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

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.

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Question

Chapter 5, Problem 51RE

To determine

To write a formula V(x) for the volume of box.

Expert Solution

Answer to Problem 51RE

The required formula is V(x)=2x3−25x2+75x .

Explanation of Solution

Given information: A piece of cardboard measures 10 in. by 15 in. Two equal sides are removed from the corners of a 15 in. sides.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5, Problem 51RE , additional homework tip 1

Calculation: The required formula is obtained as,

Refer to the diagram exercise 18 of the sction 4.4, but imagine the vertical side as 15” and the horizontal side as 10” .

The length of box be 15−2x .

The width of box be 10−2x2=5−x .

The height of box be x .

V(x)=x(15−2x)(5−x) =2x3−25x2+75x

Hence, the required formula is V(x)=2x3−25x2+75x .

To determine

To find the domain of V for the problem situation and graph V over this domain.

Expert Solution

Answer to Problem 51RE

The domain is: 0 < x< 5 and graph below,

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5, Problem 51RE , additional homework tip 2

Explanation of Solution

Given information: A piece of cardboard measures 10 in. by 15 in. Two equal sides are removed from the corners of a 15 in. sides.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5, Problem 51RE , additional homework tip 3

Calculation: The required solution is obtained as,

x must be > 0 since you have to cut something. x must be less than the smaller of 15/2 = 7.5 and 10/2 = 5, so that you don’t cut away more material than you actually have.

Hence, the domain is: 0 < x< 5 and graph below,

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5, Problem 51RE , additional homework tip 4

To determine

To find the maximum volume and the value of x that gives it, by using a graphical method.

Expert Solution

Answer to Problem 51RE

The maximum is at approximately x=1.96 and V=66.02 .

Explanation of Solution

Given information: A piece of cardboard measures 10 in. by 15 in. Two equal sides are removed from the corners of a 15 in. sides.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5, Problem 51RE , additional homework tip 5

Calculation: The required solution is obtained as,

From the graph, the maximum is at approximately x=1.96 and V=66.02 .

To determine

To confirm your result in part (C) analytically.

Expert Solution

Answer to Problem 51RE

The required solution is 1.96.

Explanation of Solution

Given information: A piece of cardboard measures 10 in. by 15 in. Two equal sides are removed from the corners of a 15 in. sides.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5, Problem 51RE , additional homework tip 6

Calculation: The required solution is obtained as,

Find V’ and where it is = 0.

V'(x)=6x2−50x+75 0=6x2−50x+75 x=−b±b2−4ac2a =−(−50)±(−50)2−4(6)(75)2(6) =50±70012 =50±10712 =25±576 =25+576≈6.37 =25−576≈1.96

Hence the required solution is 1.96.

Chapter 5, Problem 50RE

Chapter 5, Problem 52RE

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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