Question

A landscape company is hired to plant trees in three new subdivisions. The company charges the developer for each tree planted, an hourly rate to plant the trees, and a fixed delivery charge. In one subdivision it took 166 labor hours to plant 250 trees for a cost of \(\$ 7520 .\) In a second subdivision it took 124 labor hours to plant 200 trees for a cost of \(\$ 5945 .\) In the final subdivision it took 200 labor hours to plant 300 trees for a cost of \(\$ 8985 .\) Determine the cost for each tree, the hourly labor charge, and the fixed delivery charge.

Short answer

The cost for each tree is \(19.95, the hourly labor charge is \)13.75, and the fixed delivery charge is $250.

Step-by-step solution

Define Variables

Let: - x be the cost per tree,- y be the hourly rate for labor,- z be the fixed delivery charge.

Setup Equations from Given Data

From the given data, derive the three equations based on costs:1. For the 1st subdivision: \[250x + 166y + z = 7520\]2. For the 2nd subdivision: \[200x + 124y + z = 5945\]3. For the 3rd subdivision: \[300x + 200y + z = 8985\]

Eliminate One Variable

Using equations (1) and (2), eliminate variable z:\[ (250x + 166y + z) - (200x + 124y + z) = 7520 - 5945 \]\[ 50x + 42y = 1575 \]Simplify: \[ 25x + 21y = 787.5 \] Call this Equation (4).

Eliminate Same Variable Again

Using equations (2) and (3), eliminate variable z again:\[ (300x + 200y + z) - (200x + 124y + z) = 8985 - 5945 \]\[ 100x + 76y = 3040 \]Simplify: \[ 25x + 19y = 760 \] Call this Equation (5).

Solve for y

Subtract Equation (5) from Equation (4): \[ (25x + 21y) - (25x + 19y) = 787.5 - 760 \]\[ 2y = 27.5 \]\[ y = 13.75 \]

Solve for x

Substitute y back into Equation (5): \[ 25x + 19(13.75) = 760 \]\[ 25x + 261.25 = 760 \]\[ 25x = 498.75 \]\[ x = 19.95 \]

Solve for z

Substitute x and y back into any of the original equations, using Equation (1): \[ 250(19.95) + 166(13.75) + z = 7520 \]\[ 4987.5 + 2282.5 + z = 7520 \]\[ 7270 + z = 7520 \]\[ z = 250 \]

Key concepts

cost analysis

Understanding cost analysis is essential when breaking down the various expenses involved in a project into individual components. In this exercise, we have three primary cost components: the cost per tree, the hourly labor rate, and the fixed delivery charge. By examining the cost given for each subdivision, we can create a clearer picture of where the money is going.
First, we see direct costs like the price charged per tree planted. Then, there's the labor cost, which depends on the number of hours worked. Lastly, there's a fixed delivery charge, which stays the same regardless of the number of trees planted. This structured approach helps us methodically determine how each portion of the overall cost contributes to the final expense.

algebraic equations

Using algebraic equations is a powerful method to represent and solve real-world problems like this one. By defining variables for the different costs, we translate the word problem into mathematical expressions.
In this scenario, we use three variables:

• (x) for the cost per tree
• (y) for the hourly labor rate
• (z) for the fixed delivery charge.

We then derive three linear equations from the given data of cost, number of trees, and labor hours for each subdivision. Each equation relates these variables to the total cost in their respective situations.
These equations are represented as:
1. 250x + 166y + z = 7520
2. 200x + 124y + z = 5945
3. 300x + 200y + z = 8985
These algebraic equations help us systematically approach and solve the problem.

variable elimination

The variable elimination method is handy when solving a system of linear equations. It involves eliminating one variable at a time to simplify the equations progressively.

In this exercise, we start by eliminating the variable (z) from two pairs of equations. First, we subtract the second equation from the first:
(250x + 166y + z) - (200x + 124y + z) = 7520 - 5945
Simplifying this, we get:
50x + 42y = 1575
or 25x + 21y = 787.5 (Equation 4).

Next, we eliminate z from the equations 2 and 3:
(300x + 200y + z) - (200x + 124y + z) = 8985 - 5945
Simplifying, we get:
100x + 76y = 3040
or 25x + 19y = 760 (Equation 5).

This process reduces the system to two equations with two variables (x and y), making the system easier to solve.

substitution method

After simplification using variable elimination, we often use the substitution method to find the values of the remaining variables. From our earlier simplification, we have:
1. 25x + 21y = 787.5
2. 25x + 19y = 760.
Subtracting the second simplified equation from the first:
(25x + 21y) - (25x + 19y) = 787.5 - 760
results in:
2y = 27.5
which gives us:
y = 13.75.
Next, we substitute (y) back into one of the original simplified equations (Equation 5):
25x + 19(13.75) = 760
Simplifying, we find:
25x = 498.75
and:
x = 19.95.
Finally, substituting both (x) and (y) back into any original equation, we use Equation 1:
250(19.95) + 166(13.75) + z = 7520
simplifies to:
7270 + z = 7520
resulting in:
z = 250.
Thus, the substitution method allows us to find the exact values of all the variables.