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The problem is in the picture volume of box is 21 ft^3

The Problem Is In The Picture Volume Of Box Is 21 Ft3 class=

Sagot :

the width of the box is 1.21 ft

Explanation:

length of the box = 2 times the width

let the width = w

length = 2(w) = 2w

height = 6ft more than its width

height = 6ft + w = 6 + w

Volume of the rectangular box = length × width × height

Volume of the rectangular box = 2w × w × (6 + w)

Volume of the box is given as = 21ft³

21 = 2w × w × (6 + w)

21 = 2w²(6 + w)

21 = 12w² + 2w³

[tex]\begin{gathered} 2w^3+12w^2\text{ - 21 = 0} \\ U\sin g\text{ cubic formula:} \\ \text{values of w =}-5.673836,-1.53319,1.207026 \end{gathered}[/tex]

But since we can't have width as a negative number, w will be the positive number

w = 1.207206

[tex]\begin{gathered} To\text{ check:} \\ \text{volume = 2(}1.207206)\text{ }\times\text{ }1.207206\text{ }\times\text{ (6 + }1.207206) \\ \text{volume = 20.999 }\approx\text{ 21} \end{gathered}[/tex]

To two decimal place, w = 1.21

Hence, the width of the box is 1.21 ft

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