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 can apply the principles of optimization by analyzing the given information and setting up mathematical equations.
Let's assume that the length of the box is x inches and the width of the box is y inches.
We are told that the piece of cardboard used to construct the box is 10 inches long and 15 inches wide. This means that after cutting out the corners, the remaining side length will be:
x = 10 - 2 * c (where c represents the length of each cutout corner)
Similarly, the remaining width will be:
y = 15 - 2 * c
The height of the box will be the height of the cutout corners when the sides are folded up.
To maximize the volume of the box, we need to maximize the product of its dimensions.
Volume (V) = length (x) * width (y) * height (h)
V = (10 - 2c)(15 - 2c)(c)
To find the maximum volume, we can take the derivative of V with respect to c and set it equal to 0:
V'(c) = 4c³ - 50c² + 150c - 150 = 0
By solving this cubic equation, we can find the critical points. We can do this either by factoring, using synthetic division, or numerical methods.
Using numerical methods, we find that one of the critical points is c ≈ 3.318. We can discard the negative values, as they don't make physical sense in this context.
Substituting this value back into the original equation for V, we can determine the maximum volume (Vmax) and the corresponding dimensions.
Vmax ≈ (10 - 2 * 3.318)(15 - 2 * 3.318)(3.318)
Vmax ≈ 9.1414 * 8.3632 * 3.318
Vmax ≈ 242.2227 cubic inches
The dimensions of the box that maximize the volume are:
Length ≈ 10 - 2 * 3.318 ≈ 3.364 inches
Width ≈ 15 - 2 * 3.318 ≈ 8.364 inches
Height ≈ 3.318 inches
Therefore, the correct answer is not provided in the answer choices.
is that correct?
Apologies for the incorrect previous answer.
Upon reviewing the calculations, it appears there was an error. Let's reevaluate and find the correct dimensions and volume:
We are given that the length of the cardboard is 10 inches and the width is 15 inches after the corners are cut.
Let's assume that the cutout length from each corner is x inches. This means the remaining length of the box will be (10 - 2x) inches, and the remaining width will be (15 - 2x) inches.
The height of the box will be x inches since this is the portion of the cardboard folded up to form the sides.
The volume of the box is given by the equation:
V = (10 - 2x)(15 - 2x)(x)
To find the dimensions that maximize the volume, we can take the derivative of V with respect to x and set it equal to 0:
dV/dx = -8x^3 + 50x^2 - 75x = 0
We can factor out an x from this equation:
x(-8x^2 + 50x - 75) = 0
Thus, we have the critical points at x = 0 and by solving -8x^2 + 50x - 75 = 0, we find x ≈ 1.41 and x ≈ 6.09. We discard x = 0 since it does not make sense in the context of a cutout.
Substituting the approximate values of x back into the equation for V, we can find the maximum volume:
V ≈ (10 - 2 * 1.41)(15 - 2 * 1.41)(1.41)
V ≈ 9.18 * 12.18 * 1.41
V ≈ 162.53 cubic inches
Therefore, the correct answer is:
The maximum volume occurs when the box has dimensions of length ≈ 7.18 inches, width ≈ 12.18 inches, and height ≈ 1.41 inches. The maximum volume of the box is ≈ 162.53 cubic inches.
Apologies for any confusion caused by the earlier incorrect response.