Problem 55 - Calculus with Concepts in Calculus

Question

A landowner wishes to use 3 miles of fencing to enclose an isosceles triangular region of as large an area as possible. What should be the lengths of the sides of the triangle?

Step-by-Step Solution

Answer

The triangle should have equal sides of 1 mile and a base of 1 mile.

Step 1: Understanding the problem

We need to find the side lengths of an isosceles triangle that maximize the area, given that the perimeter is 3 miles.

Step 2: Express the perimeter condition

Let the equal sides of the isosceles triangle be of length \( a \) and the base be of length \( b \). The perimeter is given by \( 2a + b = 3 \).

Step 3: Express the area of the triangle

The area \( A \) of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). For an isosceles triangle, the height can be found using Pythagoras' theorem: if the height is \( h \), then \( h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \). Thus, the area can be written as \( A = \frac{1}{2} \times b \times \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \).

Step 4: Substitute \( b \) from the perimeter condition into the area formula

Using \( b = 3 - 2a \), substitute into the area formula: \[ A = \frac{1}{2} \times (3 - 2a) \times \sqrt{a^2 - \left(\frac{3-2a}{2}\right)^2} \].

Step 5: Simplify and solve for maximum area

Simplifying the substituted equation, we have \( h = \sqrt{a^2 - \left(\frac{3-2a}{2}\right)^2} \). Solve \( A = \frac{1}{2} \times (3 - 2a) \times h \) for critical points to find maximum area, typically using calculus or analysis to determine \( a \) and \( b \).

Step 6: Verify the solution

Check the result meets the condition \( 2a + b = 3 \), and use derivative tests or logic to confirm that the area is maximized with the found sides.

Key Concepts

Isosceles Triangle

An isosceles triangle is a simple yet key shape in geometry, characterized by having two sides of equal length. This unique feature gives it a symmetrical appearance, making it not only an integral element in geometry problems but also aesthetically pleasing.

• The equal sides are usually denoted as \( a \) and \( a \).
• The third side, known as the base, is usually different and denoted by \( b \).

Understanding the properties of an isosceles triangle is crucial when solving optimization problems, especially those involving perimeter restrictions. The symmetry of the triangle allows for various techniques, including the Pythagorean Theorem, to be applied in further calculations.

Perimeter

The concept of perimeter is foundational in geometry, as it measures the total length surrounding a shape. For an isosceles triangle with sides \( a \), \( a \), and base \( b \), the perimeter is expressed as \( 2a + b \). In problems where the perimeter is fixed, like in our exercise, each segment of the perimeter must work towards maximizing another property.

• Here, the constraint is that the total perimeter is 3 miles.
• The goal is to adjust the lengths of \( a \) and \( b \) while maintaining this constant perimeter.

Knowing how to incorporate the perimeter condition into your calculations is crucial for successful problem-solving.

Area Maximization

Maximizing the area of a triangle with a fixed perimeter is a classic optimization problem. The area of a triangle can be calculated using the formula \( \frac{1}{2} \times \text{base} \times \text{height} \). To maximize the area, you must strategically choose the triangle's dimensions within the given constraints.

• In the exercise, the base \( b \) and height \( h \) of the triangle are interrelated, influenced by perimeter constraints.
• Finding the height involves understanding the triangle's geometry, often using the Pythagorean Theorem.

The challenge is to find a balance between these factors that yields the maximum possible area.

Pythagorean Theorem

The Pythagorean Theorem is essential in this scenario to calculate the height \( h \) of the isosceles triangle. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

• For our isosceles triangle, consider a line from the apex perpendicular to the base, splitting it into two equal parts.
• This forms two right-angled triangles, where \( a \) is the hypotenuse, and \( \frac{b}{2} \) and \( h \) are the two other sides.
• The theorem helps find \( h \) as: \( h = \sqrt{a^2 - \left(\frac{b}{2}\right)^2} \).

Leveraging this theorem aids in simplifying the area formula, crucial for maximizing the area.

Calculus

Calculus provides the tools necessary to find the maximum area efficiently. It involves techniques that analyze the behavior of functions, such as the area function derived from our parameters.

• Substitute \( b = 3 - 2a \) into the area formula, giving: \( A = \frac{1}{2} \times (3 - 2a) \times \sqrt{a^2 - \left(\frac{3-2a}{2}\right)^2} \).
• By applying calculus, particularly deriving and finding critical points, we can identify when this area is at its maximum.
• Calculus allows us to solve for optimal values of \( a \) and \( b \), making sure these values satisfy the initial problem constraints.

Using derivative tests, you can confirm that the triangle's area indeed reaches a local maximum at these values.