Max volume = 32 cubic feet.
First, let's create an equation to express the height of the box when given the length of an edge of the base of the box. Since the box is open top, there will be 5 surfaces. The base and 4 sides. The area of the base will be x^2 and the area of each side will be xh, so the area of all 4 sides will be 4xh. And the sum of those 5 faces will be equal to 48. So
x^2 + 4xh = 48.
Now we need to solve for h, so
x^2 + 4xh = 48
4xh = 48 - x^2
h = (48 - x^2)/(4x)
Now we need to make an equation for the volume of the box. That will be hx^2 (height * width * length) where length and width are both x.
hx^2
Substitute the equation for h in terms of x we calculated earlier,
v=hx^2
v=((48 - x^2)/(4x))x^2
Simplify
v=((48 - x^2)/(4x))x^2
v=x^2((48 - x^2)/(4x))
v=x(48 - x^2)/4
v=(48x - x^3)/4
v=12x - x^3/4
v=12x - 0.25x^3
Now this function will have it's maximum or minimum value at those points where its slope is 0. And you can determine the slope of the function by calculating its first derivative. For simple equations, the first derivative is calculated by multiplying the coefficient of each term by the exponent of the term. Then subtract 1 from the exponent. So starting with
v=12x - 0.25x^3
we have 2 terms. 12x and -0.25x^2
For the 12x term, the exponent is an implied value is 1, so 12 * 1 = 12 and 1 - 1 = 0. So we get 12x^0. And since any number raised to the zeroth power is 1, we get 12*1 which is 12.
For the -0.25x^3 term, the exponent is 3, so -0.25 * 3 = -0.75, and 3 - 1 = 2. So that term becomes -0.75x^2. And the complete first derivative is
v'=12 - 0.75x^2
And we now have a quadratic equation with a=-0.75, b=0, and c=12. Using the quadratic formula we can get the roots which are -4, and 4. A negative width doesn't make sense, to the width of 4 is the one we'll use.
Now let's check if we will use all the material in the box. The base will be 4x4 = 16 square units. The height will be:
h = (48 - x^2)/(4x)
h = (48 - 4^2)/(4*4)
h = (48 - 16)/16
h = 32/16
h = 2
So each side will be 4x2 = 8 square ft. And the total area of the faces of the box will be 16 + 8*4 = 16 + 32 = 48 square ft.
So the total volume of the box will be 4 * 4 * 2 = 32 cubic ft.
Now, just as a quick spot check, let's see what the volume will be if the width is slightly smaller than 4 and if it's slightly larger than 4. Let's use a width of 3.99 and 4.01 as out spot checks.
h = (48 - x^2)/(4x)
h = (48 - 3.99^2)/(4*3.99)
h = (48 - 15.9201)/15.96
h = 32.0799/15.96
h = 2.010018797
V = 2.010018797 * 3.99 * 3.99 = 31.99970025
So a slightly smaller x gives a smaller volume. Let's try the larger x.
h = (48 - x^2)/(4x)
h = (48 - 4.01^2)/(4*4.01)
h = (48 - 16.0801)/16.04
h = 31.99969975/16.04
h = 1.990018703
V = 1.990018703 * 4.01 * 4.01 = 31.99969975
And a slightly larger X also gives a smaller volume.
So the largest possible volume is 32 cubic feet when the length and width of the box is 4 feet and the height is 2 feet.
