   Chapter 3.7, Problem 37E

Chapter
Section
Textbook Problem

# A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) A minimum?

To determine

To cut:

The wire into pieces of square and equilateral triangle so that the total area enclosed is (a) maximum (b) minimum

Explanation

1) Concept:

Divide the wire into a circle and square, and find dimensions of circle and square. Then, find the maximum and minimum area using the formula.

2) Formula:

1) Area of equilateral triangle 34 l2

Where l is a side of the triangle

2) Area of square l2 where l is side of square.

3) Calculation:

A piece of wire of length 10 m cut into two pieces.

One piece is bent into a square

Suppose length x is bent into a square

As the sides of squares are equal

Lengths of each side of squares are x4

Other is bent into an equilateral triangle

Area of the wire used = Area of square + Area of equilateral triangle

A(x)= x42+3410-x3.10-x3

A(x)=x216+310-x236

A'x=x8 +310-x18(-1)

A'x=x8-310-x18

A'x=18x-8310-x144

The total area enclosed is maximum or minimum, when A'(x)=0

0=18x-8310-x144

Multiply on both sides by 144

18x-8310-x=0

18x-803+83x=0

Combine like terms,

18+83x-803=0

18+83x=803

Divide by 18+83 on both sides,

x=80318+83

Factoring 80=2·40 and 18+83=29+43

x=2·40329+43

Cancelling out common factor,

x=4039+43=4

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 