Question

a box with a lid is to be made from a rectangular piece of cardboard 11 cm by 16 cm, as shown in the figure below. two equal squares of side are to be removed from one end, and two equal rectangles are to be removed from the other end so that the tabs can be folded to form the box with a lid. find such that the volume of the box is a maximum. the equation that gives the volume of the box in terms of the length of is: <--------- 16 cm ------->11 cm ||||||| what is the domain of the function in context of the problem? length of the equal squares that are removed: select an answer . volume of the box:

Answer 1

A box with zero dimensions would not be a box, so the practical domain of V would be considered to be 0 < V < 0.

To determine the domain of the function representing the volume of the box, we need to consider the constraints imposed by the dimensions of the cardboard piece.

Domain of the square side length (x):

The side length of the squares cannot exceed half the length of the cardboard, as this would leave no material for the base of the box. Therefore, x ≤ 11/2.

The side length of the squares cannot be negative. Therefore, x ≥ 0.

Combining these two constraints, the domain of x is:

0 ≤ x ≤ 11/2

Domain of the box volume (V):

The volume of the box depends on the side length of the squares (x). Since the domain of x is 0 ≤ x ≤ 11/2, we can substitute these values into the volume formula to determine the corresponding volume range.

For x = 0, the volume V = 0.

For x = 11/2, the volume V = 0.

Therefore, the domain of the box volume (V) is:

0 ≤ V ≤ 0

It's important to note that the volume of the box cannot be negative, and it can only be zero when the side length of the squares is zero. In reality, a box with zero dimensions would not be a box, so the practical domain of V would be considered to be:

0 < V < 0