Problem 59 - Calculus with Concepts in Calculus

Question

An isosceles triangle has base 6 and height \(12 .\) Find the maximum possible area of a rectangle that can be placed inside the triangle with one side on the base of the triangle.

Step-by-Step Solution

Answer

The maximum area of the rectangle inside the triangle is 18 square units.

Step 1: Understanding the Problem

We need to determine the maximum area of a rectangle that fits inside an isosceles triangle with a base of 6 units and a height of 12 units. The rectangle's base will be parallel to the triangle's base.

Step 2: Formula for the Area of a Rectangle

The area of a rectangle is given by the formula \( A = l \times w \), where \( l \) is the length of the base and \( w \) is the width (height). We need to find \( l \) and \( w \) that maximize this area while ensuring the rectangle fits inside the triangle.

Step 3: Equation of the Triangle's Sides

To keep the rectangle inside the triangle, the vertices of the rectangle on the triangle’s sloping sides must satisfy the equations of these sides. The triangle's left side can be represented by the line equation \( y = 4x + 12 \) and the right side by \( y = -4x + 12 \) by using the triangle's height and base endpoints.

Step 4: Finding Rectangle Dimensions Limitation

Let's set the rectangle's base from \( -a \) to \( a \). Hence, its base length \( l = 2a \). The height of the rectangle at the base will be the y-value at which this line intersects: \( w = 4(3-a) \) or \( w = 4a + 12 \). Therefore, \( w = 12 - 4a \).

Step 5: Area Function of Rectangle in Terms of "a"

The area of the rectangle, expressed as a function of \( a \), is \( A(a) = 2a(12 - 4a) \). Simplifying, we get \( A(a) = 24a - 8a^2 \).

Step 6: Maximizing the Area Function

To find the maximum area, calculate the derivative of \( A(a) \) and set it equal to zero. The derivative is \( A'(a) = 24 - 16a \). Setting \( A'(a) = 0 \) gives \( 16a = 24 \). Thus, \( a = 1.5 \).

Step 7: Calculating Maximum Area

Substitute \( a = 1.5 \) into the area formula: \[ A(1.5) = 24(1.5) - 8(1.5)^2 = 36 - 18 = 18. \] The maximum area of the rectangle is 18 square units.

Key Concepts

Isosceles Triangle

An isosceles triangle is a special type of triangle with two sides that are equal in length. Understanding its properties can be crucial in solving geometric problems like the one in our exercise.
This triangle type has several important properties:

• The angles opposite the equal sides are also equal.
• The height from the base to the apex splits the triangle into two equal right triangles.

In the context of our problem, the base is 6 units, and the height is 12 units. The isosceles property ensures symmetry, aiding us in confidently calculating dimensions for additional shapes within it, such as the maximum rectangle.
Recognizing the properties of an isosceles triangle makes it easier to visualize the spatial setup of the rectangle whose area we want to maximize.

Rectangle Area Maximization

Maximizing the area of a rectangle within an isosceles triangle involves some strategic thinking. Because the rectangle's base is parallel and coincident to the triangle's base, its only restrictions are set by the triangle's sloping sides.
The goal is to determine dimensions, represented by variables (in this exercise by 'a'), that will maximize the rectangle's area without exceeding these boundaries.

• The rectangle's length is variable and determined by the value of 'a'.
• Its height is also dependent on 'a', calculated through an equation derived from the triangle's sides.

Calculating the area of the rectangle as a function of 'a' helps us use derivatives for optimization, a technique where geometry and calculus blend to find the best solution under specific constraints.

Derivatives in Calculus

Derivatives provide a practical tool in calculus for finding maxima and minima of functions, which is essential in area maximization problems like ours.
By taking the derivative of our area function, we can determine how changes in 'a' affect the rectangle's area.

• The derivative informs us of the rate at which the area changes with respect to the rectangle's base length.
• Setting the derivative to zero helps us find critical points, candidates for maximum area.

In this exercise, the calculation's simplicity \( A'(a) = 24 - 16a \) highlights this idea. Solving \( A'(a) = 0 \) helps locate the maximum area efficiently. Thus, derivatives succinctly handle optimization, rooting geometric problems in the robust realm of calculus.