A manufacturer is building a box with no top. The box is constructed from a piece of cardboard that is 10 inches long and 15 inches wide by cutting out the corners and folding up the sides. The manufacturer wants to build a box with maximum volume. What dimensions maximize the volume of the box, and what is that volume?
A) The maximum volume occurs when the box has dimensions of length 9 inches, width 4 inches, and height 3 inches. The maximum volume of the box is 54 cubic inches.
B) The maximum volume occurs when the box has dimensions of length 9 inches, width 4 inches, and height 3 inches. The maximum volume of the box is 108 cubic inches.
C) The maximum volume occurs when the box has dimensions of length 11 inches, width 6 inches, and height 2 inches. The maximum volume of the box is 132 cubic inches.
D) The maximum volume occurs when the box has dimensions of length 11 inches, width 6 inches, and height 2 inches. The maximum volume of the box is 66 cubic inches.
To find the dimensions that maximize the volume of the box, we need to maximize the volume function V(x), where x denotes the length of the cut-out squares from each corner.
The length of the resulting box would be (10 - 2x) inches, the width would be (15 - 2x) inches, and the height would be x inches.
The volume function V(x) is given by:
V(x) = (10 - 2x)(15 - 2x)x
To find the maximum volume, we can take the derivative of V(x) with respect to x, set it equal to zero, and solve for x.
V'(x) = (15 - 2x)(10 - 2x) + x(-4) + (-2x)(10 - 2x)
Setting V'(x) equal to zero:
(15 - 2x)(10 - 2x) - 4x - 2x(10 - 2x) = 0
(150 - 20x - 30x + 4x^2) - 4x - (20x - 4x^2) = 0
150 - 20x - 30x + 4x^2 - 4x - 20x + 4x^2 = 0
8x^2 - 54x + 150 = 0
We can solve this quadratic equation by factoring:
2x^2 - 27x + 75 = 0
(x - 3)(2x - 25) = 0
Setting each factor equal to zero and solving for x:
x - 3 = 0 or 2x - 25 = 0
x = 3 or x = 25/2
However, since the length of the cardboard is 10 inches, the maximum value for x will be limited to 5 (half of the length of the cardboard). Therefore, we can discard x = 25/2 as a solution.
The dimensions that maximize the volume of the box are:
Length = 10 - 2x = 10 - 2(3) = 10 - 6 = 4 inches
Width = 15 - 2x = 15 - 2(3) = 15 - 6 = 9 inches
Height = x = 3 inches
The maximum volume of the box is:
V(x) = (10 - 2x)(15 - 2x)x = (10 - 2(3))(15 - 2(3))(3) = (4)(9)(3) = 108 cubic inches.
Therefore, the correct answer is:
B) The maximum volume occurs when the box has dimensions of length 9 inches, width 4 inches, and height 3 inches. The maximum volume of the box is 108 cubic inches.