Problem 39 - Algebra A Combined Function

Question

The maximum weight that a rectangular beam can support varies jointly as its width and the square of its height and inversely as its length. If a beam \(\frac{1}{2}\) foot wide, \(\frac{1}{3}\) foot high, and 10 feet long can support 12 tons, find how much a similar beam can support if the beam is \(\frac{2}{3}\) foot wide, \(\frac{1}{2}\) foot high, and 16 feet long.

Step-by-Step Solution

Answer

The similar beam can support 22.5 tons.

Step 1: Understand the Relationship

The maximum weight a rectangular beam can support is expressed by the formula \( W = k \times w \times h^2 / l \), where \( W \) is the weight, \( k \) is a constant of proportionality, \( w \) is the width, \( h \) is the height, and \( l \) is the length of the beam.

Step 2: Find the Constant of Proportionality

Use the information given for the first beam to find \( k \). Substitute the values into the equation: \( 12 = k \times \frac{1}{2} \times \left( \frac{1}{3} \right)^2 / 10 \). This simplifies to \( 12 = k \times \frac{1}{2} \times \frac{1}{9} / 10 \). Solving for \( k \) gives \( k = 12 \times \frac{9}{1} \times 2 \times 10 = 2160 \).

Step 3: Calculate the Supported Weight for the Second Beam

For the second beam, substitute into the formula with \( w = \frac{2}{3} \), \( h = \frac{1}{2} \), \( l = 16 \), and \( k = 2160 \). This gives \( W = 2160 \times \frac{2}{3} \times \left( \frac{1}{2} \right)^2 / 16 \). Simplify to find \( W = 2160 \times \frac{2}{3} \times \frac{1}{4} / 16 = 22.5 \).

Step 4: Interpret the Result

The calculation shows that a similar beam with the given dimensions can support a maximum weight of 22.5 tons due to its wider width and greater height compared to length, which impacts its strength positively.

Key Concepts

Understanding the Proportionality Constant

In mathematics, a constant of proportionality is a factor that relates two variables in a proportional relationship. When dealing with joint and inverse variation, a proportionality constant plays a crucial role.

For the problem of the rectangular beam, the proportionality constant, denoted as \( k \), defines the relationship between the beam's dimensions and the weight it can support. Although the dimensions of the beam change, this constant helps in maintaining a consistent relationship.

To find \( k \), use a known example. For instance, in our problem we had a beam capable of supporting 12 tons with set dimensions. With these known values, we substituted them into the formula \( W = k \times w \times h^2 / l \), allowing us to solve and isolate \( k \). In this example, \( k \) turned out to be 2160. This constant underpins how a beam's capacity changes as its dimensions vary.

The Role of the Rectangular Beam

A rectangular beam is a common structural element used in construction, designed to carry loads safely by experiencing bending. Understanding its ability to support weight is essential.

A rectangular beam’s strength derives from its width, height, and length. This exercise illustrates how these dimensions contribute to its ability to support weight.

• Width (\( w \)): contributes linearly to the supported weight.
• Height (\( h \)): its square impacts the supported weight, meaning small increases in height can significantly enhance strength.
• Length (\( l \)): inversely affects the supported weight, indicating longer beams have decreased strength.

Calculating Supported Weight

The supported weight or load capacity of a beam is the maximum weight it can bear without failing. In this problem, the formula for calculating the supported weight involves the dimensions of the beam and the proportionality constant (\( k \)).

Joint variation plays a role here. The beam’s strength varies directly with its width and the square of its height. If either the width or height of the beam increases, its ability to support weight increases as well.

For the second beam in our problem, substituting the dimensions into the formula \( W = 2160 \times \frac{2}{3} \times (\frac{1}{2})^2 / 16 \) gives a supported weight of 22.5 tons. This formula highlights the ability of joint factors to impact the overall strength of the beam.

The Concept of Inverse Variation

Inverse variation is a mathematical relationship between two variables such that as one variable increases, the other decreases. In the context of our rectangular beam problem, the length of the beam (\( l \)) inversely affects the maximum weight it can support.

This means, if the length of the beam becomes longer, its capacity to support weight goes down. Conversely, if the beam is shortened, it can carry more.

In our example, by increasing the length from 10 feet to 16 feet, the supported weight fell from 12 tons to 22.5 tons, even with increases in width and height. This demonstrates how critical inverse variation can be in determining structural integrity in beam design.