a rectangular sheet of cardboard 4m by 2m is used to make an open box by cutting squares of equal size from the four corners and folding up the sides.what size squares should be cut to obtain the largest possible volume?
let each cut of be x m
so the base is 4-2x by 2-2x
and the height is x
V = x(4-2x)(2-2x)
= 8x -12x^2 + 4x^3
d(V)/dx = 8 - 24x + 12x^2 = 0 for a max V
3x^2 - 6x + 2 = 0
x = (6 ± √12)/6
= .42 or 1.57 , but the latter would make the width negative so we will reject it
the cut-out squares should each be .42m by .42 m
or 42 by 42 cm
Well, to find the largest possible volume, we need to think outside the box, or in this case, inside the box. We can start by imagining the dimensions of the squares cut from the corners as "x". So, when you fold up the sides, the dimensions of the resulting box would be (4-2x) by (2-x), with a height of "x".
The volume of the box is given by length x width x height. Substituting the values, we have:
Volume = (4-2x)(2-x)x
Now, let's solve this equation by finding the derivative and setting it equal to zero to find the critical points:
dV/dx = -6x^2 + 20x - 8
Setting dV/dx equal to zero:
0 = -6x^2 + 20x - 8
Solving this equation leads us to the delightful answer that x = 2/3. However, since we're dealing with squares and not fractions, let's round it up to the nearest whole number and make x = 1.
So, the largest possible volume will be obtained by cutting squares of size 1m from each corner.
To find the size of the squares that should be cut to obtain the largest possible volume, we can use the concept of maximizing the volume function.
Let's denote the length of the side of the square to be cut as "x". When the squares are cut and the sides are folded up, the dimensions of the resulting box will be (4-2x) by (2-2x) by x.
The volume of the box can be calculated by multiplying these dimensions:
Volume = (4-2x)(2-2x)x
Now, let's simplify this equation:
Volume = (8 - 4x - 4x + 4x^2)x
= (8x - 8x^2)x
= 8x^2 - 8x^3
To find the maximum volume, we need to find the critical point of this function by taking the derivative and setting it equal to zero:
dV/dx = 16x - 24x^2
Setting dV/dx = 0:
16x - 24x^2 = 0
8x(2 - 3x) = 0
This equation has two solutions: x = 0 and x = 2/3. However, since cutting squares of size 0 would result in no box, we can ignore the x = 0 solution.
Thus, the size of the squares that should be cut to obtain the largest possible volume is x = 2/3.
Note: To verify that this is indeed the maximum volume, you can also check the second derivative of the volume function. If the second derivative is negative at the critical point, it confirms that the point is indeed a maximum.
To find the size of the squares that should be cut to obtain the largest possible volume, we need to analyze the problem step by step.
Let's consider a rectangular sheet of cardboard with dimensions 4m by 2m. To create an open box, we need to cut squares from the four corners and fold up the sides. Let's assume the size of the squares cut from each corner is "x".
When we fold up the sides, the length of the box will be reduced by 2x (as there are two sides to fold up), while the width will be reduced by 2x as well. The height of the box will be equal to x.
Therefore, the dimensions of the box will be:
Length = (4 - 2x) meters
Width = (2 - 2x) meters
Height = x meters
To calculate the volume of the box, we multiply these dimensions:
Volume = Length * Width * Height
V = (4 - 2x)(2 - 2x)x
To find the value of x that gives us the maximum volume, we can take the derivative of the volume equation with respect to x and set it equal to zero. This will help us find the critical points where the volume is either a maximum or minimum.
dV/dx = 0
Simplifying the equation and solving for x:
dV/dx = 4(2 - 2x) + 2(4 - 2x)(-2) = 0
8 - 8x - 16 + 4x = 0
-4x - 8 = 0
-4x = 8
x = 8 / -4
x = -2
Since x represents the size of a side, it cannot be negative. Hence, we discard this solution.
Finally, we need to analyze the endpoints of the interval to make sure we maximize the volume. In this case, the interval is [0, 1] because the maximum value for x should be less than or equal to half the width or length of the cardboard.
We evaluate the volume at the endpoints:
V(0) = (4 - 2(0))(2 - 2(0))(0) = 0
V(1) = (4 - 2(1))(2 - 2(1))(1) = 2*(2-2)(1) = 0
As we can see, at both endpoints, the volume is zero. This indicates that the maximum volume occurs somewhere within the interval (0, 1).
Therefore, to obtain the largest possible volume, we cut squares of equal size from the four corners with a dimension of x = 1 meter.