A cardboard box manufacturer wishes to make open boxes from rectangular pieces of cardboard with dimensions 40 cm by 60 cm by cutting equal squares from the four corners and turning up the sides. Find the length of the side of the cut-out square so that the box has the largest possible volume. Also, find the volume of the box

To find the length of the side of the cut-out square that maximizes the volume of the box, we can use optimization techniques.

Let's denote the side length of the cut-out square as "x" cm. After cutting out the squares from the corners, folding up the sides, and forming the box, the dimensions of the resulting box will be:
Length: (60 - 2x) cm
Width: (40 - 2x) cm
Height: x cm (as the cut-out squares are folded up to form the height)

The volume of a rectangular box is given by the formula: Volume = Length * Width * Height

So, the volume of the box in terms of x will be:
V(x) = (60 - 2x) * (40 - 2x) * x

To find the value of x that maximizes the volume, we need to find the critical points of the volume function by taking its derivative and setting it to zero:

V'(x) = 0

Let's differentiate the volume function with respect to x:

V'(x) = [(60 - 2x) * (40 - 2x) * x]' [using product rule]
= (60 - 2x)' * (40 - 2x) * x + (60 - 2x) * (40 - 2x)' * x + (60 - 2x) * (40 - 2x) * x'
= (-2) * (40 - 2x) * x + (60 - 2x) * (-2) * x + (60 - 2x) * (40 - 2x) * 1
= -4x(40 - 2x) - 2x(60 - 2x) + (60 - 2x)(40 - 2x)
= -8x^2 + 240x - 4x^2 + 120x + 2400 - 200x - 2x^2 + 80x
= 600x - 14x^2 + 2400

Setting V'(x) = 0 and solving for x:

600x - 14x^2 + 2400 = 0
-14x^2 + 600x + 2400 = 0
-7x^2 + 300x + 1200 = 0
Dividing by -3:
7x^2 - 300x - 1200 = 0

Now we can solve this quadratic equation for x using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

In our case, a = 7, b = -300, and c = -1200. Substituting these values:

x = (-(-300) ± sqrt((-300)^2 - 4 * 7 * (-1200))) / (2 * 7)
x = (300 ± sqrt(90000 + 33600)) / 14
x = (300 ± sqrt(123600)) / 14
x = (300 ± 12sqrt(686)) / 14
x = 21.18 or -8.5 (rounded to two decimal places)

Since the value of x represents side length, it cannot be negative. Therefore, we discard -8.5 as an extraneous solution.

So, the length of the side of the cut-out square that maximizes the volume of the box is approximately 21.18 cm.

To find the maximum volume, we can substitute this value of x back into the volume function:

V(x) = (60 - 2x) * (40 - 2x) * x

V(21.18) = (60 - 2 * 21.18) * (40 - 2 * 21.18) * 21.18
V(21.18) = 17.64 * 17.64 * 21.18
V(21.18) = 6453.16 cubic cm (rounded to two decimal places)

Therefore, the maximum volume of the box is approximately 6453.16 cubic cm.