We first recall the formula for the volume of a rectangular box:
[tex]V=l*w*h[/tex] where l is its length, w is the width, and h is the height.
According to the problem, the height is 3 inches less than the width therefore we can replace h with the term [tex]w-3[/tex]. Additionally, the length of the box is 2 inches more than twice the width therefore the length can also be replaced with [tex]2w+2[/tex].
We can notice that the equation will just have one unknown variable because we already know the volume of the box. We can then solve for this variable (w):
[tex]1540=(2w+2)w(w-3)[/tex]
[tex]1540=(2w+2)(w^{2}-3w)[/tex]
[tex]1540=2w^{3}-6w^{2}+2w^{2}-6w[/tex]
[tex]0=2w^{3}-4w^{2}-6w-1540[/tex]
Solving the cubic equation we'll get [tex]w=10[/tex] and two complex number solutions. We'll just need to solve for the length and height using the value of the width:
[tex]l=2w+2=2(10)+2=22[/tex]
[tex]h=w-3=10-3=7[/tex]
ANSWER:Â The width of the box is 10 inches; its length is 22 inches, and its height is 7 inches.
