Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.4, Problem 48E

To determine

To Prove that L2=2x3(2x−8.5) .

Expert Solution

Answer to Problem 48E

The answer is L2=2x3(2x−8.5) .

Explanation of Solution

Calculation: Given the dimensions of the rectangular sheet is 812 .

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.4, Problem 48E

Let QR=y and PB=8.5−x

QB=x2−(8.5−x2) =8.5(2x−8.5)

Here triangles QRS and PQB are similar.

QRRS=PQQBy8.5=x8.5(2x−8.5)

Consider the figure

QB=x2−(8.5−x2) =8.5(2x−8.5)

y2(8.5)2=x2(8.5(2x−8.5))2y2=(8.5)x2(2x−8.5)L2=x2+y2 =x2+(8.5)x2(2x−8.5) =x2(2x−8.5)+8.5x2(2x−8.5) =2x3(2x−8.5)

Thus ,the answer is L2=2x3(2x−8.5) .

To determine

Find the value of x minimizes L2 .

Expert Solution

Answer to Problem 48E

The L2 is minimum x=6.37 .

Explanation of Solution

Calculation:let ,

f(x)=L2 =2x3(2x−8.5)

For any function f(x) the condition of minimum and maximum is obtained by find f′(x)=0 ,

Critical points and end points

f(x)=L2 =2x3(2x−8.5)dfdx=(2x−8.5)(6x2)−2x3(2)(2x−8.5)2 =8x3−51x2(2x−8.5)

Critical points occur at

f′(x)=08x3−51x2(2x−8.5)=08x3−51x2=0 8x=51x=518=6.37

Critical points occur at x=6.37

Here dDdx<0forx<1anddDdx>0 for x>1 this corresponds to minimum distance.

f(x)=L2

Thus ,the L2 is minimum x=6.37 .

To determine

Find the minimum value of L .

Expert Solution

Answer to Problem 48E

The minimum value of L2

11.04 .

Explanation of Solution

Calculation:

For x=6.37

L2=2x3(2x−8.5) =2(6.37)3(2(6.37)−8.5) =11.04

Thus, the minimum value of L2

11.04 in.

Chapter 5.4, Problem 47E

Chapter 5.4, Problem 49E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

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

Ch. 5.1 - Prob. 11QR Ch. 5.1 - Prob. 12QR Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - 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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY

To Prove that L 2 = 2 x 3 ( 2 x − 8.5 ) .

Concept explainers

Videos

To Prove that L2=2x3(2x−8.5) .

Answer to Problem 48E

Explanation of Solution

Find the value of x minimizes L2 .

Answer to Problem 48E

Explanation of Solution

Find the minimum value of L .

Answer to Problem 48E

Explanation of Solution

Chapter 5 Solutions

Additional Math Textbook Solutions