a. Answer x × x , s q u a r e − u n i t Explanation Calculation: an open box from a rectangular piece of material, 15 centimetres by 9 centimetres. It is given by cutting equal squares from the corners of the box . x represent the side length of each of the squares Hence , the dimensions of the open box be x × x , s q u a r e − u n i t . b. To determine Determine the domain of the function. Answer 0 < V < 9 2 Explanation Calculation: Volume V of the box is, V = L e n g t h × W i d t h × H e i g h t V ( x ) = ( 15 − 2 x ) ( 9 − 2 x ) ( x ) Now domain of Volume V it should be positive and V > 0 , ( 15 − 2 x ) ( 9 − 2 x ) ( x ) > 0 x > 0 , 15 − 2 x > 0 , 9 − 2 x > 0 x > 0 , x < 15 2 , x < 9 2 0 < V < 9 2 Hence, the domain of V is 0 < V < 9 2 c. To determine Sketch the graph of the function and approximate the dimensions of the box that yield a maximum volume. Answer ( 15 − 2 × 1.22 ) ( 9 − 2 × 1.22 ) ( 1.22 ) = 12.56 × 6.56 × 1.22 Cubic centimetre. Explanation Calculation: Volume the graph of V is, To find the maximum value V , we have to find value of x for which d V d x = 0 we get, 135 − 96 x − 12 x 2 = 0 By solving quadratic function ,consider positive value of x = 1.22 . Hence the dimensions of the box is, ( 15 − 2 × 1.22 ) ( 9 − 2 × 1.22 ) ( 1.22 ) = 12.56 × 6.56 × 1.22 cubic centimter. d. To determine Find values of x such that V = 56 . Answer 1 2 , 7 2 Explanation Calculation: Find values of x such that V = 56 . ( 15 − 2 x ) ( 9 − 2 x ) ( x ) = 56 135 x − 18 x 2 − 30 x 2 + 4 x 3 = 56 4 x 3 − 48 x 2 + 135 x − 56 = 0 To find the value of x for function f ( x ) = 4 x 3 − 48 x 2 + 135 x − 56 = 0 by using zero rational test with leading coefficient is 4 and constant term is − 56 Possible rational zeroes, Factor of -56 Factor of 4 = ± 1 , ± 2 , ± 4 , ± 7 , ... ± 56 ± 1 , ± 2 , ± 4 = ± 1 , ± 1 2 , ± 1 4 , ± 7 , ... ± 56 Therefor lot of zeroes possibility , now we will draw the graph of , f ( x ) It is observed from the graph that, rational zeroes of f ( x ) are 1 2 , 7 2 , 8 but 8 is not in the domain. Hence, the values of x such that V = 56 is 1 2 , 7 2

3rd Edition

ISBN: 9781133947202

Author: Ron Larson

Publisher: Cengage Learning

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Question

Chapter 2.5, Problem 111E

To determine

Find dimensions of the open box.

Expert Solution

Answer to Problem 111E

x×x,square−unit

Explanation of Solution

Calculation:

an open box from a rectangular piece of material, 15 centimetres by 9 centimetres.

It is given by cutting equal squares from the corners of the box . x represent the side length of each of the squares

Hence , the dimensions of the open box be x×x,square−unit .

To determine

Determine the domain of the function.

Expert Solution

Answer to Problem 111E

0<V<92

Explanation of Solution

Calculation:

Volume V of the box is,

V=Length×Width×Height

V(x)=(15−2x)(9−2x)(x)

Now domain of Volume V it should be positive and V>0 ,

(15−2x)(9−2x)(x)>0x>0,15−2x>0,9−2x>0x>0,x<152,x<920<V<92

Hence, the domain of V is 0<V<92

To determine

Sketch the graph of the function and approximate the dimensions of the box that yield a maximum volume.

Expert Solution

Answer to Problem 111E

(15−2×1.22)(9−2×1.22)(1.22)=12.56×6.56×1.22 Cubic centimetre.

Explanation of Solution

Calculation:

Volume the graph of V is,

Precalculus with Limits, Chapter 2.5, Problem 111E , additional homework tip 1

To find the maximum value V , we have to find value of x for which dVdx=0 we get,

135−96x−12x2=0

By solving quadratic function ,consider positive value of x=1.22 .

Hence the dimensions of the box is, (15−2×1.22)(9−2×1.22)(1.22)=12.56×6.56×1.22 cubic centimter.

To determine

Find values of x such that V=56 .

Expert Solution

Answer to Problem 111E

12,72

Explanation of Solution

Calculation:

Find values of x such that V=56 .

(15−2x)(9−2x)(x)=56135x−18x2−30x2+4x3=564x3−48x2+135x−56=0

To find the value of x for function f(x)=4x3−48x2+135x−56=0 by using zero rational test with leading coefficient is 4 and constant term is −56

Possible rational zeroes,

Factor of -56Factor of 4=±1,±2,±4,±7,...±56±1,±2,±4=±1,±12,±14,±7,...±56

Therefor lot of zeroes possibility , now we will draw the graph of , f(x)

Precalculus with Limits, Chapter 2.5, Problem 111E , additional homework tip 2

It is observed from the graph that, rational zeroes of f(x) are 12,72,8 but 8 is not in the domain.

Hence, the values of x such that V=56 is 12,72

Chapter 2.5, Problem 110E

Chapter 2.5, Problem 112E

Chapter 2 Solutions

Precalculus with Limits

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Prob. 104E Ch. 2.5 - Prob. 105E Ch. 2.5 - Prob. 106E Ch. 2.5 - Prob. 107E Ch. 2.5 - Prob. 108E Ch. 2.5 - Prob. 109E Ch. 2.5 - Prob. 110E Ch. 2.5 - Prob. 111E Ch. 2.5 - Prob. 112E Ch. 2.5 - Prob. 113E Ch. 2.5 - Prob. 114E Ch. 2.5 - Prob. 115E Ch. 2.5 - Prob. 116E Ch. 2.5 - Prob. 117E Ch. 2.5 - Prob. 118E Ch. 2.5 - Prob. 119E Ch. 2.5 - Prob. 120E Ch. 2.5 - Prob. 121E Ch. 2.5 - Prob. 122E Ch. 2.5 - Prob. 123E Ch. 2.5 - Prob. 124E Ch. 2.5 - Prob. 125E Ch. 2.5 - Prob. 126E Ch. 2.5 - Prob. 127E Ch. 2.5 - Prob. 128E Ch. 2.5 - Prob. 129E Ch. 2.5 - Prob. 130E Ch. 2.5 - Prob. 131E Ch. 2.6 - Prob. 1E Ch. 2.6 - Prob. 2E Ch. 2.6 - Prob. 3E Ch. 2.6 - Prob. 4E Ch. 2.6 - Prob. 5E Ch. 2.6 - Prob. 6E Ch. 2.6 - Prob. 7E Ch. 2.6 - Prob. 8E Ch. 2.6 - Prob. 9E Ch. 2.6 - Prob. 10E Ch. 2.6 - Prob. 11E Ch. 2.6 - Prob. 12E Ch. 2.6 - Prob. 13E Ch. 2.6 - Prob. 14E Ch. 2.6 - Prob. 15E Ch. 2.6 - Prob. 16E Ch. 2.6 - Prob. 17E Ch. 2.6 - Prob. 18E Ch. 2.6 - Prob. 19E Ch. 2.6 - Prob. 20E Ch. 2.6 - Prob. 21E Ch. 2.6 - Prob. 22E Ch. 2.6 - Prob. 23E Ch. 2.6 - Prob. 24E Ch. 2.6 - Prob. 25E Ch. 2.6 - Prob. 26E Ch. 2.6 - Prob. 27E Ch. 2.6 - Prob. 28E Ch. 2.6 - Prob. 29E Ch. 2.6 - Prob. 30E Ch. 2.6 - Prob. 31E Ch. 2.6 - Prob. 32E Ch. 2.6 - Prob. 33E Ch. 2.6 - Prob. 34E Ch. 2.6 - Prob. 35E Ch. 2.6 - Prob. 36E Ch. 2.6 - Prob. 37E Ch. 2.6 - Prob. 38E Ch. 2.6 - Prob. 39E Ch. 2.6 - Prob. 40E Ch. 2.6 - Prob. 41E Ch. 2.6 - Prob. 42E Ch. 2.6 - Prob. 43E Ch. 2.6 - Prob. 44E Ch. 2.6 - Prob. 45E Ch. 2.6 - Prob. 46E Ch. 2.6 - Prob. 47E Ch. 2.6 - Prob. 48E Ch. 2.6 - Prob. 49E Ch. 2.6 - Prob. 50E Ch. 2.6 - Prob. 51E Ch. 2.6 - Prob. 52E Ch. 2.6 - Prob. 53E Ch. 2.6 - Prob. 54E Ch. 2.6 - Prob. 55E Ch. 2.6 - Prob. 56E Ch. 2.6 - Prob. 57E Ch. 2.6 - Prob. 58E Ch. 2.6 - Prob. 59E Ch. 2.6 - Prob. 60E Ch. 2.6 - Prob. 61E Ch. 2.6 - Prob. 62E Ch. 2.6 - Prob. 63E Ch. 2.6 - Prob. 64E Ch. 2.6 - Prob. 65E Ch. 2.6 - Prob. 66E Ch. 2.6 - Prob. 67E Ch. 2.6 - Prob. 68E Ch. 2.6 - Prob. 69E Ch. 2.6 - Prob. 70E Ch. 2.6 - Prob. 71E Ch. 2.6 - Prob. 72E Ch. 2.6 - Prob. 73E Ch. 2.6 - Prob. 74E Ch. 2.6 - Prob. 75E Ch. 2.6 - Prob. 76E Ch. 2.6 - Prob. 77E Ch. 2.6 - Prob. 78E Ch. 2.6 - Prob. 79E Ch. 2.6 - Prob. 80E Ch. 2.6 - Prob. 81E Ch. 2.6 - Prob. 82E Ch. 2.7 - Prob. 1E Ch. 2.7 - Prob. 2E Ch. 2.7 - Prob. 3E Ch. 2.7 - Prob. 4E Ch. 2.7 - Prob. 5E Ch. 2.7 - Prob. 6E Ch. 2.7 - Prob. 7E Ch. 2.7 - Prob. 8E Ch. 2.7 - Prob. 9E Ch. 2.7 - Prob. 10E Ch. 2.7 - Prob. 11E Ch. 2.7 - Prob. 12E Ch. 2.7 - Prob. 13E Ch. 2.7 - Prob. 14E Ch. 2.7 - Prob. 15E Ch. 2.7 - Prob. 16E Ch. 2.7 - Prob. 17E Ch. 2.7 - Prob. 18E Ch. 2.7 - Prob. 19E Ch. 2.7 - Prob. 20E Ch. 2.7 - Prob. 21E Ch. 2.7 - Prob. 22E Ch. 2.7 - Prob. 23E Ch. 2.7 - Prob. 24E Ch. 2.7 - Prob. 25E Ch. 2.7 - Prob. 26E Ch. 2.7 - Prob. 27E Ch. 2.7 - Prob. 28E Ch. 2.7 - Prob. 29E Ch. 2.7 - Prob. 30E Ch. 2.7 - Prob. 31E Ch. 2.7 - Prob. 32E Ch. 2.7 - Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.7 - Prob. 46E Ch. 2.7 - Prob. 47E Ch. 2.7 - Prob. 48E Ch. 2.7 - Prob. 49E Ch. 2.7 - Prob. 50E Ch. 2.7 - Prob. 51E Ch. 2.7 - Prob. 52E Ch. 2.7 - Prob. 53E Ch. 2.7 - Prob. 54E Ch. 2.7 - Prob. 55E Ch. 2.7 - Prob. 56E Ch. 2.7 - Prob. 57E Ch. 2.7 - Prob. 58E Ch. 2.7 - Prob. 59E Ch. 2.7 - Prob. 60E Ch. 2.7 - Prob. 61E Ch. 2.7 - Prob. 62E Ch. 2.7 - Prob. 63E Ch. 2.7 - Prob. 64E Ch. 2.7 - Prob. 65E Ch. 2.7 - Prob. 66E Ch. 2.7 - Prob. 67E Ch. 2.7 - Prob. 68E Ch. 2.7 - Prob. 69E Ch. 2.7 - Prob. 70E Ch. 2.7 - Prob. 71E Ch. 2.7 - Prob. 72E Ch. 2.7 - Prob. 73E Ch. 2.7 - Prob. 74E Ch. 2.7 - Prob. 75E Ch. 2.7 - Prob. 76E Ch. 2.7 - Prob. 77E Ch. 2.7 - Prob. 78E Ch. 2.7 - Prob. 79E Ch. 2.7 - Prob. 80E Ch. 2.7 - Prob. 81E Ch. 2.7 - Prob. 82E Ch. 2.7 - Prob. 83E Ch. 2.7 - Prob. 84E Ch. 2.7 - Prob. 85E Ch. 2.7 - Prob. 86E Ch. 2.7 - Prob. 87E Ch. 2.7 - Prob. 88E Ch. 2.7 - Prob. 89E Ch. 2.7 - Prob. 90E Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE Ch. 2 - Prob. 56RE Ch. 2 - Prob. 57RE Ch. 2 - Prob. 58RE Ch. 2 - Prob. 59RE Ch. 2 - Prob. 60RE Ch. 2 - Prob. 61RE Ch. 2 - Prob. 62RE Ch. 2 - Prob. 63RE Ch. 2 - Prob. 64RE Ch. 2 - Prob. 65RE Ch. 2 - Prob. 66RE Ch. 2 - Prob. 1CT Ch. 2 - Prob. 2CT Ch. 2 - Prob. 3CT Ch. 2 - Prob. 4CT Ch. 2 - Prob. 5CT Ch. 2 - Prob. 6CT Ch. 2 - Prob. 7CT Ch. 2 - Prob. 8CT Ch. 2 - Prob. 9CT Ch. 2 - Prob. 10CT Ch. 2 - Prob. 11CT Ch. 2 - Prob. 12CT Ch. 2 - Prob. 13CT Ch. 2 - Prob. 14CT Ch. 2 - Prob. 15CT Ch. 2 - Prob. 16CT Ch. 2 - Prob. 17CT Ch. 2 - Prob. 18CT Ch. 2 - Prob. 1PS Ch. 2 - Prob. 2PS Ch. 2 - Prob. 3PS Ch. 2 - Prob. 4PS Ch. 2 - Prob. 5PS Ch. 2 - Prob. 6PS Ch. 2 - Prob. 7PS Ch. 2 - Prob. 8PS Ch. 2 - Prob. 9PS Ch. 2 - Prob. 10PS Ch. 2 - Prob. 11PS Ch. 2 - Prob. 12PS Ch. 2 - Prob. 13PS

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