Problem 15 - Calculus Early Transcendentals

Question

A rectangle \(R\) in the plane has corners at \((\pm 8, \pm 12),\) and a 100 by 100 square \(S\) is positioned in the plane so that its csides are parallel to the coordinate axes and the lower left corner of \(S\) is on the line \(y=-3 x .\) What is the largest pos- sible area of a region in the plane that is contained in both \(R\) and \(S ?\)

Step-by-Step Solution

Answer

The largest possible overlapping area is 384 square units.

Step 1: Understand the Rectangle

The rectangle \( R \) is defined by having corners at \((\pm 8, \pm 12)\). This means the rectangle is 16 units wide and 24 units tall, vertically centered at \( y = 0 \). The equation of the rectangle's sides are \( x = \pm 8 \) and \( y = \pm 12 \).

Step 2: Understand the Square Placement

The square \( S \) is a 100 by 100 square with sides parallel to the coordinate axes. Its lower left corner follows the equation \( y = -3x \). We can express this corner as \((x, -3x)\).

Step 3: Determine Intersection Conditions

To maximize the intersection area of \( R \) and \( S \), we need \( S \) to fit within \( R \) while being positioned correctly along \( y = -3x \). Therefore, S's range \([x, x+100] \) must satisfy \(-8 \leq x \leq 8\) and \([-3x,-3x+100] \) must satisfy \(-12 \leq -3x+100 \leq 12\).

Step 4: Solve for Maximum Values of \(x\) and \(y\)

Calculate the "\( y= -3x+100 \geq -12 \)" inequality to set bound for \( x \): \[-3x + 100 \geq -12 \quad \Rightarrow \quad -3x \geq -112 \quad \Rightarrow \quad x \leq \frac{112}{3}\approx 37.33.\] For \(-3x + 100 \leq 12\) inequality:\[-3x + 100 \leq 12 \quad \Rightarrow \quad -3x \leq -88 \quad \Rightarrow \quad x \geq \frac{88}{3} \approx 29.33.\]So, for maximum intersection, \(x\) is constrained between 29.33 and 37.33.

Step 5: Optimize for Intersection Area

For the maximum area, \(x\) should be as small as possible (Closest to \(29.33\)) so that more of square fits within rectangle's left boundary of \(x=8\). Therefore, at \(x =\frac{88}{3} \), bottom line passes through \([-24,0]\) then \([0,24]\). Thus, 16-unit wide rectangle fits, hence the maximum area = 16*24=384 square units.

Key Concepts

Rectangular Coordinates

In geometry, understanding rectangular coordinates is essential for solving many problems. These coordinates are defined by a pair of numbers that specify a precise location on a plane using a horizontal (x-axis) and vertical axis (y-axis).
In the exercise, the rectangle \( R \) is situated in this 2-dimensional coordinate plane. Its corners are located at \((\pm 8, \pm 12)\).

• The notation \((\pm 8, \pm 12)\) means that the rectangle's corners have x-coordinates of +8 and -8, and y-coordinates of +12 and -12.
• This creates a rectangle that stretches horizontally between the lines \(x = -8\) and \(x = 8\).
• Vertically, the rectangle is bounded by \(y = -12\) and \(y = 12\).

With these coordinates, the entire shape and dimensions of the rectangle in the Cartesian plane can be understood. This same principle is used to position other geometric shapes, like the square \( S \), based on their defining coordinates and conditions.

Intersection Area

The intersection area in geometry refers to the overlapping region shared by two shapes. When two shapes intersect, calculating this area helps in determining how much space they collectively occupy in the same region.
In this problem, we need to find the area where the rectangle \( R \) and the square \( S \) overlap.To find the intersection, we follow these key points:

• Square \( S \) is a 100 by 100 unit square, and it is placed such that its lower-left corner adheres to the constraint \(y = -3x\).
• The positioning must be checked to ensure the square's boundaries lie within the rectangle \( R \) without exceeding its limits.
• To locate the area of intersection precisely, you solve inequalities based on \( R \)'s size and \( S \)'s constraints, specifically the boundaries \(-8 \leq x \leq 8\) and \(-12 \leq y \leq 12\).

Conceptually, analyzing how the boundaries of \( S \) relate to \( R \) helps identify where and how these two shapes meet and overlap.

Geometric Optimization

Geometric optimization involves finding the best or most efficient way to achieve a geometric goal, such as maximizing or minimizing an area.
In this exercise, the task is to maximize the area where the rectangle and the square intersect. This requires not just understanding their initial sizes and positions, but adjusting them within constraints.Key strategies used include:

• Positioning strategies like shifting \( S \) while adhering to a line constraint \(y = -3x\) to best fit \( R \) for maximum overlap.
• Setting boundaries through mathematical inequalities derived from the positions and sizes of each shape, like ensuring part of \( S \) remains within designated x and y bounds of \( R \).
• Using calculated values for \( x \) to find optimal positions, like selecting the smallest \( x \) value for maximizing intersection space without breaking \( R \)'s boundaries.

By dealing carefully with these constraints and calculations, the largest possible intersection area, which is 384 square units in this case, can be achieved efficiently.