Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 3 ft by 4 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way.
To find the volume of the largest box that can be formed, we need to determine the dimensions of the box after the squares are cut out and folded.
Let's start by visualizing the rectangular piece of cardboard and the squares cut out of each corner:
```
x x
+-----------------------------------+
| |
| 3 ft |
| |
| 4 ft |
| |
| |
+-----------------------------------+
x x
```
After cutting out the squares, the dimensions of the cardboard will change. The length will be reduced by 2x (since squares are cut out from both ends) and the width will be reduced by 2x as well.
So the dimensions of the piece of cardboard that are left after cutting out the squares are:
Length = 3 ft - 2x
Width = 4 ft - 2x
Now, let's fold the resulting piece of cardboard into a box without a lid. This means the height of the box will be equal to the depth of the original piece of cardboard.
The depth of the box is equal to the side length of the squares that were cut out, which is x.
Now we have all the dimensions we need to calculate the volume of the box. The volume of a rectangular box is given by the formula: Volume = Length * Width * Depth.
Substituting the values in, we have:
Volume = (3 ft - 2x) * (4 ft - 2x) * x
To find the maximum volume, we need to maximize this function. We can do this by taking the derivative with respect to x, setting it equal to zero, and solving for x.
Once we find the value of x that maximizes the volume, we can substitute it back into the volume equation to find the maximum volume.
Let's differentiate the volume equation and solve for x.
dV/dx = (3 - 4x) * (4 - 2x) * x - (3 - 2x) * (4 - 2x) = 0
Simplifying this equation and solving it will give us the value of x that maximizes the volume.
After finding x, we can substitute it back into the volume equation to calculate the maximum volume of the box.
To find the largest volume of the box, we need to determine the size of the squares cut out of the corners. Let's denote the side length of the squares as "x".
When the squares are cut out, the dimensions of the rectangular cardboard remaining are as follows:
Length: 3 ft - 2x
Width: 4 ft - 2x
Height: x ft
The volume of the box is calculated by multiplying these dimensions:
Volume = (3 ft - 2x) * (4 ft - 2x) * x ft
To maximize the volume, we need to find the value of x that maximizes this expression.
Differentiating the volume expression with respect to x:
dV/dx = (4 - 4x) * (3 - 2x) - 2 * (3 - 2x) * (4 - 2x)
Setting the derivative equal to zero to find the critical points:
(4 - 4x) * (3 - 2x) - 2 * (3 - 2x) * (4 - 2x) = 0
Expanding and simplifying:
12 - 8x - 6x + 4x^2 - 8 + 8x - 6x^2 = 0
Combining like terms:
-2x^2 - 2x + 4 = 0
Dividing through by -2:
x^2 + x - 2 = 0
Factoring or using the quadratic formula, we find that:
(x - 1)(x + 2) = 0
So, there are two critical points: x = 1 and x = -2.
Since we can't have a negative side length, we discard x = -2.
Thus, the only critical point is x = 1.
To confirm that this is a maximum, we can take the second derivative of the volume expression:
d^2V/dx^2 = -2 - 2
d^2V/dx^2 = -4
Since the second derivative is negative, this confirms that x = 1 corresponds to the maximum volume.
To find the volume, we substitute x = 1 back into the volume expression:
Volume = (3 ft - 2(1)) * (4 ft - 2(1)) * 1 ft
Volume = (3 ft - 2 ft) * (4 ft - 2 ft) * 1 ft
Volume = 1 ft * 2 ft * 1 ft
Volume = 2 ft^3
Therefore, the largest volume of the box that can be formed is 2 cubic feet.
v = x(3-2x)(4-2x)
dv/dx = 4(3x^2 - 7x + 3)
dv/dx=0 when x = (7-√13)/6 = 0.566
v = 3.03