Holt Mcdougal Larson Algebra 2: Student Edition 2012
Holt Mcdougal Larson Algebra 2: Student Edition 2012
1st Edition
ISBN: 9780547647159
Author: HOLT MCDOUGAL
Publisher: HOLT MCDOUGAL
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Chapter 2.4, Problem 60PS
Solution

(a)

Find the volume, in terms of x , of each of three level of platform.

The volume of the bottom platform is given by 48x3 .

The volume of the middle platform is given by 24x3 .

The volume of the upper platform is given by 8x3 .

Given:

  Holt Mcdougal Larson Algebra 2: Student Edition 2012, Chapter 2.4, Problem 60PS , additional homework tip 1

Calculation:

The length, width and height of the bottom platform are 8x,6x,x respectively.

The length, width and height of the middle platform are 6x,4x,x respectively.

The length, width and height of the upper platform are 4x,2x,x respectively.

Therefore,

The volume of the bottom platform is given by 8x⋅6x⋅x = 48x3 .

The volume of the middle platform is given by 6x⋅4x⋅x = 24x3 .

The volume of the upper platform is given by 4x⋅2x⋅x = 8x3 .

(b)

Write an equation for the total volume.

  48x3+24x3+8x3=1250

Given:

The total volume is given by 1250 cubic feet.

  Holt Mcdougal Larson Algebra 2: Student Edition 2012, Chapter 2.4, Problem 60PS , additional homework tip 2

Calculation:

The length, width and height of the bottom platform are 8x,6x,x respectively.

The length, width and height of the middle platform are 6x,4x,x respectively.

The length, width and height of the upper platform are 4x,2x,x respectively.

Therefore,

The volume of the bottom platform is given by 8x⋅6x⋅x = 48x3 .

The volume of the middle platform is given by 6x⋅4x⋅x = 24x3 .

The volume of the upper platform is given by 4x⋅2x⋅x = 8x3 .

Total volume is given as 1250 cubic feet.

Hence, the equation for the total volume is given by

  48x3+24x3+8x3=1250

(c)

Solve the equation 48x3+24x3+8x3=1250 and then find the dimensions of the three levels of platform.

The length, width and height of the bottom platform are given by 20feet,15feet,2.5feet .

The length, width and height of the middle platform are 15feet,10feet,2.5feet .

The length, width and height of the upper platform are =10feet,5feet,2.5feet .

Given:

  48x3+24x3+8x3=1250

The total volume is given by 1250 cubic feet.

  Holt Mcdougal Larson Algebra 2: Student Edition 2012, Chapter 2.4, Problem 60PS , additional homework tip 3

Calculation:

The equation for the total volume is given by

  48x3+24x3+8x3=1250

Solve this equation

  80x3=1250

  x3=1258

  x=52=2.5 .

Therefore,

The length, width and height of the bottom platform are given by

  8⋅2.5,6⋅2.5,2.5

  =20feet,15feet,2.5feet

The length, width and height of the middle platform are

  6⋅2.5,4⋅2.5,2.5

  =15feet,10feet,2.5feet

The length, width and height of the upper platform are

  4⋅2.5,2⋅2.5,2.5

  =10feet,5feet,2.5feet

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Chapter 2.4, Problem 61PS

Chapter 2 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 2.1 - Prob. 1GPCh. 2.1 - Prob. 2GPCh. 2.1 - Prob. 3GPCh. 2.1 - Prob. 4GPCh. 2.1 - Prob. 5GPCh. 2.1 - Prob. 6GPCh. 2.1 - Prob. 7GPCh. 2.1 - Prob. 8GPCh. 2.1 - Prob. 1ECh. 2.1 - Prob. 2E
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Prob. 9TCh. 2 - Prob. 10TCh. 2 - Prob. 11TCh. 2 - Prob. 12TCh. 2 - Prob. 13TCh. 2 - Prob. 14TCh. 2 - Prob. 15TCh. 2 - Prob. 16TCh. 2 - Prob. 17TCh. 2 - Prob. 18TCh. 2 - Prob. 19TCh. 2 - Prob. 20TCh. 2 - Prob. 21TCh. 2 - Prob. 22TCh. 2 - Prob. 23TCh. 2 - Prob. 24TCh. 2 - Prob. 25TCh. 2 - Prob. 26TCh. 2 - Prob. 27TCh. 2 - Prob. 28TCh. 2 - Prob. 29TCh. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 1TPCh. 2 - Prob. 2TPCh. 2 - Prob. 3TPCh. 2 - Prob. 4TPCh. 2 - Prob. 5TPCh. 2 - Prob. 6TPCh. 2 - Prob. 7TPCh. 2 - Prob. 8TPCh. 2 - Prob. 9TPCh. 2 - Prob. 10TPCh. 2 - Prob. 11TPCh. 2 - Prob. 12TPCh. 2 - Prob. 13TPCh. 2 - Prob. 14TPCh. 2 - Prob. 15TPCh. 2 - Prob. 16TPCh. 2 - Prob. 17TPCh. 2 - Prob. 18TPCh. 2 - Prob. 19TPCh. 2 - Prob. 20TPCh. 2 - Prob. 21TP
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Publisher:Wellesley-Cambridge Press
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College Algebra (Collegiate Math)
Algebra
ISBN:9780077836344
Author:Julie Miller, Donna Gerken
Publisher:McGraw-Hill Education
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