a piece of cardboard is twice as it is wide. It is to be made into a box with an open top by cutting 2-in squares from each corner and folding up the sides. Let x represent the width (in inches) of the original piece of cardboard.

a.Represent the length of the original piece of cardboard in terms of x
b.What will be the dimension of the bottom rectangular base of the box? Give the restrictions on x.
c. Determine function V that represents the volume of the box in terms of x.
d. For what dimension of the bottom of the box will the volume be 320in?
e. Find the values of x (to the nearest tenth of an inch) if such a box is to have a volume between 400 and 500 in.

a. To represent the length of the original piece of cardboard in terms of x, we can use the given information that the length is twice the width. Therefore, the length would be 2 times x.

b. To find the dimensions of the bottom rectangular base of the box, we need to consider the effect of cutting 2-inch squares from each corner. Each side of the base will have the original width reduced by 4 inches (2 inches from each end). So the width of the base would be x - 4 inches. The length would be the original length reduced by 4 inches on each end as well, resulting in a length of (2x - 4) inches. Therefore, the dimensions of the bottom rectangular base of the box would be (x - 4) inches by (2x - 4) inches.

To find the restrictions on x, we need to ensure that the resulting dimensions of the base are positive. This means x - 4 > 0 and 2x - 4 > 0. Solving these inequalities, we get x > 4 and x > 2, respectively. Therefore, the restrictions on x are x > 4 and x > 2 (or simply x > 4, since x > 2 is already implied by x > 4).

c. The volume of the box can be found by multiplying the dimensions of the rectangular base (x - 4) inches by (2x - 4) inches, and multiplying the result by the height of the box (which is 2 inches, since we cut squares with side length 2 inches from each corner). Therefore, the volume V, in terms of x, is given by:

V = (x - 4)(2x - 4)(2)

d. To find the dimension of the bottom of the box for a volume of 320 cubic inches, we need to set up an equation using the volume function from part c.

320 = (x - 4)(2x - 4)(2)

Then, we can solve this equation for x using algebraic methods to find the corresponding dimension of the bottom of the box.

e. To find the values of x (to the nearest tenth of an inch) for a box with a volume between 400 and 500 cubic inches, we need to set up a similar inequality using the volume function from part c.

400 ≤ (x - 4)(2x - 4)(2) ≤ 500

We can then solve this inequality for x to find the valid range of dimensions for the box.