A box with a square base and an open top is constructed from 5400 cm^2 of cardboard. Find the dimensions of the largest possible box.

I know the answer is :
base lenght - 42.4 cm
height- 21.2 cm

please help me, thank you so much :)

It depends on the shape of the 5400cm^2 cardboard.

If it is square, it is sqrt5400 or 73.4cm to a side.

Now set it down, call the sides s, and the height h.
You have to cut out h^2 from each of the four corners.

Area=b^2 h
but 2h+b=73.4 or h=(73.4-b)/2

Area=b^2(73.4-b)/2
dArea/db= 2b(73.4-b)/2 -b^2/2=0
solve for b

Then use that to go back and solve for h.

If this is the typical questions, identical squares are to be cut out of each of the corners of the cardboard, and then the sides folded up to form the box.

Let each side of the cutout be x cm
so each side of the base of the box will be √5400 - 2x
volume of box
= x(√5400-2x)(√5400-2x)
= 5400x-4√5400x^2 + 4x^3

d(volume)/dx = 5400 - 8√5400x + 12x^2 = 0 for a max of volume

x^2 - 20√6 + 450 = 0

using the quadratic formula

x = (20√6 ± √(2400 -4(1)(450))/2
= (20√6 ± √600)/2
= 36.74 or 12.25

The first answer gives us a volume of zero, the minimum, and the second gives us the maximum value

height = 12.25
base = √5400-2x = 48.99

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