To find the dimensions of the largest possible box, we need to maximize the volume of the box given the surface area constraint.
Let's assume the base length of the square box is "x" cm and the height of the box is "h" cm.
The surface area of the box can be divided into two parts: the base (which is a square) and the four sides.
The area of the base is the square of the base length, so it is equal to x^2 cm^2.
The area of the four sides can be calculated by multiplying the perimeter of the base by the height, as each side has the same height "h". The perimeter of the base is 4 times the base length, so it is equal to 4x cm. Therefore, the area of the four sides is 4xh cm^2.
Given that the total surface area of the box is 5400 cm^2, we can write the following equation:
x^2 + 4xh = 5400
Now, let's isolate one of the variables in the equation. I'll solve for "h" in terms of "x":
4xh = 5400 - x^2
h = (5400 - x^2) / 4x
Now, we need to find the maximum volume of the box. The volume of a rectangular prism (box) is given by the formula:
Volume = base area * height
So, the volume of this square-based box is:
V = x^2 * h
V = x^2 * ((5400 - x^2) / 4x)
We can simplify the equation further:
V = 5400x - x^3 / 4
Now, let's find the derivative of the volume with respect to "x" and set it equal to zero to find the maximum volume:
dV/dx = 5400 - 3x^2 / 4
Setting dV/dx equal to zero, we get:
5400 - 3x^2 / 4 = 0
5400 = 3x^2 / 4
x^2 = (4 * 5400) / 3
x^2 = 7200
x = √(7200)
x ≈ 84.85 cm
Since the dimensions of a square base must be equal, the base length of the square box is approximately 84.85 cm.
Now, substitute this value of "x" back into the equation for "h" that we derived earlier:
h = (5400 - x^2) / 4x
h = (5400 - (84.85)^2) / (4 * 84.85)
h ≈ 21.21 cm
Therefore, the dimensions of the largest possible box are approximately:
Base length = 84.85 cm
Height = 21.21 cm