A cardboard box manufacturer makes open boxes from rectangular pieces of cardboard of size 30cm by 40cm by cutting squares from the four corners and turning up the sides.
A) find a mathematical model expressing the volume of the box as a function of the length of the side of the square to be cut out.
B) what is the domain of your function in part a?
let the length of the side to be cut out be x cm
length = 40-2x
width = 30-2x
height = x
Volume = x(40-2x)(30-2x)
b) clearly, all of the factors must be positive number
so 30-2x>0
-2x > -30
x < 15 and of course x > 0
so 0 < x < 15
A) Well, isn't this box manufacturer just putting a little geometry twist on things? Let's break it down. By cutting squares from the corners of a 30cm by 40cm piece of cardboard, we effectively reduce the length, width, and height of the box by twice the side length of the square cut out.
So, if we let "x" represent the length of the side of the square cut out, the length of the box will be (30 - 2x) cm, the width will be (40 - 2x) cm, and the height (since it's the turned-up sides) will be x cm. The volume of the box is given by the product of these three dimensions:
Volume = (30 - 2x) * (40 - 2x) * x
B) Now, let's talk about the domain of this function. Since we're cutting squares out of the cardboard, the side length of the square cannot be greater than the dimensions of the cardboard itself. In other words, x ≤ 30cm and x ≤ 40cm. We also have to consider that the square must have a positive length, so x > 0.
Taking all these into account, the domain of the function is 0 < x ≤ 30 and 0 < x ≤ 40.
A) To find the mathematical model expressing the volume of the box as a function of the length of the side of the square to be cut out, let's call the side length of the square to be cut out "x".
When the square is cut out from each corner, the dimensions of the resulting box will be (30 - 2x) cm by (40 - 2x) cm by x cm.
Therefore, the volume of the box can be calculated as the product of these three dimensions:
Volume = (30 - 2x)(40 - 2x)(x).
B) The domain of the function can be determined by considering the physical constraints of the problem. Since the length of the side of the square to be cut out cannot exceed the smallest side of the original piece of cardboard, we have the following restrictions:
0 ≤ x ≤ min(30, 40),
since both 30 and 40 represent the dimensions of the original piece of cardboard.
Thus, the domain of the function is 0 ≤ x ≤ 30.
A) To find a mathematical model expressing the volume of the box as a function of the length of the side of the square to be cut out, we need to determine the dimensions of the resulting box after cutting and folding the cardboard.
Let's denote the side length of the square to be cut out as "x". When we cut out the square from each corner, the length of the resulting sides of the box will be reduced by 2x (since we cut out two squares from each dimension). Thus, the dimensions of the resulting box will be:
Length: 30cm - 2x
Width: 40cm - 2x
Height: x (as the squares will be folded up to form the box)
The volume of the box can be found by multiplying these dimensions together:
Volume = (30cm - 2x) * (40cm - 2x) * x
Therefore, the mathematical model expressing the volume of the box as a function of the length of the side of the square to be cut out is:
V(x) = (30 - 2x)(40 - 2x)x
B) The domain of the function V(x) in part A represents the allowable values for the side length of the square to be cut out. In this case, we need to consider that both the length and width of the cut-out square cannot exceed half the dimensions of the original cardboard piece.
Since the original dimensions of the cardboard are 30cm by 40cm, the maximum length of the cut-out square (x) should not exceed half of the minimum dimension:
x ≤ min(30cm, 40cm)/2
x ≤ 15cm
Therefore, the domain of the function is x ≤ 15cm.