Problem 26 - Precalculus Essentials

Question

You have 1200 feet of fencing to enclose a rectangular region and subdivide it into three smaller rectangular regions by placing two fences parallel to one of the sides. Express the area of the enclosed rectangular region, \(A,\) as a function of one of its dimensions, \(x\). (GRAPH CANT COPY)

Step-by-Step Solution

Answer

The area \(A\) of the enclosed rectangular region expressed as a function of one of its dimensions (\(x\)) is \[A(x) = x * \frac{1200 - 2x}{3}\]

Step 1: Denote the sides

Denote the length of the rectangular region as \(x\) (in feet). Hence, the width as \(y\) (in feet). The total fencing 1200 feet includes two lengths and three widths, giving the formula: \(2x + 3y = 1200\).

Step 2: Express y in terms of x

From the equation in step 1, express \(y\) in terms of \(x\) by isolating \(y\): \[y = \frac{1200 - 2x}{3}\]

Step 3: Formulate the function for area

The total area \(A\) of a rectangle is given by length multiplied by the width (\(A=xy\)). Substitute \(y\) from step 2 into the function for \(A\): \[ A(x) = x * \frac{1200 - 2x}{3}\] This equation expresses the area \(A\) of the enclosed rectangular region as a function of one of its dimensions, \(x\).

Key Concepts

Rectangular Region

A rectangular region is a flat shape where opposite sides are parallel and equal in length. It resembles a rectangle, which is one of the most fundamental geometric shapes. Rectangles are essential in both geometry and everyday life. We find them in architecture, design, and even in solving math problems.
In this exercise, the rectangular region is enclosed by a fence and divided into three smaller rectangles. This division is made by placing two additional fences parallel to one of the sides. The goal is to optimize the allocation of the fencing material to cover maximum area, considering the constraints imposed by fence lengths.

Area Function

The area of a rectangle is calculated by multiplying its length by its width. However, in optimization problems, we often express the area as a function of a single variable. This simplifies the calculations and helps in determining the extreme values (maximum or minimum) of the area.
In the context of this problem, we denote one of the dimensions of the rectangular region as \( x \), and express the width \( y \) in terms of \( x \). Consequently, the area function \( A(x) \) is formulated. This function assists in analyzing how the area changes with different values of \( x \). Here, the area function is given as:

• \( A(x) = x \cdot \frac{1200 - 2x}{3} \)

This equation is crucial for determining the appropriate configuration of the rectangular region to maximize the area covered by the available fencing.

Fencing Constraints

Constraints are conditions that must be satisfied in an optimization problem. In this case, the primary constraint involves the available fencing material. The problem specifies 1200 feet of fencing, which must be used to form one large rectangle and two additional dividers. Therefore, we need to consider both the length \( x \) and the width \( y \) of the rectangle as the fencing needs to cover two lengths and three widths.
The fencing constraint can be expressed as the equation:

• \( 2x + 3y = 1200 \)

This equation ensures that the total fence length does not exceed the given limit. By working with this constraint, it's possible to find a balance between the dimensions \( x \) and \( y \), allowing optimization of the rectangular area's size.

Algebraic Expressions

Optimizing a rectangular region's area often involves manipulating algebraic expressions. In this exercise, algebraic expressions are used to express one variable in terms of another, simplify equations, and formulate the area function.
Starting with the equation for fencing constraints \( 2x + 3y = 1200 \), algebra allows us to solve for \( y \) in terms of \( x \):

• \( y = \frac{1200 - 2x}{3} \)

This expression is crucial because it replaces one variable in the area calculation, allowing us to single out \( x \) as the main variable. Algebraic manipulation like this helps solve for extreme areas by simplifying expressions to a manageable form suitable for calculus and other optimization techniques. Overall, understanding and using algebraic expressions is vital in efficiently solving such problems.