Answer:
The volume of the open box is [tex]V = 108\cdot x -42\cdot x^{2}+4\cdot x^{3}\,\,\,[in^{3}][/tex].
Step-by-step explanation:
Geometrically speaking, the volume of the open box ([tex]V[/tex]), measured in cubic inches, is described by the volume of a right parellelepiped. That is:
[tex]V = w\cdot l \cdot h[/tex] (1)
Where [tex]w[/tex], [tex]l[/tex] and [tex]h[/tex] are the width, length and height of the open box, respectively, measured in inches. Please notice that width of the open box is the width of the piece of cardboard ([tex]w'[/tex]), measured in inches, minus two times the side of the cut squares ([tex]x[/tex]), measured in inches, length of the open box is the length of the piece of cardboard ([tex]l'[/tex]), measured in inches, minus two times the side of the cut squares and the height of the open box is the side of the cut squares. That is:
[tex]w = w' -2\cdot x[/tex] (2)
[tex]l = l' - 2\cdot x[/tex] (3)
[tex]h = x[/tex] (4)
By applying (2), (3), (4) in (1), we expand and simplify the volume function:
[tex]V = (w'-2\cdot x)\cdot (l'-2\cdot x)\cdot x[/tex] (5)
If we know that [tex]w' = 9\,in[/tex] and [tex]l' = 12\,in[/tex], then the equation above is:
[tex]V = (9-2\cdot x)\cdot (12-2\cdot x)\cdot x\,\,\,[in^{3}][/tex]
[tex]V = (108-24\cdot x-18\cdot x+4\cdot x^{2})\cdot x\,\,\,[in^{3}][/tex]
[tex]V = (108-42\cdot x +4\cdot x^{2})\cdot x \,\,\,[in^{3}][/tex]
[tex]V = 108\cdot x -42\cdot x^{2}+4\cdot x^{3}\,\,\,[in^{3}][/tex]
The volume of the open box is [tex]V = 108\cdot x -42\cdot x^{2}+4\cdot x^{3}\,\,\,[in^{3}][/tex].
