Problem 2 - Thomas Calculus

Question

Show that among all rectangles with an 8 -m perimeter, the one with largest area is a square.

Step-by-Step Solution

Answer

The rectangle with largest area and an 8m perimeter is a square with 2m sides.

Step 1: Understand the Problem

We need to find a rectangle with a perimeter of 8 meters that has the maximum possible area. Then, we need to demonstrate that this rectangle is a square.

Step 2: Express the Perimeter Equation

For a rectangle with length \( l \) and width \( w \), the perimeter \( P \) is given by the formula: \[ P = 2(l + w) \]. Based on the problem, set the perimeter to 8 meters: \[ 2(l + w) = 8 \], which simplifies to \[ l + w = 4 \].

Step 3: Express Area in Terms of One Variable

The area \( A \) of the rectangle is given by \( A = l \times w \). Using the equation \( l + w = 4 \), we can express \( w \) as \( w = 4 - l \). Substitute \( w \) in the area formula: \[ A = l \times (4 - l) = 4l - l^2 \].

Step 4: Find the Maximum Area Using Derivatives

The area function \( A = 4l - l^2 \) is a quadratic function. To find the maximum area, first find the derivative with respect to \( l \): \( A'(l) = 4 - 2l \). Set the derivative equal to zero and solve for \( l \): \[ 4 - 2l = 0 \] \[ 2l = 4 \] \[ l = 2 \].

Step 5: Find Corresponding Width and Verify Square

When \( l = 2 \), the width is \( w = 4 - l = 4 - 2 = 2 \). Since both dimensions are equal, the rectangle is a square with side length 2 meters.

Step 6: Conclusion on the Maximum Area

The maximum area is when the rectangle is a square with side length 2, and the area is \( 4 \text{ square meters} \), since \( A = l \times w = 2 \times 2 = 4 \). This confirms that the rectangle with the largest area given a fixed perimeter is indeed a square.

Key Concepts

Quadratic Functions

Quadratic functions play a vital role in solving optimization problems in calculus, particularly when trying to find the maximum or minimum values of a function. A quadratic function is generally expressed in the form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. In our exercise, the area of the rectangle is expressed as a quadratic function \( A(l) = 4l - l^2 \). Here, the area function is a downward-facing parabola because the coefficient of \( l^2 \) is negative. This means that there is a maximum point on the graph.
To find this maximum, we examine the vertex of the parabola, which, in mathematical terms, represents the maximum point of the function. The vertex formula \( x = -\frac{b}{2a} \) helps determine the \( l \) value at which the function reaches its peak. By setting the derivative of the area function to zero and solving for \( l \), we monitored where this maximum occurs, ensuring that the largest area for a fixed perimeter results in a square with sides equal.

Derivative Applications

In calculus, derivatives provide critical insights into the behavior of functions, such as identifying local maxima and minima. The derivative of a function shows how quickly it is changing at any point. For our rectangle's area function \( A(l) = 4l - l^2 \), finding the derivative, \( A'(l) = 4 - 2l \), was essential in identifying the value of \( l \) that maximizes the area.
When solving optimization problems, setting the derivative equal to zero determines the critical points of the function. These are the points where the function could potentially have maxima, minima, or points of inflection. After clearing out the algebra, we found that \( l = 2 \) is critical. Plugging back into the equation also confirmed it leads to both equal side lengths, indicating a square. The process of differentiation is thus fundamental in optimizing geometric figures, allowing us to ascertain that a square provides the optimal area given a fixed perimeter.

Geometric Properties of Rectangles

The geometric properties of rectangles help us understand their behavior in optimization problems. A rectangle has two pairs of parallel sides, with opposite sides being equal in length. The perimeter, the total distance around the rectangle, is calculated by adding up all side lengths. In our problem with a perimeter of 8 meters, we represented this constraint with \( l + w = 4 \), allowing us to express one variable in terms of the other.
This relationship between length and width simplified our exploration of the rectangle's potential maximum area. When these dimensions were calculated to be equal, the rectangle naturally formed a square. A square, a special type of rectangle, is both a geometric and algebraic solution here because it maximizes area under a given perimeter. This is a classic demonstration in calculus and geometry of how equal distribution of dimensions often yields optimal results.