1. Write the polynomial function that models the given situation. A rectangle has a length of 12 units and a width of 11 units. Squares of x by x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume V of the box as a polynomial function in terms of x. 2. Write the polynomial function that models the given situation. A square has sides of 24 units. Squares x + 1 by x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume V of the box as a function in terms of x.
3. Write the polynomial function that models the given situation. A cylinder has a radius of x + 6 units and a height 3 units more than the radius. Express the volume V of the cylinder as a polynomial function in terms of x.

1.  (12 - 2x)(11 - 2x)x

2. 4(11 - 2x)²(x + 1)

3. π(x³ + 15x² + 63x + 81)

Step-by-step explanation:

1. Write the polynomial function that models the given situation.

A rectangle has a length of 12 units and a width of 11 units. Squares of x by x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume V of the box as a polynomial function in terms of x.

Since the length of the rectangle is 12 units and its width 11 units and squares of x by x units are cut from its corners, it implies that a length x is cut from each side. So, the length of the open box is L = 12 - 2x and its width is w = 11 - 2x.

Since the cut corners of the rectangle are folded, the side x which is cut represents the height of the open box, h. so, h = x

So, the volume of the open box V = LWh = (12 - 2x)(11 - 2x)x

2. Write the polynomial function that models the given situation. A square has sides of 24 units. Squares x + 1 by x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume V of the box as a function in terms of x.

Since the square has sides of 24 units and squares of x + 1 by x + 1 units are cut from its corners, it implies that a length x + 1 is cut from each corner and the length 2(x + 1) is cut from each side. So, the length of side open box is L = 24 - 2(x + 1) = 24 - 2x - 2 = 24 - 2 - 2x = 22 - 2x = 2(11  - x)

Since the cut corners of the square are folded, the side x + 1 which is cut represents the height of the open box, h. so, h = x + 1

Since the area of the base of the pen box is a square, its area is L² = [2(11 - 2x)]²

So, the volume of the open box V = L²h = [2(11 - 2x)]²(x + 1) = 4(11 - 2x)²(x + 1)

3. Write the polynomial function that models the given situation. A cylinder has a radius of x + 6 units and a height 3 units more than the radius. Express the volume V of the cylinder as a polynomial function in terms of x.

The volume of a cylinder is V = πr²h where r = radius and h = height of cylinder.

Given that r = x + 6 and h is 3 units more than r, h = r + 3 = x + 6 + 3 = x + 9

So, V = πr²h

V = π(x + 3)²(x + 9)

V = π(x² + 6x + 9)(x + 9)

V = π(x³ + 6x² + 9x + 9x² + 54x + 81)

V = π(x³ + 15x² + 63x + 81)

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