Problem 50 - Algebra 2

Question

Work begins on digging a moat on the perimeter of a property. After the first weekend, the first worker recruits friend to help. After every succeeding weekend, each moat-digger recruits another friend. One person can dig 15 \(\mathrm{m}^{3}\) of dirt per weekend. The moat should be 4 \(\mathrm{m}\) wide and 3 \(\mathrm{m}\) deep, and it must lie entirely within the \(60 \mathrm{m}-\mathrm{by}-70 \mathrm{m}\) property. a. Geometry Determine the volume of dirt that must be removed for the moat. b. Write an exponential function to model the volume of dirt remaining to be shoveled after \(x\) weekends. c. On which weekend will the moat be completed?

Step-by-Step Solution

Answer

The moat will be completed on the 9th weekend.

Step 1: Determine the volume of the moat

The perimeter of the property is \(60 + 70 + 60 + 70 = 260\,m\). The moat is 4m wide and 3m deep, therefore, the volume is \(260 * 4 * 3 = 3120\,m^3\).

Step 2: Write an exponential function to model the volume of dirt remaining

Every weekend, the number of workers doubles and each worker can dig 15 \(m^3\). The function modeling the amount of dirt dug is \(F(x) = 15 * 2^x\), where \(x\) is the number of weekends.

Step 3: Find out when the moat will be completed

This will take place when the volume dug is equal to the volume of the moat. Solving the equation \(3120 = 15 * 2^x\) for \(x\) gives \(x \approx 8\). Therefore, the moat will be completed on the 9th weekend, because \(x\) cannot be a fraction of a weekend.

Key Concepts

Geometry

Geometry helps us to understand space and shapes, such as lines, angles, and surfaces. In the case of digging a moat, we use geometry to calculate its size and space. A moat typically surrounds a property; understanding its dimensions is crucial.
A moat is like a large trench, and for our situation involves calculating the perimeter of a large rectangular property. This property measures 60 meters by 70 meters. The perimeter of a rectangle is the sum of all its sides:

• Perimeter = 2 * (Length + Width)
• Thus, Perimeter = 2 * (60m + 70m) = 260m

The moat's dimensions further include its width and depth. Knowing these dimensions helps us understand the space it occupies and the sheer volume of dirt that needs to be moved.

Volume Calculation

Volume calculations are essential for understanding the size of the moat not only on the surface but also at its depth. We need to compute the amount of dirt that must be removed to make the moat come into being.

The volume of the moat is calculated by multiplying its dimensions:

• Length = Perimeter, which we calculated as 260m
• Width = 4 meters
• Depth = 3 meters

The formula for calculating the volume of the moat is:

\[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Depth} = 260m \times 4m \times 3m = 3120 \text{m}^3 \]
This means 3120 cubic meters of dirt must be removed. Such volume calculations are important for construction and planning.

Exponential Function

An exponential function helps model scenarios where growth occurs at a constant percentage rate, which can rapidly increase over time. In our moat digging problem, we use it to determine how quickly the work is completed.

Initially, there's only one worker. But each weekend, the workforce doubles as each moat-digger recruits a friend. The amount of dirt dug is expressed by an exponential function. Each worker digs 15 cubic meters per weekend; thus, the function becomes:
\[ F(x) = 15 \times 2^x \]

• \( F(x) \) represents the volume of dirt dug over \( x \) weekends

This exponential increase illustrates rapid growth in the workers' collective effort, which matches real-world scenarios where resources or conditions grow consistently faster over time.

Problem Solving

Problem solving is the process of finding solutions to complex or challenging issues. With our moat problem, it involves calculating the duration needed to finish the digging using the previous concepts.
We know the total volume is 3120 cubic meters and modeled the digging effort with an exponential function. The objective is to find when the remaining dirt will be zero.

Setting our function equal to the moat volume, we solve:

• \( 3120 = 15 \times 2^x \)
• Solving for \( x \), which represents the number of weekends, we find:
  \[ 2^x = \frac{3120}{15} = 208 \]
  Approximately, \( x = 8 \)

This answer implies the moat will be completed after not 8 full weekends, but by the 9th weekend, as \( x \) cannot be fractional in practice. Understanding these steps is key to overcoming any kind of real-life project planning and execution challenges.