Problem 112 - College Algebra with Modeling and Visualization

Question

Aframe for a picture is 2 inches wide. The picture inside the frame is 4 inches longer than it is wide. See the figure. If the area of the picture is 320 square inches, find the outside dimensions of the picture frame. (PICTURE NOT COPY)

Step-by-Step Solution

Answer

The outside dimensions are 20 inches by 24 inches.

Step 1: Understand the Problem

We are given a picture frame that is 2 inches wide all around a picture. The picture itself is 4 inches longer than it is wide, and its area is 320 square inches. We need to find the outside dimensions of the frame.

Step 2: Define Variables

Let the width of the picture be \( x \) inches. Then, the length of the picture is \( x + 4 \) inches.

Step 3: Set Up the Equation for the Area

The area of the picture is given by the formula for the area of a rectangle: \( \text{Area} = \text{Width} \times \text{Length} = x(x + 4) \). We know this area is 320 square inches, so we set up the equation: \[ x(x + 4) = 320 \]

Step 4: Solve the Quadratic Equation

Expand the equation: \[ x^2 + 4x = 320 \]Rearrange it to standard form:\[ x^2 + 4x - 320 = 0 \] Now, we use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 4 \), and \( c = -320 \).

Step 5: Apply the Quadratic Formula

Calculate the discriminant:\[ b^2 - 4ac = 4^2 - 4(1)(-320) = 16 + 1280 = 1296 \] Now, solve for \( x \):\[ x = \frac{-4 \pm \sqrt{1296}}{2} \]Since \( \sqrt{1296} = 36 \), the solutions are:\[ x = \frac{-4 + 36}{2} = 16 \] and \[ x = \frac{-4 - 36}{2} = -20 \]Since width cannot be negative, \( x = 16 \).

Step 6: Calculate Picture Dimensions

Now that we have \( x = 16 \), the width of the picture is 16 inches and the length is \( x + 4 = 20 \) inches.

Step 7: Compute the Outside Dimensions of the Frame

The frame adds 2 inches to each side of the picture, so it adds 4 inches total to both the width and the length. Thus, the outside dimensions of the frame are: Width = 16 + 4 = 20 inches, Length = 20 + 4 = 24 inches.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). Quadratic equations are fundamental in algebra because they describe relationships involving squares of unknown quantities.
This means that the highest degree of the variable is 2. In our exercise, the quadratic equation was derived from the area of a rectangle relating to the picture's dimensions.

To solve a quadratic equation, we commonly use the quadratic formula:

• Given as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• The discriminant \( b^2 - 4ac \), under the square root sign, tells us the nature of the roots. If it’s positive, there are two different real roots.
• If it's zero, the roots are real and equal.
• If it's negative, the roots are complex numbers.

In this case, the discriminant (1296) was positive, which resulted in two possible real roots: 16 and -20. Since width cannot be negative, only \( x = 16 \) makes sense in the context of the problem. It's important when solving to interpret the results logically based on the real-world scenario being modeled.

Rectangular Area Calculation

Calculating the area of a rectangle is crucial in many geometric applications. For a rectangle, the area \( A \) is determined using the formula \( A = \, \text{width} \times \text{length} \). This formula allows us to find the size of the flat surface that the rectangle covers.

In the problem, the picture itself is defined by its width \( x \) and its length \( x + 4 \). Thus, the area can be defined as \( x(x + 4) \), which equals 320 square inches. Understanding this formula helps to set up our equation correctly, ensuring that we are capturing the entire surface area in question.

• Width (\( x \)): the shorter dimension of the rectangle.
• Length (\( x + 4 \)): given as 4 inches longer than the width.

It’s helpful to visualize why multiplying these values gives the area. Imagine the length as extending a uniform band of width over the entire length, which fills the rectangle exactly.

Problem Solving Steps

Approaching algebra problems systematically helps students tackle complex questions efficiently. Breaking the problem into digestible steps as done in the exercise provides a clearer path to the solution. Let's recap the steps used:

• Understand the problem: Clearly define what is being asked. In this case, the dimensions of the outside frame of a picture.
• Define the variables: Assign letters to unknown quantities, such as \( x \) for the width of the picture, making expressions easier to handle.
• Set up the equation: Use known formulas or relationships. Identify that the area equation for this problem is a quadratic one.
• Solve the equation: Apply mathematical techniques like the quadratic formula. Ensure calculations consider the context, dismissing non-physical solutions.
• Interpret the results: Translate the solution into a real-world context, calculating all dimensions originally sought after.

Each step is integral because it builds on the previous, ensuring that no aspect is overlooked or incorrectly assumed. This methodical approach is a blueprint for solving a wide array of algebra problems beyond this exercise.