A manufacturer of open tin boxes wishes to use a piece of tin with dimensions 8 in. by 15 in. by cutting equal squares from the four corners and turning up the sides.
Find a mathematical model expressing the volume of the box as a function of the length of the side of the square cut out.
What is the domain of the function in part A?
since the square of side x is cut from all corners, the volume v is
v = x(8-2x)(15-2x)
as with all polynomials, the domain is all reals.
To find a mathematical model expressing the volume of the box as a function of the length of the side of the square cut out, we need to visualize the box after the corners are cut and the sides are turned up.
Let's assume we cut out a square of side length x from each corner. After folding up the sides, the dimensions of the resulting box will be:
Length = 8 - 2x (since we cut x from both ends)
Width = 15 - 2x (again, x from both ends)
Height = x (this is the height of the sides we turned up)
To calculate the volume, we multiply these dimensions:
Volume = Length * Width * Height
= (8 - 2x) * (15 - 2x) * x
So, the mathematical model expressing the volume of the box as a function of the length of the side of the square cut out (x) is:
V(x) = (8 - 2x) * (15 - 2x) * x
Now, let's consider the domain of the function V(x). The domain refers to the possible values that x can take. In this case, the side length of the square cut out cannot be negative, and it also cannot be greater than the length or width of the original piece of tin.
From the given information, the dimensions of the original piece of tin are 8 in. by 15 in., so the maximum side length of the cut out square should be less than or equal to half of the minimum of those two dimensions.
Therefore, the domain of the function V(x) is the set of all real numbers such that 0 ≤ x ≤ 4 (since the minimum of 8 and 15 is 8, and half of 8 is 4).
So, the domain of the function V(x) is [0, 4].