Problem 82 - Calculus Single Variable

Question

Let \(x\) denote the perimeter of a rectangle, and let \(y\) denote the area. Calculate \(x\) and \(y\) for the five rectangles with (width, height) \(=(7,42), \quad(8,23), \quad(11,13), \quad(18,9), \quad\) and (26,8) . Plot the five ordered pairs \((x, y)\) that you have calculated. The plot is nearly linear. Find the least squares line through the origin. For a square of side length \(2,\) the perimeter \(x\) is 8 . What is the regression line approximation of the area \(y\) ? What is the relative error? What went wrong?

Step-by-Step Solution

Answer

The regression approximation for the square's area is 26.96. The relative error is 524%. The data did not account for squares with equal side lengths.

Step 1: Calculate Perimeter and Area

For each rectangle, calculate the perimeter \(x = 2( ext{width} + ext{height})\) and area \(y = ext{width} imes ext{height}\). For the given dimensions:1. \((7, 42)\): - Perimeter: \(x = 2(7 + 42) = 98\) - Area: \(y = 7 imes 42 = 294\)2. \((8, 23)\): - Perimeter: \(x = 2(8 + 23) = 62\) - Area: \(y = 8 imes 23 = 184\)3. \((11, 13)\): - Perimeter: \(x = 2(11 + 13) = 48\) - Area: \(y = 11 imes 13 = 143\)4. \((18, 9)\): - Perimeter: \(x = 2(18 + 9) = 54\) - Area: \(y = 18 imes 9 = 162\)5. \((26, 8)\): - Perimeter: \(x = 2(26 + 8) = 68\) - Area: \(y = 26 imes 8 = 208\)Thus, the pairs \((x, y)\) are \((98, 294), (62, 184), (48, 143), (54, 162), (68, 208)\).

Step 2: Plot Ordered Pairs

Plot the five ordered pairs \((x, y)\): \((98, 294), (62, 184), (48, 143), (54, 162), (68, 208)\). You will observe that the plot points form a pattern that is almost linear, suggesting a linear relationship between the perimeter and area of the rectangles.

Step 3: Find Least Squares Line Through Origin

To find the least squares line \(y = mx\) through the origin, calculate the slope\[ m = \frac{\sum x_{i}y_{i}}{\sum x_{i}^2} \]Calculate the sums:1. \(\sum x_i y_i = 98 \times 294 + 62 \times 184 + 48 \times 143 + 54 \times 162 + 68 \times 208 = 82616\)2. \(\sum x_i^2 = 98^2 + 62^2 + 48^2 + 54^2 + 68^2 = 24488\)Now the slope:\[ m = \frac{82616}{24488} \approx 3.37 \]Thus, the line is \(y = 3.37x\).

Step 4: Regression Line Approximation for a Square

For a square with side 2, the perimeter \(x = 8\). Using the regression line \(y = 3.37x\), estimate the area:\[ y = 3.37 \times 8 = 26.96 \]

Step 5: Calculate Actual Area and Relative Error

The actual area of a square with side 2 is \(y = 2 \times 2 = 4\). Calculate the relative error:\[ \text{Relative Error} = \frac{|26.96 - 4|}{4} \approx 5.24 \text{ or } 524\% \]This large error indicates a significant deviation between the predicted and actual values.

Step 6: Identifying the Reason for Error

The error occurred because the data used to derive the regression line primarily involved rectangles with varying width-to-height ratios, whereas the square used in the test case has equal sides. This results in a substantial mismatch in the relationship between perimeter and area in the test square compared to the original data set.

Key Concepts

Perimeter and Area Calculation

Understanding how to calculate the perimeter and area of shapes, like rectangles, is a fundamental concept in geometry. For rectangles:

• The **perimeter**, denoted as \( x \), is the total distance around the rectangle. It is computed using the formula: \( x = 2(\text{width} + \text{height}) \).
• The **area**, represented by \( y \), is the space enclosed within the rectangle and is calculated as \( y = \text{width} \times \text{height} \).

Consider a rectangle with a width of 7 and a height of 42. Its perimeter would be \( 98 \), calculated as \( x = 2(7 + 42) \), while the area would be \( 294 \), using \( y = 7 \times 42 \). Performing these calculations for various rectangles gives us pairs of perimeter and area, such as \((98, 294)\). Understanding these computations allows us to explore more complex relationships, like those found in linear regression.

Least Squares Method

The Least Squares Method is a statistical approach used to find the best-fitting line through a set of points. It's especially useful in identifying trends and relationships in data. In this context, the objective is to find a line that correlates the perimeter \( x \) with the area \( y \) of rectangles, optimizing the fit for the data:

• Start by calculating the slope \( m \) of the line using the formula: \[ m = \frac{\sum x_{i} y_{i}}{\sum x_{i}^2} \] where \( x_i \) and \( y_i \) are the perimeter and area values for each rectangle.
• For the given rectangle data, \( \sum x_i y_i = 82616 \) and \( \sum x_i^2 = 24488 \), yielding a slope \( m \approx 3.37 \).

So, the equation of our line becomes \( y = 3.37x \), showing the linear relationship derived through least squares. This line is an approximation of how area changes with perimeter based on the sample rectangles.

Relative Error

Relative error is a measure of how inaccurate a prediction is relative to the true value. It compares the difference between the predicted and actual values to the actual value itself:
\[ \text{Relative Error} = \frac{|\text{Predicted value} - \text{Actual value}|}{\text{Actual value}} \]
In our regression example, for a square with a side length of 2, the predicted area using the regression line is \( 26.96 \) while the actual area is \( 4 \). Calculating the relative error gives:
\[ \frac{|26.96 - 4|}{4} \approx 5.24 \text{ or } 524\% \]
Such a high relative error percentage indicates a poor prediction. This suggests that the regression model does not perform well for squares, possibly due to the initial data set consisting mainly of non-square rectangles.

Linear Relationship in Geometry

Linear relationships are essential in understanding and predicting how one quantity varies with another. In geometry, such relationships can often be observed through measurements like perimeter and area. For rectangles, you might expect that as the perimeter increases, the area should also increase.

• The ordered pairs \((x, y)\) derived from perimeter and area calculations can be plotted to visualize this relationship.
• If the line connecting these points is straight, it indicates a linear relationship, suggesting that changes in perimeter directly influence changes in area to an extent.

However, complexities arise when dealing with different shapes, like squares with equal sides, compared to varied-width rectangles, as seen with the regression example. This reminds us that the linear relationship found may not hold universally across all geometric figures. Understanding these nuances helps in accurately applying mathematical models to predict outcomes.