Problem 75 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that, among all rectangles with given perimeter \(p\), the square has the largest area. Hint: If \(a\) and \(b\) denote the lengths of adjacent sides of a rectangle of perimeter \(p\), then the area is \(a b\), and for the square the area is \(a^{2}=[(a+b) / 2]^{2}\). Now see Problem 74 .

Step-by-Step Solution

Answer

The square has the largest area among rectangles with a fixed perimeter.

Step 1: Understand the Problem

We need to show that among all rectangles with a fixed perimeter \( p \), a square has the maximum area. We'll start by setting up an expression for the area of the rectangle, then derive a relationship between the sides \( a \) and \( b \), and prove that the area is maximized when the rectangle is a square.

Step 2: Express the Perimeter

The perimeter \( p \) of the rectangle is given by the equation \( 2(a + b) = p \). Solving for \( b \), we have \( b = \frac{p}{2} - a \).

Step 3: Write the Area Expression

The area \( A \) of the rectangle is given by \( A = a \times b = a \left( \frac{p}{2} - a \right) \). Simplifying, we get \( A = \frac{p}{2}a - a^2 \).

Step 4: Use Calculus to Find Maximum Area

To find the value of \( a \) that maximizes the area, take the derivative of \( A \) with respect to \( a \): \( \frac{dA}{da} = \frac{p}{2} - 2a \). Set the derivative equal to zero to find the critical points: \( \frac{p}{2} - 2a = 0 \), so \( a = \frac{p}{4} \).

Step 5: Ensure Maximum Area with Second Derivative Test

Check the second derivative \( \frac{d^2A}{da^2} = -2 \). Since this is negative, the function has a maximum at \( a = \frac{p}{4} \). Thus, \( b = \frac{p}{2} - a = \frac{p}{4} \), meaning the rectangle is a square.

Step 6: Conclusion

Thus, the area is maximized when both sides of the rectangle are equal, which means it is a square. Therefore, among all rectangles with a given perimeter, the square has the largest area.

Key Concepts

Rectangular Area

In optimization problems involving geometry, calculating the area of shapes is crucial. For a rectangle, the area is determined by its length and width. If we let \( a \) and \( b \) represent the lengths of the rectangle's sides, the area \( A \) is simply calculated as \( A = a \cdot b \). This means the area varies depending on the dimensions of the rectangle.
Understanding how to manipulate these dimensions under specific conditions, like with a fixed perimeter, is key to solving optimization problems in calculus. The goal is to maximize this area within given constraints.

Perimeter Constraint

When dealing with rectangles that must satisfy a certain perimeter, this constraint significantly affects the possible configurations. The perimeter \( p \) of a rectangle, defined by \( 2(a + b) = p \), indicates that the sum of twice the lengths of the adjacent sides is constant.
By solving this equation for one variable, say \( b \), in terms of the other \( a \), we obtain \( b = \frac{p}{2} - a \). This relationship shows how adjusting one side impacts the other due to the fixed perimeter. It is a critical factor when using calculus to find maximum or minimum values of geometric properties like area.

Critical Points

In calculus, critical points occur where the derivative of a function is zero or undefined. These points are potential locations for local maxima or minima. In our rectangle problem, we differentiate the area function \( A = \frac{p}{2}a - a^2 \) to find its critical points. The derivative \( \frac{dA}{da} \) simplifies to \( \frac{p}{2} - 2a \).
Setting this derivative to zero gives the critical point \( a = \frac{p}{4} \). Finding critical points is an essential step to determine where the maximum area may occur under the given constraints.

Second Derivative Test

Once a critical point is identified, the second derivative test helps ascertain whether this point is a maximum or a minimum. For a function \( f(x) \), if the second derivative \( f''(x) \) is positive at a critical point, the function has a local minimum there. Conversely, if \( f''(x) \) is negative, the point is a local maximum.
For our area function \( A \), the second derivative \( \frac{d^2A}{da^2} = -2 \) is negative, confirming a maximum at \( a = \frac{p}{4} \). This result, combined with the perimeter constraint, shows that the rectangle with maximum area is a square, as \( a = b \). Testing the second derivative solidifies our conclusion in optimization tasks.