The Mathematical Analysis of a Conference Hall Floor Area

One of the basic requirements for designing a conference hall involves determining its dimensions to fit specific space constraints. For instance, if a conference hall floor is to be covered with tiles and its length is 36 feet longer than its width, while its area is constrained to be less than 240 square feet, we can set up a set of mathematical equations and inequalities to determine the possible dimensions.

Understanding the Problem

Given that the floor of a conference hall is to be completely covered with tiles, and knowing that its length is 36 feet longer than its width, let's denote the width of the conference hall as W (for simplicity, we will continue to use w as the variable). The length, therefore, can be expressed as:

Equation 1: Length (L) W 36

Furthermore, we know that the area of the floor must be less than 240 square feet. The area of a rectangle is given by multiplying its length by its width. Hence, we can write the area as:

Equation 2: Area (A) L × W

Substituting the expression for L from Equation 1 into Equation 2, we get:

Equation 3: A (W 36) × W W^2 36W

Setting Up the Inequality

Given that the area must be less than 240 square feet, we can set up the inequality:

Inequality: W^2 36W

To clear the inequality, we can rearrange it to form a standard quadratic equation:

Standard Quadratic Equation: W^2 36W - 240

Solving the Quadratic Equation

First, let's solve the equation W^2 36W - 240 0 to find the roots.

Using the quadratic formula, where a 1, b 36, c -240,

Quadratic Formula: W [-b ± √(b^2 - 4ac)] / 2a

Substitute the values:

Solution: W [-36 ± √(36^2 - 4 × 1 × -240)] / 2 × 1

Calculating the discriminant:

36^2 - 4 × 1 × -240 1296 960 2256

Thus,

W [-36 ± √2256] / 2

W [-36 ± 47.5] / 2

This provides two solutions:

W (-36 47.5) / 2 11.5 / 2 5.75

W (-36 - 47.5) / 2 -83.5 / 2 -41.75

Since the width cannot be negative, we discard the negative root. Hence, the width W must be:

W 5.75 feet

Subsequently, the length L is:

L W 36 5.75 36 41.75 feet

Verification and Range of Values

Given that we want the area to be less than 240 square feet, we can check if the calculated dimensions meet this requirement:

A 5.75 × 41.75 239.3125 240

Finally, to ensure the width is positive and within the constraints provided, we use the quadratic inequality:

W^2 36W - 240 0

By solving the quadratic inequality, we find the range of W that satisfies the condition. Using the roots we calculated, we get:

5.75 W -41.75

Since -41.75 is not a valid solution, the valid range for W is:

0 W 5.75

Conclusion

The mathematical analysis of a conference hall's floor area, given the constraints, indicates that the width of the hall must be between 0 and 5.75 feet, and the length must be between 36 and 41.75 feet. This ensures that the area is less than 240 square feet, providing a useful framework for designing halls within specified dimensions.