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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

Solve Exercise 37 if one piece is bent into a square and the other into a circle.

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 square and circle pieces so that the total area enclosed is (a) maximum (b) minimum

Explanation

1) Concept:

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

2) Formula:

1) Area of circle πr2

Where r is radius of the circle

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 the squares are equal

Length of each side of square is x4

The remaining length (10-x) is bent into a circle.

Therefore, circumference of circle is 10-x

2πr=(10-x)

Divide by 2π

r=10-x2π

Area of the wire used = Area of square + Area of circle

By using formula above,

A(x)= x42+π10-x2π2

A(x)=x216+10-x24π

A'x=x8 +10-x2π(-1)

A'x=x8-10-x2π

A'x=πx-410-x8π

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

0=πx-410-x8π

Multiply on both sides by 8π

πx-410-x=0

πx-40+4x=0

Combine like terms,

π+4x-40=0

Adding 40 from both sides,

π+4x=40

Divide by π+4 on both sides,

x=40π+4=5

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