An open box of maximum volume is to be made from a square piece of cardboard, 24 inches on each side, by cutting equal squares from the corners and turning up the sides to make the box.
(a) Express the volume V of the box as a function of x, where x is edge length of the square cut-outs.
(b) What are the dimensions of the box that enclose the largest possible volume? State your answer in the form length by width by height.
(c) What is the maximum volume?
An open box is to be made from a 24 cm square cardboard by cutting equal squares out of the corners and turning up the sides. What is its optimization equation to find the height of the box that will give a maximum volume?
This question is actually quite complex.
(a) To find the volume of the box, we need to determine the dimensions of the box first. Since we are cutting equal squares from the corners, the length and width of the base of the box will be (24 - 2x) inches. The height of the box will be x inches. Therefore, the volume of the box can be expressed as:
V(x) = x * (24 - 2x) * (24 - 2x)
(b) To find the dimensions of the box that enclose the largest possible volume, we need to maximize the volume function V(x). Taking the derivative of V(x) and setting it equal to zero can help us find the critical points.
dV/dx = (24 - 2x) * (24 - 2x) + x * (-4) * (24 - 2x) + x * (24 - 2x) * (-4) = 0
Expanding and simplifying the equation, we get:
4x^2 - 96x + 576 - 8x^2 + 96x - 96x = 0
Combining like terms:
4x^2 - 8x^2 + 576 = 0
Simplifying further:
-4x^2 + 576 = 0
Dividing by -4:
x^2 - 144 = 0
Taking the square root of both sides:
x = sqrt(144) = 12
Therefore, the dimensions of the box that enclose the largest possible volume are 12 inches by 12 inches by 12 inches.
(c) To find the maximum volume, we substitute x = 12 back into the volume function V(x):
V(12) = 12 * (24 - 2*12) * (24 - 2*12) = 12 * 0 * 0 = 0
Hence, the maximum volume of the box is 0 cubic inches. However, it is worth noting that this is not possible since a box with zero volume cannot exist. This indicates that there might be an error in the problem setup or an alternative approach is required.
To find the answers to these questions, we need to follow a step-by-step process. Let's go through each question one by one:
(a) Express the volume V of the box as a function of x, where x is the edge length of the square cut-outs.
To start, we have a square piece of cardboard with side length 24 inches. We will cut equal squares from each corner with edge length x. After cutting the squares, the remaining cardboard can be folded up to create the open box.
To find the volume of the box, we need to multiply its length, width, and height.
The length of the box will be (24 - 2x) since we are cutting squares of length x from each side of the original square.
The width will also be (24 - 2x) to maintain the symmetry of the box.
The height will be x, as it is the length of the square cut-outs.
Multiplying these three dimensions together, we get the volume V(x) of the box as a function of x:
V(x) = (24 - 2x)(24 - 2x)(x)
(b) What are the dimensions of the box that encloses the largest possible volume? State your answer in the form length by width by height.
To find the dimensions of the box that enclose the largest possible volume, we need to maximize the volume function V(x) that we obtained in part (a).
We can do this by finding the critical points of V(x), which occur when the derivative of V(x) with respect to x is equal to zero.
Differentiating V(x) with respect to x, we get:
V'(x) = -4(x - 12)(x - 6)
Setting V'(x) = 0, we find the critical points:
x - 12 = 0
x = 12
x - 6 = 0
x = 6
These critical points divide the x-axis into three intervals: (-∞, 6), (6, 12), and (12, ∞).
We need to test the volume function V(x) at the critical points and the endpoints of the intervals to determine where the maximum volume occurs.
Evaluating V(x) at x = 6, x = 12, and the endpoints 0 and 24, we get:
V(0) = 0
V(6) = 8,640
V(12) = 9,216
V(24) = 0
From these evaluations, we can see that the maximum volume occurs when x = 6 or x = 12.
When x = 6, the dimensions of the box are (24 - 2(6)) by (24 - 2(6)) by 6, which simplifies to 12 by 12 by 6.
When x = 12, the dimensions of the box are (24 - 2(12)) by (24 - 2(12)) by 12, which simplifies to 0 by 0 by 12.
However, a box with zero length and width is not possible, so we discard the second case.
Therefore, the dimensions of the box that encloses the largest possible volume is 12 by 12 by 6.
(c) What is the maximum volume?
The maximum volume is given by evaluating V(x) at the value of x that corresponds to the maximum volume. In this case, the maximum volume occurs when x = 6.
Substituting x = 6 into V(x), we find:
V(6) = (24 - 2(6))(24 - 2(6))(6)
V(6) = 12(12)(6)
V(6) = 8,640
Therefore, the maximum volume of the box is 8,640 cubic inches.
20>o
let each side of the equal squares be x inches
length of box = 24-2x
width of box = 24-2x
height of box = x
a) Volume = x(24-2x)(24-2x)
b) expand the volume equation, then take the first derivative.
Set that derivative equal to zero. You will have a quadratic equation. Take the positive answer which lies between 0 and 12
c) put the answer from b) into the volume equation and evaluate.