Blaze L.

asked • 11/08/17# Story problem in algebra

The length of a toy box is two feet more than the width, and the height is 1 foot less than the width. If the volume of the toy box is 8 cubic feet, what are it’s dimensions?

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## 2 Answers By Expert Tutors

So in general we know that the volume of some box is V = (length)(width)(height). I'm going to call these l, w, and h respectively so that V = lwh. Now we know that the Height is one foot less than the width, in math terms this means

h = w-1 (where I've imagined the word "is" as implying an equals sign). In a similar sense we get from the other statement that l = w + 2. These equations together with the total volume being 8 cubic feet gives

8 = (w+2)w(w-1)

multiplying out and moving all terms to one side of the equation gives:

w^3 + w^2 -2w - 8 = 0

using synthetic division to factor we get:

(w-2)(w^2 + 3x + 4) = 0

next use the quadratic equation to get the possible solutions:

w = 2, w = Imaginary solutions

therefore the width must be 2, then the height is 1 and the length is 4

width=w

length=w+2

height=w-1

V=lwh

V=w(w+2)(w-1)

V=w^3+w^2-2w

8=w^3+w^2-2w

w^3+w^2-2w-8=0

solutions must be factors of 8 (+or -)

2 is a solution

w=2, w-1=2-1=1, w+2=2+2=4

width=2 feet

length=4 feet

height=1 foot

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Kym M.

^{3}=51211/08/17