A rectangular box opens Ata the top is to be formed from a rectangular piece of cardboard which is 3 m * 8m. What size of square should be cut from each corner to form the box with maximum volume.?
Let each side of the cut-out square be x m
Volume of box = (3-2x)(8-2x)(x) = 24x - 22x^2 + 4x^3
dV/dx = .....
set equal to zero and solve for x
To find the size of the square that should be cut from each corner to form the box with maximum volume, we can follow these steps:
Step 1: Visualize the cardboard and determine the dimensions of the box.
The original dimensions of the cardboard are 3m by 8m.
By cutting squares from each corner, the length and width of the resulting box will be reduced.
Let's assume that the size of the squares to be cut is "x" meters.
After cutting the squares, the length of the box will be reduced by 2x (since there are squares cut from both ends),
Similarly, the width of the box will be reduced by 2x as well.
Hence, the length of the box will be (8 - 2x) meters, and the width will be (3 - 2x) meters.
Step 2: Express the volume of the box in terms of "x".
The volume of a rectangular box is given by V = length * width * height.
In this case, we are looking to maximize the volume. The height of the box will be the size of the squares cut from each corner, which is "x".
So, the volume of the box can be expressed as:
V = (8 - 2x) * (3 - 2x) * x
Step 3: Simplify the volume equation.
Expand the expression:
V = (24 - 16x + 4x^2) * x
V = 4x^3 - 16x^2 + 24x
Step 4: Find the critical points.
To find the critical points, take the derivative of the volume equation and set it to zero.
dV/dx = 12x^2 - 32x + 24 = 0
Step 5: Solve for x.
To solve the equation above, we can factorize it or use the quadratic formula. In this case, let's use the quadratic formula.
x = (-(-32) ± √((-32)^2 - 4 * 12 * 24)) / (2 * 12)
x = (32 ± √(1024 - 1152)) / 24
x = (32 ± √(-128)) / 24
Since the value under the square root is negative, there are no real solutions. This means there are no critical points in the interval (0, 4), which means the function is continuous and increasing in this interval.
Step 6: Determine the endpoints.
To find the maximum volume, we need to consider the endpoints of the interval (0, 4).
When x = 0, there are no squares cut, so the volume is 0.
When x = 4, the size of squares cut would be the maximum possible (half of the smaller dimension of the cardboard).
Step 7: Calculate the volumes at the critical points and endpoints.
V(0) = 4(0)^3 - 16(0)^2 + 24(0) = 0
V(4) = 4(4)^3 - 16(4)^2 + 24(4) = 256 - 256 + 96 = 96
Step 8: Compare the volumes and determine the maximum.
Comparing the volumes obtained, we see that V(0) = 0 and V(4) = 96.
Therefore, the maximum volume is 96 cubic meters, which corresponds to cutting squares with sides of 4 meters from each corner.
To find the size of the square that should be cut from each corner to form the box with maximum volume, we need to consider the dimensions of the rectangular piece of cardboard.
Given:
Length of the cardboard = 8m
Width of the cardboard = 3m
Let's assume that a square of side length x is cut from each corner of the cardboard. This means that the length and width of the resulting box would be decreased by 2x.
Length of the box = 8m - 2x
Width of the box = 3m - 2x
Height of the box = x (since the square cut from each corner forms the height of the box)
The volume of the box can be calculated using the formula:
Volume = Length * Width * Height
Substituting the given values, we have:
Volume = (8 - 2x) * (3 - 2x) * x
To find the maximum volume, we need to take the derivative of the volume function with respect to x, set it equal to zero, and solve for x. However, it's important to note that the resulting x-value should lie within the feasible range (0 < x < 1.5) to ensure that the sides of the square cut are not longer than the width or length of the cardboard.
Alternatively, we can use trial and error to find the value of x that maximizes the volume within the given feasible range. We can incrementally test different values of x (e.g., 0.1, 0.2, 0.3, ...) and calculate the corresponding volumes until we find the maximum volume.
Therefore, to find the size of the square that should be cut from each corner to form the box with maximum volume, we need to calculate the volume for various values of x within the feasible range (0 < x < 1.5) and determine the value of x that yields the maximum volume.