Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 37ft by 20 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.

b) Suppose that in part a the original piece of cardboard is a square with side lengths k. Find the volume of the largest box that can be formed in this way.

c)Suppose that in part (a) the original piece of cardboard is a rectangle with sides of length k and L. Holding k fixed, find the size of the corner squares x that maximizes the volume of the box as W→∞.

a) To find the volume of the largest box that can be formed, we need to determine the dimensions of the folded piece of cardboard.

First, we need to find the length and width of the folded piece after the squares are cut out. Since the squares are cut out from each corner, the resulting length of the folded piece will be (37 - 2x) ft, and the resulting width will be (20 - 2x) ft.

To form a box, we need to fold the cut-out piece along the sides and glue or tape it together. The height of the box will be equal to the length of the removed squares, which is x ft.

So, the volume of the box is given by:
Volume = length * width * height
Volume = (37 - 2x) * (20 - 2x) * x
Volume = (x^3 - 57x^2 + 740x)

To find the maximum volume, we need to find the critical points of this function. We can do this by taking the derivative of the volume function and setting it equal to zero:

d(Volume)/dx = 3x^2 - 114x + 740 = 0

Solving this equation will give us the values of x that correspond to maximum or minimum volume. However, since we are looking for the maximum volume, we can use the vertex formula to find the x-coordinate of the vertex of this quadratic function:

x = -b / (2a)
x = -(-114) / (2 * 3)
x = 19 ft

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

Volume = (19^3 - 57 * 19^2 + 740 * 19) ft^3
Volume ≈ 16244 ft^3

Therefore, the volume of the largest box that can be formed is approximately 16244 cubic feet.

b) If the original piece of cardboard is a square with side lengths k, the dimensions of the folded piece after the squares are cut out will be (k - 2x) by (k - 2x), and the height of the box will still be x. The volume of the box is given by the same formula as before:

Volume = (k - 2x)^2 * x

To find the maximum volume, we can follow the same steps as in part a. Take the derivative of the volume function, set it equal to zero, solve for x, and substitute it back into the volume function to get the maximum volume.

c) If the original piece of cardboard is a rectangle with sides of length k and L, and we are holding k fixed, we want to find the size of the corner squares x that maximizes the volume of the box as L approaches infinity.

The dimensions of the folded piece after the squares are cut out will be (k - 2x) by (L - 2x), and the height of the box will still be x. The volume of the box is given by:

Volume = (k - 2x)(L - 2x)x

We can follow the same steps as before to find the maximum volume. Take the derivative of the volume function, set it equal to zero, solve for x, and substitute it back into the volume function. However, since L is approaching infinity, we can simplify the problem by taking the limit of the volume function as L approaches infinity.

By finding the maximum value of the simplified volume function, we can determine the size of the corner squares x that maximizes the volume of the box as L approaches infinity.

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