Solution:

To find the value of x that maximizes the volume of the box, we need to consider the dimensions of the box and how they are related to the side length x.

Step 1: Determine the dimensions of the box

When the four squares are removed from the corners of the rectangular sheet, the resulting box will have a length, width, and height.

Length of the box = length of the rectangular sheet - 2x
Width of the box = width of the rectangular sheet - 2x
Height of the box = x

Step 2: Express the volume of the box in terms of x

The volume of the box can be calculated by multiplying the length, width, and height together.

Volume of the box = (length of the box) * (width of the box) * (height of the box)
= (30 cm - 2x) * (80 cm - 2x) * x
= x(30-2x)(80-2x)

Step 3: Find the critical points of the volume function

To find the value of x that maximizes the volume, we need to find the critical points of the volume function. We can do this by taking the derivative of the volume function with respect to x and setting it equal to zero.

dV/dx = (30-2x)(80-2x) + x(-2)(80-2x) + x(30-2x)(-2)
= (30-2x)(80-2x) - 2x(80-2x) - 2x(30-2x)
= (30-2x)(80-2x) - 4x(80-2x)

Setting dV/dx = 0 and solving for x:
(30-2x)(80-2x) - 4x(80-2x) = 0

Step 4: Solve for x

Expanding the equation and combining like terms:
2400 - 140x + 4x^2 - 320x + 8x^2 - 4x^2 = 0
2400 - 460x + 8x^2 = 0

Simplifying the equation:
2(4x^2 - 230x + 1200) = 0
4x^2 - 230x + 1200 = 0

Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values:
x = (-(-230) ± √((-230)^2 - 4(4)(1200))) / (2(4))
x = (230 ± √(52900 - 19200)) / 8
x = (230 ± √33700) / 8

Calculating the square root and simplifying:
x ≈ (230 ± 183.6) / 8

x ≈ 413.6 / 8 or x ≈ 46.4 / 8

Therefore, we have two possible values for x:
x ≈ 51

Is it 20/3 cm ??