Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Of all rectangles with a given diagonal, find the one with the maximum area.

Step-by-Step Solution

Answer

The rectangle with maximum area is a square with sides \(\frac{d}{\sqrt{2}}\).

Step 1: Understanding the Problem

To find which rectangle with a given diagonal has the maximum area, we will relate the sides of the rectangle to its diagonal using the Pythagorean theorem. Given that the diagonal length is constant.

Step 2: Express Constraints Mathematically

If the sides of the rectangle are denoted by lengths \(x\) and \(y\), and the diagonal is denoted by \(d\), the relationship is given by the Pythagorean theorem: \[ x^2 + y^2 = d^2 \] The area of the rectangle \(A\) is \( A = x \cdot y \).

Step 3: Use a Trigonometric Approach

Consider the angle \(\theta\) between the diagonal \(d\) and side \(x\). Using trigonometric identities, we have:\[ x = d \cos\theta \] \[ y = d \sin\theta \]

Step 4: Formulate the Area in Terms of Trigonometric Functions

Substitute the expressions for \(x\) and \(y\) from the previous step into the area formula:\[ A = x \cdot y = d \cos\theta \cdot d \sin\theta = d^2 \cos\theta \sin\theta \]

Step 5: Maximize the Area Using Trigonometric Identity

Recognize that \( \cos\theta \sin\theta = \frac{1}{2} \sin(2\theta) \). Thus, the area becomes:\[ A = \frac{1}{2}d^2 \sin(2\theta) \] To maximize \(A\), \(\sin(2\theta)\) needs to be maximized, which occurs at \(2\theta = \frac{\pi}{2}\) or \(\theta = \frac{\pi}{4}\).

Step 6: Finding Dimensions and Conclusion

When \(\theta = \frac{\pi}{4}\), both \(\cos\theta\) and \(\sin\theta\) equal \(\frac{\sqrt{2}}{2}\). Thus, \(x = d\frac{\sqrt{2}}{2}\) and \(y = d\frac{\sqrt{2}}{2}\). Therefore, the rectangle is a square with sides \(\frac{d}{\sqrt{2}}\), and it has the maximum area.

Key Concepts

Pythagorean Theorem

When tackling optimization problems involving geometric shapes, like finding the rectangle with the maximum area given a fixed diagonal, the Pythagorean theorem is often the go-to mathematical tool. In a right-angled triangle, the Pythagorean theorem relates the three sides: the square of the hypotenuse (diagonal in our case) is equal to the sum of the squares of the other two sides.For a rectangle with sides lengths \(x\) and \(y\) and diagonal \(d\), this relationship is expressed mathematically as:

• \(x^2 + y^2 = d^2\)

This equation is pivotal as it sets the foundation for determining other necessary elements like area, using these side lengths.In real-world applications, the Pythagorean theorem is used to establish these crucial relationships between sides and angles, especially when calculating lengths that aren't directly measurable. Understanding this theorem is key to understanding many geometric and optimization problems in calculus.

Trigonometry in Calculus

Trigonometry in calculus plays a vital role when dealing with optimization problems, especially those involving areas of cyclical shapes or angles, such as our rectangle example. By incorporating trigonometric identities, we can translate the problem from terms of coordinates to angles, which often simplifies the calculus involved.In our rectangle problem, to express the sides \(x\) and \(y\) in terms of the diagonal \(d\) and an angle \(\theta\), we use the following trigonometric identities:

• \(x = d \cos\theta\)
• \(y = d \sin\theta\)

This transformation allows us to use the interrelations between sine and cosine functions, effectively reducing the steps needed to maximize the area with respect to the angle. Trigonometry offers a bridge between geometric intuition and calculus strategies, providing tools like the product-to-sum formulas to easily solve these kinds of optimization problems.Thus, learning how to apply trigonometric concepts within calculus greatly enhances one's ability to solve complex problems efficiently.

Maximizing Area

The ultimate goal in many calculus problems is optimization, such as maximizing an area given certain constraints. The rectangle problem is a great illustration of applying calculus to achieve this.Once we communicate the area \(A\) of the rectangle as \(A = x \times y\), substituting our trigonometric expressions gives:

• \(A = d^2 \cos\theta \sin\theta\)

By utilizing the trigonometric identity \(\cos\theta \sin\theta = \frac{1}{2} \sin(2\theta)\), the area becomes \(A = \frac{1}{2}d^2 \sin(2\theta)\). This expression is key to identifying the maximum area, which occurs when \(\sin(2\theta)\) reaches its maximum value.The function \(\sin(2\theta)\) is maximized when \(2\theta = \frac{\pi}{2}\), meaning \(\theta = \frac{\pi}{4}\). At this angle, both sides of the rectangle equate to \(d/\sqrt{2}\), making the rectangle a square and achieving the maximum potential area under the constraints. This problem demonstrates how calculus and optimization techniques are often used to find values that extremes like maximum and minimum values.