a . Answer V ( x ) = x 3 + 9 x 2 + 26 x + 24 = 120 Explanation Given: Current dimensions of bin: 2 feet by 3 feet by 4 feet It is required to increase the volume of the bin 5 times. Calculation: Let each dimension is increased by the same amount x. Then, new dimension of the bin is: ( x + 2 ) feet by ( x + 3 ) feet by ( x + 4 ) feet Then, new volume of the bin is: V ( x ) = l e n g t h × w i d t h × h e i g h t = ( x + 2 ) ( x + 3 ) ( x + 4 ) = ( x 2 + 5 x + 6 ) ( x + 4 ) = x 3 + 9 x 2 + 26 x + 24 The original volume of bin, V = l e n g t h × w i d t h × h e i g h t = 2 × 3 × 4 = 24 feet 3 It is given that new volume is 5 timed the original volume. So, V ( x ) = 5 × V x 3 + 9 x 2 + 26 x + 24 = 5 × 24 x 3 + 9 x 2 + 26 x + 24 = 120 b. To determine To find: the dimensions of the new bin. Answer 4 feet by 5 feet by 6 feet Explanation Formula Used: Quadratic Formula : The zeros of the polynomial a x 2 + b x + c are given by: x = − b ± b 2 − 4 a c 2 a Calculation: From part (a), the volume of new bin is given by: x 3 + 9 x 2 + 26 x + 24 = 120 Consider, x 3 + 9 x 2 + 26 x + 24 = 120 x 3 + 9 x 2 + 26 x + 24 − 120 = 0 x 3 + 9 x 2 + 26 x − 96 = 0 ( x − 2 ) ( x 2 + 11 x + 48 ) = 0 Now find zeros of x 2 + 11 x + 48 by using quadratic formula. x = − 11 ± 1 − 4 ( 1 ) ( 48 ) 2 ( 1 ) = − 11 ± 1 − 192 2 = − 11 ± − 191 2 = − 11 ± 191 i 2 So, x 2 + 11 x + 48 has no real solution. Thus, x 3 + 9 x 2 + 26 x − 96 = 0 has only one solution x = 2 . Hence, the dimension of new bin: length = x + 2 = 2 + 2 = 4 feet width = x + 3 = 2 + 3 = 5 feet height = x + 4 = 2 + 4 = 6 feet Thus, dimension of new bin: 4 feet by 5 feet by 6 feet

3rd Edition

ISBN: 9781285607160

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 2.5, Problem 113E

a .

To determine

To write: a function that represents the volume V of the new bin.

a .

Expert Solution

Answer to Problem 113E

V(x)=x3+9x2+26x+24=120

Explanation of Solution

Given:

Current dimensions of bin: 2 feet by 3 feet by 4 feet

It is required to increase the volume of the bin 5 times.

Calculation:

Let each dimension is increased by the same amount x.

Then, new dimension of the bin is: (x+2) feet by (x+3) feet by (x+4) feet

Then, new volume of the bin is:

V(x)=length×width×height=(x+2)(x+3)(x+4)=(x2+5x+6)(x+4)=x3+9x2+26x+24

The original volume of bin,

V=length×width×height=2×3×4=24 feet3

It is given that new volume is 5 timed the original volume. So,

V(x)=5×Vx3+9x2+26x+24=5×24x3+9x2+26x+24=120

To determine

To find: the dimensions of the new bin.

Expert Solution

Answer to Problem 113E

4 feet by 5 feet by 6 feet

Explanation of Solution

Formula Used:

Quadratic Formula: The zeros of the polynomial ax2+bx+c are given by:

x=−b±b2−4ac2a

Calculation:

From part (a), the volume of new bin is given by: x3+9x2+26x+24=120

Consider,

x3+9x2+26x+24=120x3+9x2+26x+24−120=0x3+9x2+26x−96=0(x−2)(x2+11x+48)=0

Now find zeros of x2+11x+48 by using quadratic formula.

x=−11±1−4(1)(48)2(1)=−11±1−1922=−11±−1912=−11±191i2

So, x2+11x+48 has no real solution.

Thus, x3+9x2+26x−96=0 has only one solution x=2 .

Hence, the dimension of new bin:

length=x+2=2+2=4 feet

width=x+3=2+3=5 feet

height=x+4=2+4=6 feet

Thus, dimension of new bin: 4 feet by 5 feet by 6 feet

Chapter 2.5, Problem 112E

Chapter 2.5, Problem 114E

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

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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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