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Magazine Chapter 10: Problem 60
Sign Dimensions. Alison's Advertising is building a rectangular sign with an area of 2 yd \(^{2}\) and a perimeter of 6 yd. Find the dimensions of the sign.
Short Answer
Expert verified
The dimensions are 1 yd by 2 yd or 2 yd by 1 yd.
Step by step solution
01
Define Variables
Let the length of the sign be denoted by \( l \) and the width be denoted by \( w \). We have two equations based on the given information: the area and the perimeter.
02
Write Equations
The area of the rectangle is given by: \( lw = 2 \). The perimeter of the rectangle is given by: \( 2l + 2w = 6 \).
03
Solve for One Variable
First, solve the perimeter equation for one variable. From \( 2l + 2w = 6 \), we get \( l + w = 3 \). Hence, \( l = 3 - w \).
04
Substitute in Area Equation
Substitute \( l = 3 - w \) into the area equation: \( (3 - w)w = 2 \).
05
Form and Solve Quadratic Equation
Expand and form the quadratic equation: \( 3w - w^2 = 2 \). This can be rewritten as \( w^2 - 3w + 2 = 0 \).
06
Factor the Quadratic Equation
Factor the quadratic equation: \( (w - 1)(w - 2) = 0 \). Thus, the solutions are \( w = 1 \) and \( w = 2 \).
07
Find Corresponding Length
For \( w = 1 \), substitute back into \( l = 3 - w \): \( l = 3 - 1 = 2 \). For \( w = 2 \), substitute back into \( l = 3 - w \): \( l = 3 - 2 = 1 \).
08
Conclude Dimensions
The dimensions of the sign are either \(1 \text{ yd} \times 2 \text{ yd}\) or \( 2 \text{ yd} \times 1 \text{ yd} \). Both sets of dimensions satisfy the given conditions for area and perimeter.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rectangular area calculation
Understanding how to calculate the area of a rectangle is key to solving many problems in geometry. The area represents the amount of space within the boundary of the rectangle, and it is given by the formula: \ \ \( A = l \times w \).
In this formula, \( l \) is the length and \( w \) is the width of the rectangle. For Alison's advertising sign, the area is provided as 2 square yards (\( \text{yd}^2 \)). Hence, the equation becomes: \ \ \( l \times w = 2 \).
To interpret this:
In this formula, \( l \) is the length and \( w \) is the width of the rectangle. For Alison's advertising sign, the area is provided as 2 square yards (\( \text{yd}^2 \)). Hence, the equation becomes: \ \ \( l \times w = 2 \).
To interpret this:
- If the length is known, the width can be found by dividing the area by the length.
- If the width is known, the length can be found by dividing the area by the width.
Perimeter equations
The perimeter of a rectangle is the total length around its border. This is important for determining the total boundary length. The formula for the perimeter is: \ \ \( P = 2l + 2w \) or \( P = 2(l + w) \).
In Alison's problem, the perimeter is specified as 6 yards. So, the equation we use is: \ \ \( 2l + 2w = 6 \).
We can simplify this equation to find the relationship between length and width: \ \ \( l + w = 3 \).
By isolating one variable, like the length \( l \), we can express it in terms of the width \( w \): \ \ \( l = 3 - w \).
This form is useful for substituting into other equations, like our area equation, to find consistent dimensions for the rectangle.
In Alison's problem, the perimeter is specified as 6 yards. So, the equation we use is: \ \ \( 2l + 2w = 6 \).
We can simplify this equation to find the relationship between length and width: \ \ \( l + w = 3 \).
By isolating one variable, like the length \( l \), we can express it in terms of the width \( w \): \ \ \( l = 3 - w \).
This form is useful for substituting into other equations, like our area equation, to find consistent dimensions for the rectangle.
Solving quadratic equations
Sometimes, finding the dimensions of a rectangle involves solving a quadratic equation. In our problem, we reach a quadratic equation by substituting into the area formula: \ \ \( (3 - w)w = 2 \).
This expands to: \ \ \( 3w - w^2 = 2 \) and then we rearrange it to: \ \ \( w^2 - 3w + 2 = 0 \).
A quadratic equation in the form of \( ax^2 + bx + c = 0 \) can often be factored to find its roots. For our equation, factoring yields: \ \ \( (w - 1)(w - 2) = 0 \).
Solving this, we find the width values: \ \ \( w = 1 \) and \( w = 2 \).
Substituting these back into the equation \( l = 3 - w \) gives us the corresponding lengths. It's important to remember:
This expands to: \ \ \( 3w - w^2 = 2 \) and then we rearrange it to: \ \ \( w^2 - 3w + 2 = 0 \).
A quadratic equation in the form of \( ax^2 + bx + c = 0 \) can often be factored to find its roots. For our equation, factoring yields: \ \ \( (w - 1)(w - 2) = 0 \).
Solving this, we find the width values: \ \ \( w = 1 \) and \( w = 2 \).
Substituting these back into the equation \( l = 3 - w \) gives us the corresponding lengths. It's important to remember:
- Solving stress proper equation formation,
- Factoring,
- Checking roots,
- Plugging back to check consistency.
