Mcdougal Littell Algebra 2: Student Edition (c) 2004 2004
Mcdougal Littell Algebra 2: Student Edition (c) 2004 2004
4th Edition
ISBN: 9780618250202
Author: Ron Larson, Laurie Boswell, Timothy D. Kanold, Lee Stiff
Publisher: Mcdougal Littell / Houghton Mifflin
Expert Solution & Answer
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Chapter 5.7, Problem 28PS
Solution

(a)

To find: Compare the minimum surface areas and corresponding values of d .

The surface area of cylinder is less than the prism and diameter of prism is lee than the cylinder.

Given information:

Given that the ordered pair of surface area and diameter for prism is given as:

  9.4,687.4,9.8,681.9,10.2,678.7,10.6,677.6,11,678.4,11.4,681.0

The graph of the surface area corresponding to diameter of cylinder is given as:

  Mcdougal Littell Algebra 2: Student Edition (c) 2004 2004, Chapter 5.7, Problem 28PS

Calculation:

It is observed from the given data for prism the minimum surface area is 677.6 and corresponding diameter is 10.6.

It is observed from the graph that the minimum surface area of cylinder is 6250cm2 and corresponding diameter is 11 .

The surface area of cylinder is less than the prism and diameter of prism is lee than the cylinder.

(b)

To find: Compare the height that corresponds to the minimum surface area of the containers.

The height of cylinder is greater than the height of prism.

Given information:

The height of the prism is h2=1200d2 and the height of the cylinder is h2=4800πd2 .

The surface area of cylinder is 2πrh+2πr2 which can be written as:

  S=2πh d2+2πd22S=πhd+πd22........(1)

The minimum surface area of the cylinder Is 6250 and the diameter is 11 .

Substitute the value of diameter in equation h2=4800πd2 :

  h2=4800π112

Further simplifying the above expression:

  h2=4800π121

Substitute the value of π=3.14 gives:

  h2=48003.14⋅121h2=4800379.94h2=12.63h=3.55

The height of the cylinder is 3.55 cm.

The height of prism is given by h2=1200d2 .The minimum surface area of prism is is 677.6 and corresponding diameter is 10.6.

Substitute d=10.6 in h2=1200d2 gives:

  h2=120010.62h2=1200112.36h2=10.67h=3.22

The height of cylinder is greater than the height of prism.

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Chapter 5.7, Problem 27PS
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Chapter 5.7, Problem 29PS

Chapter 5 Solutions

Mcdougal Littell Algebra 2: Student Edition (c) 2004 2004

Ch. 5.1 - Prob. 1GPCh. 5.1 - Prob. 2GPCh. 5.1 - Prob. 3GPCh. 5.1 - Prob. 4GPCh. 5.1 - Prob. 5GPCh. 5.1 - Prob. 6GPCh. 5.1 - Prob. 7GPCh. 5.1 - Prob. 8GPCh. 5.1 - Prob. 9GPCh. 5.1 - Prob. 10GP
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