Problem 66 - College Algebra

Question

You have 200 feet of fencing to enclose a rectangular plot that borders on a river. If you do not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed? GRAPH CANNOT COPY.

Step-by-Step Solution

Answer

The maximum area that can be enclosed is 5000 square feet. The plot dimensions are 50 feet width and 100 feet length along the river.

Step 1: Set Up The Equations

Let us consider the length of the rectangle along the river as \(L\) and the width (the side opposite to the river) as \(W\). As the total length of the fence is 200 feet, we can say \(2W + L = 200\). To maximize the area, express the area \(A = LW\), then substitute \(L = 200 - 2W\) into the area equation to get a function of \(W\), \(A = W(200 - 2W)\).

Step 2: Find the Maximum of the Function

Differentiate the area \(A\) with respect to \(W\), set this derivative equal to zero and solve for \(W\). The derivative of \(A\) is \(A' = 200 - 4W\). Solving \(A' = 0\) gives \(W = 50\). Apply the first derivative test or second derivative test to prove \(W = 50\) results in maximum area.

Step 3: Find The Dimensions of the Rectangle

Substitute \(W = 50\) into the equation \(2W + L = 200\) gives \(L = 200 - 2W = 200 - 2*50 = 100\). So, the dimensions that give maximum area are \(W = 50\) and \(L = 100\). These are the two required dimensions of the plot.

Step 4: Find The Maximum Area

Substitute \(W = 50\) and \(L = 100\) into the area function \(A = LW\) to find the maximum area. \(A = L*W = 50*100 = 5000\) square feet. So, the largest area that can be enclosed is 5000 square feet.

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with the study of change and motion. It provides tools for analyzing problems involving dynamic systems.
Calculus covers two main areas:

• Differential Calculus: This focuses on the concept of a derivative, which measures how a function changes as its input changes.
• Integral Calculus: This deals with the accumulation of quantities and areas under curves.

In optimization problems like finding maximum or minimum values, calculus is invaluable because it helps us understand how functions behave by using techniques like differentiation. This makes it possible to solve real-world problems efficiently.

Derivatives

Derivatives are fundamental concepts in calculus that describe the rate at which a quantity changes.
In mathematical terms, the derivative of a function at any point is the slope of the tangent line at that point. This means it tells us how steep the function is at a certain point and in which direction it is moving.

• For optimization tasks, derivatives help find critical points.
• By setting the derivative equal to zero, we can find points where the function might have a maximum or minimum value.

In our fencing problem, by differentiating the function representing the area of the rectangle, we pinpoint where the area might be the largest, leading us to find the optimal dimensions of the plot.

Rectangular Area

When addressing problems involving rectangular areas, understanding the basic geometry is crucial. The area of a rectangle is calculated by multiplying its length ( L ) by its width ( W ).
In optimization problems such as the one in the exercise, the objective is often to maximize or minimize this area under given constraints.

• Constraints can be anything, like the amount of fencing available in our problem.
• This concept helps set up the main function that requires optimization.

By expressing the dimensions in terms of one variable (like expressing L in terms of W or vice versa using constraints), it becomes easier to derive a single-variable function to solve.

Fencing Problem

In a fencing problem, you are tasked with maximizing or minimizing an area using a fixed length of fence. These problems often feature constraints where one side of a plot may not need fencing, like a river or a wall.
For instance, if a rectangle borders a river, you only need three sides of fencing.

• Here, the total length of the fence forms part of the equations and inequalities we need to consider.
• The goal is to use the fence optimally to get the largest area possible.

This type of problem is practical in agriculture and architecture, where maximizing land use or space is important.