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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 ft^3-example-1
User Tardomatic
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1 Answer

7 votes

the width of the box is 1.21 ft

Step-by-step 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


\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}

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

w = 1.207206


\begin{gathered} To\text{ check:} \\ \text{volume = 2(}1.207206)\text{ }*\text{ }1.207206\text{ }*\text{ (6 + }1.207206) \\ \text{volume = 20.999 }\approx\text{ 21} \end{gathered}

To two decimal place, w = 1.21

Hence, the width of the box is 1.21 ft

User Asgerhallas
by
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