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.

base will be 12 - 2x by 12-2x, height will be x

V = x(12-2x)^2

range of x: 0 < x < 6

To express the volume V as a function of x, let's break down the problem step by step:

1. Start with a square tin of side length 12 cm.

2. From each corner, remove small squares of side length x. This will leave us with a square base for the open box.

3. The height of the open box will be x, as the removed squares are turned up to form the sides.

4. The dimensions of the base of the open box will be (12 - 2x) cm by (12 - 2x) cm, as the x-length squares are removed from each corner.

5. Therefore, the volume V of the open box can be calculated by multiplying the area of the base by the height: V = (12 - 2x) * (12 - 2x) * x cm^3.

To determine the range of x, we need to consider the limitations imposed by the problem:

1. The side length of the square tin is 12 cm, so the removed squares cannot be larger than half of the side length, i.e., x ≤ 6 cm.

2. Additionally, to form a box, the height cannot exceed the available material. So 0 < x ≤ 12 cm.

Combining these conditions, the range of x is 0 < x ≤ 6 cm.

To find the volume of the open box formed by removing squares of side x from each corner of the square tin, we need to calculate the total volume of the original square tin and subtract the volume of the removed squares.

The original square tin has sides measuring 12 cm, so its volume is:
V_original = (12 cm) * (12 cm) * (12 cm) = 1728 cm^3

The removed squares from the corners form the bottom and the four sides of the box. Each removed square has side length x, so the volume of one removed square is:
V_removed = x * x * x = x^3

Since we are removing four squares from the corners of the original square tin, the total volume of the removed squares is:
V_total_removed = 4 * V_removed = 4 * x^3

Therefore, the volume V(x) of the open box is given by subtracting the total volume of the removed squares from the original volume of the square tin:
V(x) = V_original - V_total_removed = 1728 cm^3 - 4 * x^3

The range of x depends on the dimensions of the original square tin. In this case, the side length of the original square tin is 12 cm. Since we are removing squares from each corner, the maximum possible value for x is half of the side length of the square tin:
x ≤ 12 cm / 2 = 6 cm

Therefore, the range of x is 0 ≤ x ≤ 6 cm.