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Maximizing Area of a Rectangle with Given Fencing Constraints, Study Guides, Projects, Research of Pre-Calculus

A problem statement and solution for finding the maximum area of a rectangle using a fixed amount of fencing. The problem involves creating the rectangle with the maximum area by determining the length and width, given the number of lengths and widths of fencing used. Examples and instructions for students to complete the problem as a group project and individual extra credit.

Typology: Study Guides, Projects, Research

Pre 2010

Uploaded on 08/19/2009

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THE ULTIMATE RECTANGLE PROBLEM
Suppose that a fixed amount of fencing "f" is available. The goal is to create the rectangle with the
maximum area, under the given conditions. The length of the rectangle will be called "x", and the width "y".
The large rectangle may have some smaller rectangles created within it. If so, all vertical fences are parallel
and all horizontal fences are parallel. All interior fences go completely across the large rectangle. The
widths of interior rectangles do not have to be equal; neither do the lengths.
The number of lengths of fencing required is "a"; "a" will be one or more. The number of widths of fencing
required is "b"; "b" will be one or more.
Examples:
existing barn neighbor’s
fence
river
existing
fence
a = 1, b = 2 a = 1, b = 1 a = 2, b = 1 a = 4, b = 4 a = 6, b = 3
Example: Alex has 1000 feet of fencing with which to enclose garden
plots as illustrated. Determine the length and width of the
outside dimensions so that the total area enclosed is
maximized.
Step 1: Choose a variable, identify it completely; identify other unknowns in terms of same variable.
Let x = length of fenced area
Then = width of fenced area
1000 2
6
x
Step 2: Write an equation in terms of your chosen variable which summarizes the given information.
The number of lengths used is 2, i.e. a = 2
The number of widths used is 6, i.e. b =6
Since the goal of the problem is to maximize area, the equation is based on the area of the overall
rectangle : Ax x x x x
( )
=
=
1000 2
6
1000 2
6
2
Step 3: Solve your equation.
In this problem, you need to maximize the area. Since the area function is quadratic with a negative
leading coefficient, an absolute maximum exists.
and xb
a
= = =
2
10006
23
250 1000 2 250
683 1
3
=
( )
Step 4: Answer the question(s) with complete sentences, including units.
The overall rectangle should be 250 ft by ft, with each interior fence being ft long. The total
83 1
383 1
3
area enclosed will be .
20833 1
32
, ft
pf2

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THE ULTIMATE RECTANGLE PROBLEM

Suppose that a fixed amount of fencing "f" is available. The goal is to create the rectangle with the maximum area, under the given conditions. The length of the rectangle will be called "x", and the width "y". The large rectangle may have some smaller rectangles created within it. If so, all vertical fences are parallel and all horizontal fences are parallel. All interior fences go completely across the large rectangle. The widths of interior rectangles do not have to be equal; neither do the lengths.

The number of lengths of fencing required is "a"; "a" will be one or more. The number of widths of fencing required is "b"; "b" will be one or more.

Examples:

existing barn neighbor’s fence

river

existing fence a = 1, b = 2 a = 1, b = 1 a = 2, b = 1 a = 4, b = 4 a = 6, b = 3

Example: Alex has 1000 feet of fencing with which to enclose garden plots as illustrated. Determine the length and width of the outside dimensions so that the total area enclosed is maximized.

Step 1: Choose a variable, identify it completely; identify other unknowns in terms of same variable.

Let x = length of fenced area

Then = width of fenced area

− x

Step 2: Write an equation in terms of your chosen variable which summarizes the given information.

The number of lengths used is 2, i.e. a = 2 The number of widths used is 6, i.e. b = Since the goal of the problem is to maximize area, the equation is based on the area of the overall

rectangle : (^) A x( ) = x  − x^ x^ x ^

^

2

Step 3: Solve your equation.

In this problem, you need to maximize the area. Since the area function is quadratic with a negative leading coefficient, an absolute maximum exists.

x b and a

Step 4: Answer the question(s) with complete sentences, including units.

The overall rectangle should be 250 ft by 83 13 ft, with each interior fence being 83 13 ft long. The total

area enclosed will be (^) 20 833 , 13 ft^2.

GROUP PROJECT DIRECTIONS:

  1. Use pencil and organize all work neatly on one side of notebook paper. No credit will be given unless all supporting work is present!!
  2. Rework the example above for each of the five diagrams given. Summarize your results in this table:
  3. All measurements must be exact!
  4. Each student is responsible for contributing to the project and must be able to work all problems.
  5. Each student must turn in an evaluation form to order to receive any credit for this project. Each student’s grade will be determined by the quality of the project submitted and the amount of the student’s contribution to the project. The extra credit portion below will be graded separately and individually.

Amount of fencing available = f = 1000

Number of lengths = a Length Number of widths = b Widt h

Area

3 20 833^

1 , 3

2 1

1 1

2 1

4 4

6 3

INDIVIDUAL EXTRA CREDIT: do not work together on this portion!

Write up your work separately from the group project. Find a formula for x ( in terms of “f” and “a” only ) and a formula for y ( in terms of “f” and “b” only ) that will enable you to predict the optimum dimensions of the rectangle for any given set of numbers f, a, and b. Find a formula for the maximum area of the large rectangle ( in terms of “f”, “a”, and “b” only ). Explain how you discovered your formulas. After you have found your formulas, use them to complete the following table. All measurements must be exact! [Copy the table on your paper.] Choose one additional combination of values for “f”, “a”, and “b” and use them to complete the last row in the table.

Fencing available = f

Number of lengths = a

Length Number of widths = b

Width Area