \( G_2 \) holds 20% nitrogen.- Fertilizer \( G_3 \) contains 15% nitrogen.For a mixture, calculating the overall nitrogen content involves considering each component's contribution proportionate to its weight. The supplier wants to achieve an overall nitrogen content of 25% in a 600-pound mixture.
This requires setting up a linear equation to ensure the sum of nitrogen contributions from each fertilizer type equals 150 pounds (25% of 600). Understanding how to distribute and mix different fertilizers involves striking the right equilibrium of their individual nitrogen percentages.

Mixtures

A mixture, in mathematical terms, involves combining different elements to form a whole. When we talk about fertilizer mixtures, we deal with combining specific amounts to achieve desired properties, such as nutrient content. In this exercise, the aim was to mix three different fertilizers to get a combined 600-pound blend.
Furthermore, conditions like having 100 pounds more of fertilizer \( G_3 \) than \( G_2 \) add constraints to our mixture. Such constraints are algebraically represented in equations and are crucial for determining the exact quantities of each component.
As seen in the solution, you begin by defining variables (in this case, \( x, y, \text{ and } z \)) as unknowns representing the weight for each fertilizer type \( G_1 \), \( G_2 \), and \( G_3 \). Then:

• You set up equations based on total weight: \( x + y + z = 600 \).
• You check for nitrogen content correctness: \( 0.30x + 0.20y + 0.15z = 150 \).
• You factor constraints like \( z = y + 100 \).

This example illustrates not only the precision required in mathematical mixtures but also how systems of equations elegantly solve real-world problems involving multiple constraints and objectives.

Linear Equations

Linear equations are pivotal tools in solving systems like these, where multiple conditions must be met simultaneously. They express relationships with constant coefficients, forming straight-line graphs in two variables. In this problem, linear equations assist in determining how to split 600 pounds of fertilizers while meeting specified nitrogen content.
Each equation formed addresses one aspect of the problem:

• The sum of all fertilizer weights is constrained: \( x + y + z = 600 \).
• The nitrogen content requirement is met with: \( 0.30x + 0.20y + 0.15z = 150 \).
• The relationship defining the difference in weight between \( G_3 \) and \( G_2 \): \( z = y + 100 \).

These equations work jointly to provide solutions, revealing the amounts of each fertilizer needed. By manipulating these equations through substitution and elimination methods, we efficiently find the values of \( x \), \( y \), and \( z \).
It's the systematic approach of linear algebra that brings clarity and precision to the problem, making it possible to achieve the desired mixture of fertilizers correctly.