An open box is to be made from a 21 ft by 56 ft rectangular piece of sheet metal by cutting out squares of equal size from the four corners and bending up the sides. Find the maximum volume that the box can have.

To find the maximum volume of the box, we need to determine the size of the square that will maximize the volume.

Let's assume that the side length of each square cut from the corners is x ft.

The dimensions of the resulting box will be (21-2x) ft by (56-2x) ft by x ft.

The volume of the box is given by the expression V = (21-2x)(56-2x)x.

To find the maximum volume, we need to find the value of x that maximizes this expression.

We can start by finding the critical points of V by taking the derivative.

dV/dx = (56-2x)(21-2x) + (21-2x)(-2x) + (56-2x)(-2x)

Expanding the equation, we get:

dV/dx = 4x^3 - 154x^2 + 1176x

To find the critical points, we set the derivative equal to zero and solve for x:

4x^3 - 154x^2 + 1176x = 0

We can factor out 4x to simplify the equation:

4x(x^2 - 38.5x + 294) = 0

Setting each factor equal to zero, we have:

4x = 0,
x^2 - 38.5x + 294 = 0

From the first equation, we find x = 0. However, this does not make sense in the context of our problem since a box with zero size does not exist.

We can solve the second equation using the quadratic formula:

x = (-(-38.5) ± √((-38.5)^2 - 4*1*294)) / (2*1)

Simplifying the expression inside the square root, we have:

x = (38.5 ± √(1482.25 - 1176)) / 2

x = (38.5 ± √(306.25)) / 2

x ≈ (38.5 ± 17.497) / 2

This gives us two possible values for x:

x ≈ (38.5 + 17.497) / 2 ≈ 28.998
or
x ≈ (38.5 - 17.497) / 2 ≈ 10.501

Since the size of the square cannot be negative, we can discard the second value.

Thus, the side length of the square cut from the corners should be approximately 28.998 ft.

To find the maximum volume, we substitute this value back into the expression for V:

V = (21 - 2*28.998)(56 - 2*28.998)(28.998)

V ≈ 29 ft * 22 ft * 28.998 ft

V ≈ 18243.72 ft^3

Therefore, the maximum volume that the box can have is approximately 18243.72 ft^3.

To find the maximum volume of the box, we need to determine the dimensions of the box that will maximize the volume.

Let's assume that the size of each square cut from the corners is "x" ft.

When we cut out squares from each corner and fold up the sides, the dimensions of the resulting box are as follows:

Length = (56 - 2x) ft
Width = (21 - 2x) ft
Height = x ft

Thus, the volume of the box can be calculated by multiplying the length, width, and height together:

Volume = (56 - 2x)(21 - 2x)(x) cubic ft

To find the maximum volume, we need to find the value of x that maximizes the volume function.

Taking the derivative of the volume function with respect to x, setting it equal to zero, and solving for x will give us the optimal value for x:

dV/dx = (21 - 2x)(56 - 2x) + (56 - 2x)(x) + (21 - 2x)(x) = 0

Simplifying the equation yields:

(21 - 2x)(56 - 2x) + 56x - 2x^2 + 21x - 2x^2 = 0

Expanding and combining like terms:

(1176 - 114x + 4x^2) + 77x - 4x^2 = 0

1176 - 37x = 0

Solving for x:

37x = 1176
x = 1176/37
x ≈ 31.78 ft

Since we cannot have a fraction of a foot for the length or width, we need to choose the closest whole number for x. In this case, we will choose x = 31 ft.

Now that we have the value of x, we can substitute it back into the original volume formula to find the maximum volume of the box:

Volume = (56 - 2(31))(21 - 2(31))(31) cubic ft

Volume ≈ 25740 cubic ft

Therefore, the maximum volume that the box can have is approximately 25740 cubic ft.

Let x be the length of the side of the square taken off each corner. The open box will have length 56-2x, width 21-2x and height x.

The volume of the box is
V(x) = x(56-2x)(21-2x)

Set the derivative dV/dx equal to zero and compute the corresponding x. Use that to compute the maximum volume.

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