AP Calculus AB Exam Question

An open box is to be made from a rectangular piece of cardboard measuring 12 inches by 20 inches. The box is formed by cutting squares with sides of length x from each corner of the cardboard and bending up the sides.

1. Write an equation for the volume, V, of the box in terms of x.
2. Determine the domain of x that makes sense in the context of this problem.
3. Find the dimensions of the box that result in the maximum possible volume.
4. Calculate the maximum volume of the box.

Provide step-by-step explanation for each part above.

Answer

1. Writing the equation for the volume, V, of the box in terms of x:

Let the width of the box be W and the height of the box be H.

By cutting squares with sides of length x from each corner of the cardboard and bending up the sides, the width of the resulting box will be reduced by 2x (x from each side), and the height will be reduced by 2x as well.

Therefore, the width and height of the box will be: Width = 20 - 2x inches Height = 12 - 2x inches

The length of the box, L, will be x inches.

The volume, V, of the box can be calculated by multiplying the width, length, and height: V = Width * Length * Height V = (20 - 2x) * x * (12 - 2x)

2. Determining the domain of x:

In the context of this problem, x represents the length of the square cut from each corner.

To ensure that the resulting dimensions of the box make sense, the following conditions must be met:

• The length of the square cut from each corner cannot exceed half the length or width of the cardboard: x ≤ 6 (half of 12) and x ≤ 10 (half of 20).
• The resulting width and height of the box after cutting cannot be negative: 20 - 2x > 0 and 12 - 2x > 0.

Combining these conditions, the domain of x that makes sense is: 0 < x ≤ 6.

3. Finding the dimensions of the box that result in the maximum volume:

To find the dimensions of the box that result in the maximum volume, we need to find the critical points of the volume function V(x) = (20 - 2x) * x * (12 - 2x) within the domain 0 < x ≤ 6.

Taking the derivative of V(x) with respect to x:

V'(x) = -8x^2 + 64x - 240

Setting V'(x) = 0 to find critical points:

-8x^2 + 64x - 240 = 0

Dividing by -8: x^2 - 8x + 30 = 0

Using the quadratic formula: x = (8 ± √[64 - 4(1)(30)]) / 2 x = (8 ± √[64 - 120]) / 2 x = (8 ± √[-56]) / 2

Since the discriminant is negative (√[-56]), there are no real solutions to this equation. Therefore, there are no critical points within the domain 0 < x ≤ 6.

4. Calculating the maximum volume of the box:

Since there are no critical points within the given domain, we need to consider the endpoints.

When x = 0, the volume V = (20 - 2(0))(0)(12 - 2(0)) = 0, which is the minimum possible volume.

When x = 6, the volume V = (20 - 2(6))(6)(12 - 2(6)) = 96 cubic inches.

Therefore, the maximum volume of the box is 96 cubic inches when x = 6.

In summary:

• The equation for the volume, V, of the box in terms of x is V = (20 - 2x) * x * (12 - 2x).
• The domain of x that makes sense is 0 < x ≤ 6.
• The dimensions of the box that result in the maximum possible volume are: width = 20 - 2(6) = 8 inches, length = 6 inches, and height = 12 - 2(6) = 0 inches.
• The maximum volume of the box is 96 cubic inches.

Note: While the height of the box turning out to be 0 may seem counterintuitive, it occurs because the squares cut from the corner result in no vertical sides when their length is equal to half the side length of the cardboard.