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Chapter 2: Problem 64

A 24-by-36-inch sheet of paper is to be used for a poster, with the shorter side at the bottom. The margins at the sides and top are to have the same width, and the bottom margin is to be twice as wide as the other margins. Find the width of the margins if the printed area is to be \(661.5 \mathrm{in}^{2}\).

Short Answer

Expert verified
The width of the margins is 1.5 inches.

Step by step solution

01

Define the Variables

Let \( x \) be the width of the side and top margins in inches. This means the bottom margin is \( 2x \).
02

Express Dimensions of the Printed Area

The dimensions of the printed area are reduced by the margins: the horizontal dimension will be \( 24 - 2x \) (subtracting the margins on both sides), while the vertical dimension will be \( 36 - x - 2x = 36 - 3x \) (top and bottom margins summed up).
03

Set Up the Equation for Printed Area

The printed area is given by the product of its dimensions, which is \( (24 - 2x)(36 - 3x) = 661.5 \).
04

Expand and Simplify the Equation

Expand the expression: \[ (24 - 2x)(36 - 3x) = 24 \times 36 - 24 \times 3x - 2x \times 36 + 6x^2 \]Simplifying, we have:\[ 864 - 72x - 72x + 6x^2 = 661.5 \]This reduces to: \[ 6x^2 - 144x + 864 = 661.5 \].
05

Solve the Quadratic Equation

Subtract 661.5 from both sides to set the equation to zero:\[ 6x^2 - 144x + 202.5 = 0 \]Divide each term by 6 to simplify:\[ x^2 - 24x + 33.75 = 0 \]Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -24 \), \( c = 33.75 \).
06

Calculate with Quadratic Formula

\( b^2 - 4ac = 576 - 135 = 441 \)\( x = \frac{24 \pm \sqrt{441}}{2} = \frac{24 \pm 21}{2} \)The solutions are \( x = \frac{45}{2} = 22.5 \) and \( x = \frac{3}{2} = 1.5 \). Due to the constraints of the problem (positive and feasible width for margins), \( x = 1.5 \) inches.
07

Verify the Solution

Substituting \( x = 1.5 \) back into the dimensions of the printed area:Horizontal: \( 24 - 2(1.5) = 21 \)Vertical: \( 36 - 3(1.5) = 31.5 \)So, the area of the printed part is \( 21 \times 31.5 = 661.5 \). This is consistent with the problem statement, confirming the solution is correct.

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Key Concepts

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

Poster Dimensions
When creating a poster from a sheet of paper, understanding its dimensions is crucial. In this exercise, a 24-by-36-inch sheet is used with the short side at the bottom. Modifying the dimensions involves adjustments for the margins, which is a vital part of our solution. The dimensions of a poster can be affected by:
  • Side margins - which are equal on both vertical sides.
  • Top margin - which is equal to the side margins.
  • Bottom margin - which is twice as wide as the side and top margins.
Figuring out these margins allows us to understand how much space is left for printing. For example, a side margin adds twice its width when considering the entire horizontal width, while top and bottom margins combine to reduce the vertical print space.
Printed Area Calculation
The printed area is the part of the poster where information or images are visible. It's critical to determine the appropriate margins first, as these subtract from the overall dimensions.The printed area calculation involves:
  • Subtracting the total of both side margins from the width of the poster:
    \( 24 - 2x \)
  • Subtracting the total of the top and bottom margins from the height of the poster:
    \( 36 - (x + 2x) = 36 - 3x \)
  • Multiplying these dimensions to get the area:
    \((24 - 2x)(36 - 3x) = 661.5\)
Setting up this equation allows for calculating exactly how dimensions affect the available space for printing.
Algebraic Problem Solving
This problem requires solving a quadratic equation to find the margin width that allows for a specific printed area size. Here, we explore how algebra aids in converting a practical problem into a solvable mathematical one.First, express the equation for the printed area:
  • The area formula set to the given target:
    \((24 - 2x)(36 - 3x) = 661.5\)
  • Expand and simplify:
    \[6x^2 - 144x + 864 = 661.5\]
  • Subtract the constant term from both sides to form a standard quadratic:
    \[6x^2 - 144x + 202.5 = 0\]
Using the quadratic formula, you solve for \(x\):
  • The discriminant, \(b^2 - 4ac\), confirms real roots.
  • The formula is used: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
From here, only the realistic solution for poster margins, \(x = 1.5\), is chosen, given the problem's constraints.

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