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Magazine Chapter 13: Problem 46
Solve. A rectangular holding pen for cattle is to be designed so that its perimeter is 92 feet and its area is 525 feet. Find the dimensions of the holding pen.
Short Answer
Expert verified
The dimensions are 25 feet by 21 feet.
Step by step solution
01
Define the Variables
Let the length of the pen be denoted as \( l \) and the width be denoted as \( w \). The perimeter \( P \) of a rectangle is given by the formula \( P = 2l + 2w \).
02
Set Up the Perimeter Equation
The problem states that the perimeter is 92 feet. Substitute into the perimeter formula: \( 2l + 2w = 92 \). Simplify this equation to get \( l + w = 46 \).
03
Set Up the Area Equation
The area \( A \) of a rectangle is given by \( A = lw \). We know the area is 525 square feet, so we set up the equation \( lw = 525 \).
04
Express Width in Terms of Length
From the equation \( l + w = 46 \), express \( w \) in terms of \( l \): \( w = 46 - l \).
05
Substitute and Solve
Substitute \( w = 46 - l \) into the area equation \( lw = 525 \), resulting in \( l(46 - l) = 525 \). Simplify to get the quadratic equation: \( l^2 - 46l + 525 = 0 \).
06
Solve the Quadratic Equation
Use the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -46, c = 525 \). Calculate the discriminant: \( (-46)^2 - 4 \cdot 1 \cdot 525 = 2116 - 2100 = 16 \). Thus, \( l = \frac{46 \pm 4}{2} \). Calculate the roots: \( l = 25 \) or \( l = 21 \).
07
Find the Corresponding Widths
For \( l = 25 \), using \( w = 46 - l \), the width \( w = 46 - 25 = 21 \). For \( l = 21 \), the width \( w = 46 - 21 = 25 \).
08
State the Dimensions
The dimensions of the rectangular holding pen are 25 feet by 21 feet or 21 feet by 25 feet. Either set of dimensions satisfies the problem conditions for perimeter and area.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations play an important role when dealing with shapes like rectangles, especially when both area and perimeter are involved. A quadratic equation is any equation that can be rewritten in the standard form:
To solve this quadratic equation, we use the quadratic formula:
- \( ax^2 + bx + c = 0 \)
To solve this quadratic equation, we use the quadratic formula:
- \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Perimeter Calculation
The perimeter of a rectangle is essential when solving area problems. It is the total length around a rectangle and is calculated using the formula:
- \( P = 2l + 2w \)
- \( 2l + 2w = 92 \)
- \( l + w = 46 \)
Area Calculation
Calculating the area of a rectangle involves understanding the product of its length and width. The area formula is:
- \( A = lw \)
- \( lw = 525 \)
Variables Substitution
Substitution of variables is a core technique in solving equations, particularly when dealing with more than one variable. In this problem, after finding that \( l + w = 46 \), we can solve for one variable:
This substitution leads to a single equation in terms of one variable, simplifying the process. It transforms the situation into one solvable with a quadratic equation. Mastering variable substitution is often the key to unlocking complex problems efficiently.
- \( w = 46 - l \)
- \( l(46 - l) = 525 \)
This substitution leads to a single equation in terms of one variable, simplifying the process. It transforms the situation into one solvable with a quadratic equation. Mastering variable substitution is often the key to unlocking complex problems efficiently.
