Problem 35 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

OPTIMAL DESIGN A farmer wishes to enclose a rectangular pasture with 320 feet of fence. What dimensions give the maximum area if a. the fence is on all four sides of the pasture? b. the fence is on three sides of the pasture and the fourth side is bounded by a wall?

Step-by-Step Solution

Answer

a. Dimensions for maximum area are 80 feet by 80 feet. b. Dimensions for maximum area are 80 feet by 160 feet.

Step 1 - Define Variables

Let the length of the rectangular pasture be denoted as L, and the width be W.

Step 2 - Express Constraint for Part a

For a fence on all four sides: \[ 2L + 2W = 320 \] Simplify to get: \[ L + W = 160 \]

Step 3 - Express Area as Function of One Variable

The area is given by A = L \times W. Substitute W from the constraint: \[ W = 160 - L \] so, \[ A = L(160 - L) \]

Step 4 - Find Maximum Area Using Derivative

Differentiate A with respect to L: \[ \frac{dA}{dL} = 160 - 2L \] Set the derivative to zero to find the critical points: \[ 160 - 2L = 0 \] Solve for L: \[ L = 80 \] Then, \[ W = 160 - 80 = 80 \]

Step 5 - Express Constraint for Part b

For a fence on three sides with the fourth side bounded by a wall: \[ 2L + W = 320 \] Simplify to get: \[ W = 320 - 2L \]

Step 6 - Express Area as Function of One Variable for Part b

The area is given by A = L \times W. Substitute W from the constraint: \[ W = 320 - 2L \] so, \[ A = L(320 - 2L) \] Simplify to: \[ A = 320L - 2L^2 \]

Step 7 - Find Maximum Area Using Derivative for Part b

Differentiate A with respect to L: \[ \frac{dA}{dL} = 320 - 4L \] Set the derivative to zero to find the critical points: \[ 320 - 4L = 0 \] Solve for L: \[ L = 80 \] Then, \[ W = 320 - 2(80) = 160 \]

Key Concepts

rectangular enclosure

When you need to design a rectangular enclosure, the most common task is to determine the dimensions that either maximize or minimize a particular measurement.
In the example provided, a farmer wants to create a pasture with a specific amount of fencing. This involves understanding and applying constraints related to the perimeter or total fencing available.
First, define your variables:

• Let length be denoted as \(L\) and width as \(W\).

Depending on the problem setup, you create an equation that represents the boundary constraints:

• For a fence on all four sides, the equation is \(2L + 2W = 320\).
• For a fence on three sides with the fourth side being a wall, the equation becomes \(2L + W = 320\).

This initial step sets the stage for further calculations and ensures you are working within the given limits.

area maximization

The goal of area maximization is to find the dimensions that yield the largest possible area for the enclosed space.
Using the constraints from the previous section, you can express one variable in terms of the other to simplify the equation.
For example, in part (a) where \(L + W = 160\), express \(W\) as \(W = 160 - L\), and then substitute into the area formula so it reads \(A = L(160 - L)\).
In part (b), where \(W = 320 - 2L\), substitute to get \(A = L(320 - 2L)\).
The next step is to find the dimensions that maximize this area function. This is achieved by finding the critical points using calculus.

critical points

Critical points are necessary to determine the maximum or minimum values of a function. In this context, we use derivatives to find these points.
Take the derivative of the area functions with respect to \(L\):

• For part (a): \(\frac{dA}{dL} = 160 - 2L\).
• For part (b): \(\frac{dA}{dL} = 320 - 4L\).

Set these derivatives equal to zero and solve for \(L\):

• For part (a):\[ 160 - 2L = 0 \rightarrow L = 80 \]
• For part (b):\[ 320 - 4L = 0 \rightarrow L = 80 \]

After finding these critical values, substitute back into the constraint equations to find the corresponding \( W \) values:

• For part (a): \( W = 160 - 80 = 80 \).
• For part (b): \( W = 320 - 2(80) = 160 \).

These dimensions provide the maximum areas for the respective scenarios.