An open box is to be made from a flat square piece of material 20 inches in length and width by cutting equal squares of length x from the corners and folding up the sides.
(a) Write the volume V of the box as a function of x. Leave it as a product of factors; you do not have to multiply out the factors.
V(x)=
(b) Give the domain of this volume function in interval notation:
base will be 20-2x by 20-2x, and the height will be x
Volume = x(20-2x)(20-2x) or x(20-2x)^2
clearly each of the dimensions must be positive, so 20-2x > 0
-2x > -20
x < 10
In my notation, 0 < x < 10, I will let you change it to yours.
(a) V(x) = x(20-2x)(20-2x)
(b) The domain of this volume function is [0, 10].
(a) The volume V of the box can be found by multiplying the length, width, and height of the box.
Since the length and width of the flat square piece of material are both 20 inches, and we are cutting squares of length x from each corner, the dimensions of the resulting box would be (20-2x) inches by (20-2x) inches by x inches (where 20-2x represents the length and width of the box after folding up the sides).
Therefore, the volume V(x) of the box can be written as:
V(x) = (20-2x) * (20-2x) * x.
(b) The domain of this volume function depends on the physical requirements of the problem.
In this case, the length and width of the box cannot be negative, so we need to ensure that 20-2x is greater than zero.
Furthermore, the height of the box (represented by x) also cannot be negative.
Hence, the domain of the volume function V(x) is:
0 ≤ x ≤ 10 (since 0 ≤ 20-2x ≤ 20 and 0 ≤ x ≤ infinity).
Therefore, the domain of the volume function can be expressed in interval notation as [0, 10].
To find the volume of an open box formed by cutting equal squares from the corners of a flat square piece of material, we need to determine the dimensions of the box and then calculate its volume.
(a) Let's start by visualizing the box. The length and width of the flat square piece of material are both 20 inches. By cutting equal squares of length x from each corner and folding up the sides, the resulting box will have a length of 20 - 2x (since two squares are removed from both ends), a width of 20 - 2x (again, two squares are removed from both ends), and a height of x (as the squares are folded to form the sides).
To calculate the volume V of the box, we multiply its length, width, and height:
V(x) = (20 - 2x) * (20 - 2x) * x
(b) Now let's determine the domain of this volume function. In order for the box to be formed, the length, width, and height all need to be positive values.
The length and width are given by (20 - 2x) and will be positive as long as 20 - 2x > 0. Solving this inequality, we get 20 > 2x, which simplifies to 10 > x.
The height is given by x, which needs to be positive as well. Therefore, the domain of the volume function is restricted by two conditions:
1) x must satisfy 10 > x, which implies x < 10, and
2) x must be positive, so x > 0.
Combining these conditions, the domain of the volume function is given in interval notation as (0, 10).