Exercise 3 Page 662

Practice makes perfect

a We are given that a crayon can be approximated by a composite solid made from a cylinder and a cone. Let's draw each part of the crayon separately. All measures are given in millimeters.

Notice that the volume of the crayon will be the sum of the volume of a cylinder and the volume of a cone. Let's write an equation using appropriate formulas. Remember that a radius is one half of the diameter.

V=π (8.5/2)^2( 80)+1/3π (6.5/2)^2 ( 10)

CalcQuot

Calculate quotient

V=π(4.25)^2(80)+1/3π(3.25)^2(10)

CalcPow

Calculate power

V=π(18.0625)(80)+1/3π(10.5625)(10)

Multiply

V=1445π+105.625/3π

UseCalc

Use a calculator

V=4650.2116...

RoundInt

Round to nearest integer

V≈ 4650.2

The volume of the crayon is approximately 4650.2 cubic millimeters.

b In this part we want to find the amount of space within the crayon box not taken up by the crayons. This means that we need to evaluate the difference between the volume of the crayon box and the volumes of the crayons.

V_(box)-V_(crayons)

First let's find the volume of the box that is a rectangular prism. Recall that the volume of a prism is the product of its dimensions. Therefore, let's multiply the dimensions of the crayon box. V_(box)=94*28*71=186 872 The volume of the box is 186 872 cubic millimeters. Since we found in Part A that the volume of one crayon is approximately 4650.2 mm^2 and we have 24 crayons in the box, the volume of the crayons will be 24 times 4650.2. V_(box)-V_(crayons) ⇓ 186 872- 24* 4650.2=75 267.2 The amount of space within the crayon box not taken up by the crayons is approximately 75 267.2 cubic millimeters.

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Exercises p.662-663

Exercise 3 Page 662

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