From each corner of a square of tin, 12 cm on a side, small squares of side x are removed and the edges are turned up to form an open box. Express the volume V
(cm3) as a function of x. Determine the range of x.
see related questions below.
To find the volume of the open box, we need to subtract the volume of the cut-out squares from the original square.
The volume of the original square is given by:
V_original = (side length)^2 = (12 cm)^2 = 144 cm^2
To find the volume of the cut-out squares, we need to calculate the individual volumes of each square and then subtract them from the original square.
Let's consider one corner of the square. We can remove a square with side length x from each corner. Therefore, the total volume of the cut-out squares is given by:
V_cut-out = 4 * (side length of each cut-out square)^2 = 4 * x^2
Now, we can determine the volume of the open box:
V = V_original - V_cut-out
= 144 cm^2 - 4x^2 cm^2
= 144 - 4x^2 cm^2
The range of x would typically be limited by the size of the original square. Since each cut-out square has a side length of x, the maximum value for x would be half the side length of the square, which is 6 cm in this case. Therefore, the range for x would be 0 ≤ x ≤ 6 cm.
To find the volume of the open box, we need to calculate the area of the base and the height.
The base area is formed by the sides of the square after the small squares are removed.
Since the original square is 12 cm on each side, when we remove a small square with side length x from each corner, the remaining side length of the base will be (12 - 2x) cm.
So, the base area is (12 - 2x) * (12 - 2x).
The height of the box will be the side length of the small squares that we cut out, which is x cm.
Therefore, the volume of the open box, V, is given by multiplying the base area by the height:
V = (12 - 2x) * (12 - 2x) * x
= (144 - 48x - 48x + 4x^2) * x
= (144 - 96x + 4x^2) * x
= 4x^3 - 96x^2 + 144x
To determine the range of x, we need to consider the restrictions of the problem. Firstly, x should be greater than zero since we cannot remove negative squares. Secondly, the side length of the small squares should be less than half of the side length of the original square (12 cm) in order for them to fit inside the corners. Therefore, x should be less than 6 cm.
So, the range of x is: 0 < x < 6 cm.