Problem 52 - Algebra 2

Question

Construction A concrete supplier sells premixed concrete in \(300-f t^{3}\) truckloads. The area \(A\) that the concrete will cover is inversely proportional to the depth \(d\) of the concrete. a. Write a model for the relationship between the area and the depth of a truckload of poured concrete. b. What area will the concrete cover if it is poured to a depth of 0.5 \(\mathrm{ft}\) ? A depth of 1 \(\mathrm{ft}\) ? A depth of 1.5 \(\mathrm{ft}\) ? c. When the concrete is poured into a circular area, the depth of the concrete is inversely proportional to the square of the radius \(r\) . Write a model for this relationship.

Step-by-Step Solution

Answer

a. The model representing the relationship between the area and depth is \(A=300\). b. The areas covered for depths of 0.5ft, 1ft and 1.5ft are 600ft², 300ft², and 200ft² respectively. c. The model representing the relationship between concrete depth and radius when poured in a circular area is \(d=300/(πr^{2})\).

Step 1: Part a - Formulating the model

The problem statement mentions that the area \(A\) is inversely proportional to the depth \(d\), which mathematically can be represented as: \(A=k/d\), where \(k\) is the constant of variation. Given that a truckload is \(300 \, ft^{3}\), we can assign this as the volume \(V\) which is the product of area and depth - \(A*d=V\). Substituting \(A=k/d\) in the volume equation gives: \(k=V*d\), which upon substitution of given values yields: \(k=300*d\). Substituting this in the \(A=k/d\) equation gives final model as: \(A=300\).

Step 2: Part b - Calculate the areas

For \(d=0.5ft\), substituting \(d=0.5\) in the model \(A=300\), we get \(A=600ft^{2}\). For \(d=1ft\), likewise, we get \(A=300ft^{2}\). And for \(d=1.5ft\), we get \(A=200ft^{2}\).

Step 3: Part c - Formulate the concrete depth and radius model

The area of the circular area is given by \(A=πr^{2}\), and from part a, we have \(A=300/d\). Equating both equations, we get \(πr^{2}=300/d\), which simplifies to \(πr^{2}d=300\). Rearranging we get \(d=300/(πr^{2})\). Hence, the depth \(d\) is inversely proportional to the square of radius \(r\).

Key Concepts

Area and Depth Relationship

When discussing the concept of inverse variation, the area and depth relationship is a classic example. In the context of pouring concrete, if you want the concrete to spread across a greater area, you’ll have to pour it at a shallower depth, and vice versa. The inverse relationship is mathematically represented by the equation:

• \( A = \frac{k}{d} \)

where \(A\) is the area the concrete covers, \(d\) is the depth, and \(k\) is a constant. For a fixed volume, like the 300 cubic feet in a concrete truckload, this relationship holds true because deepening the depth reduces the area covered. The constant \(k\) can be found by multiplying area and depth to give the total volume, leading, in this case, to \(A = 300/d\). This formula is practical in construction, as knowing any two of these variables - volume, area, or depth - can help you identify the third.

Concrete Volume Calculation

Understanding how to calculate the volume of concrete is essential in construction. Here, the concrete's volume is pre-set at 300 cubic feet per truck. This volume doesn't change, which means that any increase in the depth of the concrete must result in a decrease in the area it covers, and vice versa. To calculate the volume, use:

• Volume = Area \( \times \) Depth

This equation showcases how for a fixed volume, either the depth or the area can shift, but not both together without adjustment. Given the problem at hand, when depth changes to various values like 0.5 feet, 1 foot, or 1.5 feet, the area will accordingly adjust as 600 ft², 300 ft², and 200 ft² respectively keeping the volume constant at 300 cubic feet. This gives a practical blueprint for deciding how much area you can cover depending on your desired depth.

Radius and Depth Relationship

In the case of circular areas, the radius plays a crucial role in determining the depth of concrete that can be applied. Here, the inverse variation is a bit different because the area of a circle is related to the square of the radius \(r\) by the formula:

• \( A = \pi r^2 \)

Given the same constant volume, we use this to define how depth \(d\) varies inversely with \(r^2\). The relationship is expressed mathematically as:

• \( d = \frac{300}{\pi r^2} \)

This shows that as the radius increases, the depth must decrease to maintain the volume constant at 300 cubic feet. This relationship allows builders to predict how changing the radius of their pour will affect the concrete depth. It's a vital calculation in planning projects involving circular concrete slabs, such as pools or fountains, ensuring that the application stays within design parameters.