/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 A manufacturer wants to enlarge ... [FREE SOLUTION] | 91Ó°ÊÓ

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A manufacturer wants to enlarge an existing manufacturing facility such that the total floor area is 1.5 times that of the current facility. The floor area of the current facility is rectangular and measures 250 feet by 160 feet. The manufacturer wants to increase each dimension by the same amount. (a) Write a function that represents the new floor \(\operatorname{area} A\). (b) Find the dimensions of the new floor. (c) Another alternative is to increase the current floor's length by an amount that is twice an increase in the floor's width. The total floor area is 1.5 times that of the current facility. Repeat parts (a) and (b) using these criteria.

Short Answer

Expert verified
In the first scenario, the new floor dimensions will be approximately 284.42 by 194.42 feet. In the second scenario, where the width increases at twice the rate of the length, the new dimensions will be approximately 421.29 by 502.58 feet.

Step by step solution

01

Function For The New Floor Area

Let \( x \) represent the additional feet for both the length and the width the manufacturer wants to add. This means the length of the new manufacturing facility becomes \( 250 + x \) feet, and the width becomes \( 160 + x \) feet. Consequently, the area of the enhanced facility, \( A \), is \((250+x)(160+x)\), or \( A = 40000 + 410x + x^2 \).
02

Find The Dimensions of The New Floor

The problem says the new area will be 1.5 times the old one, so \( A = 1.5 \times 250 \times 160 = 60000 \). Substitute this new value into the equation established in Step 1. The equation becomes \( x^2 + 410x + 40000 - 60000 = 0 \), or simplifying it to \( x^2 + 410x - 20000 = 0 \). Solving this quadratic equation using the quadratic formula \( x = [-b ± sqrt(b^2 - 4ac)] / (2a) \), we have \( x approx 34.42 or -582.42 \). A negative measurement doesn't make sense in this context, so the extra amount to be added to both dimensions is \( 34.42 feet \), which means the new dimensions will now be approximately \( 284.42 feet by 194.42 feet \).
03

Function For The New Floor Area With Different Increase

In this scenario, the width increases twice as much as the length. So if \( y \) is the amount the length increases, the width increases by \( 2y \), giving us dimensions of \( 250 + y \) and \( 160 + 2y \). Consequently, the area of the enhanced facility, \( A \), is \((250+y)(160+2y) = 40000 + 570y + 2y^2 \).
04

Find The Dimensions of The New Floor With Different Increase

Again, the new area is 1.5 times the old, so \( A = 60000 \). Substitute into the equation from Step 3 results in \( 2y^2 + 570y + 40000 - 60000 = 0 \), simplified to \( 2y^2 + 570y - 20000 = 0 \). Solving this quadratic equation gives us two potential solutions: \( y approx -58.29 or 171.29 \). Discarding the negative value means the length increases by \( 171.29 feet \), and the width increases by \( 2 \times 171.29 = 342.58 feet \). Therefore, the new dimensions will now be approximately \( 421.29 feet by 502.58 feet \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dimensions of Rectangles
Understanding the dimensions of rectangles is crucial when tackling problems involving floor areas and manufacturing expansions. A rectangle is a four-sided figure with opposite sides that are equal and parallel.
When discussing dimensions, we focus on two main aspects: the length and the width. In many real-world applications, such as expanding a facility, knowing how to adjust these dimensions can help achieve desired outcomes, like increasing floor space.
  • The length is typically considered the longer side, while the width is the shorter side. However, this distinction can be interchangeable.
  • In mathematical problems, we often add or subtract a variable from both the length and width to obtain new dimensions.
  • The resulting area of a rectangle is simply the product of its length and width: \( \text{Area} = \text{Length} \times \text{Width} \).
For example, if a facility's original dimensions are 250 by 160 feet and we wish to increase each dimension by \( x \) feet, the new dimensions become \( 250 + x \) and \( 160 + x \), respectively. Knowing how to manipulate these dimensions helps solve complex problems like facility expansion.
Floor Area
The floor area of a building or a space is critical information for both functionality and measurement purposes. It represents the total surface covered by floors within a boundary of walls.
Measuring the floor area accurately is essential for various reasons:
  • It determines the available usable space and impacts the layout.
  • Affects the valuation in real estate and construction projects.
  • Guides specific requirements for construction, like materials needed.
For rectangles, you calculate the floor area by multiplying the length by the width. In this problem, the original area was \( 250 \times 160 = 40,000 \) square feet. However, the new floor area needs to be 1.5 times the original, equating to 60,000 square feet. By knowing the current and desired area, equations can be formulated to determine necessary changes in dimensions for achieving desired expansions.
Manufacturing Facility Expansion
Expanding a manufacturing facility entails planning the adjustment of existing spaces to meet growing demands. This expansion often involves increasing the floor area while maintaining or optimizing functional layouts.
For this problem, two approaches were analyzed:
  • Equally increasing both the length and width.
  • Increasing the width twice as much as the length.
The goal was to achieve a new floor area that is 1.5 times the current, necessitating precise calculations of possible dimension increases.
For the first scenario, adding 34.42 feet to both dimensions resulted in new dimensions of roughly \( 284.42 \) (length) by \( 194.42 \) (width) feet.
The second scenario gave different increases: the length increased by \( 171.29 \) feet and width by \( 342.58 \) feet, leading to larger overall dimensions. These solutions utilized the quadratic formula, demonstrating how math can help solve real-life problems on facility adjustments.

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