when folded up and welded what is the maximum volume of the box, and the width, length and height?

Question

An open metal box is made of a sheet 1m by 1.5m that has four equal squares cut out of the corners.

Answer 1

Start with your sheet. Out of each corner you will cut a square of x by x meters. When you fold this you will have an open top box.

The dimensions of this box will be a height of x, a width of (1 - 2x) and a length of (1.5 - 2x). The reason will be pretty clear if you draw a diagram, but basically the width and length are reduced by the length of two of the squares.

So that creates an equation for the volume. Remember that volume is H x W x L. So in your case it becomes:

f(x) = x * (1 - 2x) * (1.5 - 2x)

If you multiply this out you get:
f(x) = x * (1.5 - 3x - 2x + 4x²)
f(x) = x * (1.5 - 5x + 4x²)
f(x) = 4x^3 - 5x² + 1.5x

Now if you know calculus, you can take the derivative:
f'(x) = 12x² - 10x + 1.5

The maximum volume will be when this is equal to zero:
12x² - 10x + 1.5 = 0

Using the quadratic formula:
x = [10 ± sqrt(10² - 4(12)(1.5)) ] / 2*12
x = [10 ± sqrt(100 - 72) ] / 24
x = (10 ± sqrt(28))/24

If you evaluate this with your calculator you get:
x ≈ 0.637145943 and x ≈ 0.196187391

Now obviously you can't cut four squares with dimensions of 0.637m when the metal is only 1 m wide.

So the only reasonable value is x ≈ 0.196187391

You can just plug this value back into the volume formula to get the maximum volume.
H = x = 0.196187391
W = 1 - 2x = 1 - 2(0.196187391) = 0.607625218
L = 1.5 - 2x = 1.5 - 2(0.196187391) = 1.107625218

H x W x L = 0.196187391 x 0.607625218 x 1.107625218
= 0.132038237

So the maximum volume is approximately, 0.132 cubic meters.

Answer 2

call the length of one side of your cutout 'c'

then,

Volume = c x (1-2c) x (1.5-2c) or

V = 1.5c - 5c^2 + 4c^3

To find the extrema, we set the derivative to zero:

dV/dc = 1.5 - 10c + 12 c^2 = 0

solving the quadratic equation, c=0.6371 or c=0.1962

lets try both and see which gives the largest volume

of course since c=0.6371 is not physically possible... the 0.1962 must be the correct answer

0.1962(1-2x0.1962)(1.5-2x0.1962) =

0.6076 x 1.1076 x 0.1962 = 0.132 m^3

Answer 3

H: .333 m
W:.333 m
L : .834m
V : .0925m3

The maximum volume would be derived if height and width were equal based on "equal squares cut out" (hehehe tricky)
(RATS I dropped a decimal point... Leonardo wins, but.. more than one way to skin a cat)

Answer 4

No. the quantity is a level of the area interior the field. Supposing which you became the field the different way up, or stood it on one end, or became it sideways... the area interior the field may be the comparable.