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Magazine Chapter 14: Problem 32
Set up systems of equations and solve them graphically. A rectangular security area is to be enclosed by fencing and divided in two equal parts of \(1600 \mathrm{m}^{2}\) each by a fence parallel to the shorter sides. Find the dimensions of the security area if the total amount of fencing is \(280 \mathrm{m}\).
Short Answer
Expert verified
The security area dimensions can be either \(40 \times 80\) \(\mathrm{m}^2\) or \(53.33 \times 60\) \(\mathrm{m}^2\).
Step by step solution
01
Defining Variables
Let's define two variables to represent the dimensions of the rectangular security area. Let \( x \) be the length of the shorter side and \( y \) be the length of the longer side.
02
Setting Up Equations
We know the total area of the security area is \( 2 \times 1600 = 3200 \) \(\mathrm{m}^2\), so we set up the equation \( x \, y = 3200 \). Additionally, including the dividing fence, the total amount of fencing needed is \( 3x + 2y = 280 \).
03
Solving for One Variable
From the area equation \( x \, y = 3200 \), solve for \( y \): \( y = \frac{3200}{x} \).
04
Substituting into Perimeter Equation
Substitute \( y = \frac{3200}{x} \) into the perimeter equation to get \( 3x + 2 \left( \frac{3200}{x} \right) = 280 \). Simplify the equation to \( 3x + \frac{6400}{x} = 280 \).
05
Multiplying to Clear the Fraction
Multiply through by \( x \) to clear the fraction: \( 3x^2 + 6400 = 280x \).
06
Rearranging into a Quadratic Equation
Rearrange the equation to the standard quadratic form: \( 3x^2 - 280x + 6400 = 0 \).
07
Solving the Quadratic Equation
Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = -280 \), \( c = 6400 \), find the values of \( x \). Compute the discriminant \( b^2 - 4ac = 78400 - 76800 = 1600 \).
08
Finding Roots of the Quadratic
The roots are \( x = \frac{280 \pm 40}{6} \). Calculate: \( x = 53.33\, \mathrm{m} \) or \( x = 40\, \mathrm{m} \).
09
Determining Corresponding "y" Values
Substitute each value of \( x \) back into the equation \( y = \frac{3200}{x} \). For \( x = 40 \), \( y = 80 \) and for \( x = 53.33\), \( y = 60 \).
10
Verification of Solutions
Verify that both (\( x, y \)) pairs satisfy the original equations. Both solutions lead to the perimeter equation being satisfied, thus both are valid.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). This equation includes variables, coefficients, and constants, and it's called "quadratic" because it has a degree of 2. In our exercise, the quadratic equation is formed when seeking the dimensions of a rectangular area using the relationship between the perimeter and area.
- The coefficients \(a\), \(b\), and \(c\) determine the shape and position of the parabola when graphed.
- In solving our problem, we ended up with the quadratic equation \( 3x^2 - 280x + 6400 = 0 \). Here, \(a = 3\), \(b = -280\), and \(c = 6400\).
- To find the solutions, we utilize the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Graphical Solution
Finding a graphical solution involves plotting equations on a graph to visually identify their points of intersection. For our problem, this would mean plotting the equations for the rectangle's perimeter and area to see where they intersect.
- The solution to the system of equations occurs where the graphs intersect, representing the values of \(x\) and \(y\) that simultaneously satisfy both equations.
- The system of equations was \( xy = 3200 \) for area and \( 3x + 2y = 280 \) for the perimeter with dividing fence.
- By substituting \( y = \frac{3200}{x} \) into the perimeter equation and simplifying to the quadratic form, the graphical representation would involve solving this new equation graphically as a parabola.
- Graphically, the points \( x = 40 \) and \( x = 53.33 \) provide the intersection points, confirming our algebraic solutions.
Rectangular Dimensions
Rectangular dimensions refer to the measurements of the length and width of a rectangle, which dictate its shape and size. In our exercise, deciding on the dimensions of the rectangular security area was essential to meet the given conditions.
- The shorter side is represented as \(x\) and the longer side as \(y\).
- The product of these dimensions, \(xy\), gives the total area of the rectangle, which was specified as \(3200 \mathrm{m}^2\).
- Using the dimensions and the total fence length equation, we calculated possible values for \(x\) and \(y\).
- The final dimensions derived, \(x = 40, y = 80\) and \(x = 53.33, y = 60\), highlight that there can be more than one valid configuration meeting the problem鈥檚 conditions.
Perimeter Calculation
Perimeter refers to the total length of the boundary around a shape. Calculating it for a rectangle involves adding up the lengths of all sides. In contexts where a fence is involved, like our exercise, knowing the perimeter is crucial for determining how much fencing material is needed.
- The perimeter of a rectangle is given by the formula \( P = 2x + 2y \).
- In this exercise, because the rectangle is divided by a fence parallel to its shorter side, the formula adapted to include this additional length is \( 3x + 2y = 280 \).
- This equation was key in determining the possible dimensions by narrowing down appropriate \(x\) and \(y\) pairs that would comply with the perimeter restrictions.
- Solving the system of equations helped in ensuring that these dimensions not only match the area requirements but also fit within the available fencing.
