A square sheet of metal of side a has squares cut on all its corners and the sides bent to form an open box. What is the maximum volume of the box?

The sheet of metal we have has sides equal to a. Let the
sides of the squares that are cut from all the corners be
x.

Now we have a length equal to a – 2x, to fold on all the
sides and create the box.

The volume of the box is (a –
2x)^2*x = (a^2 + 4x^2 – 4ax)*x

V = a^2x + 4x^3 –
4ax^2

To maximize V, we find its
derivative.

V’ = a^2 + 12x^2 –
8ax

This is equated to
0

=> 12x^2 – 8ax + a^2=
0

=> 12x^2 – 6ax -2ax + a^2 =
0

=> 6x (2x – a) – a (2x – a) =
0

=> (6x – a) (2x – a) =
0

=> x = a/6 and x =
a/2

Now, we have two values of x and we have to determine
which value provides the maximum volume.

V’’ = 24x –
8a,

at x = a/2, V’’ = 12a – 8a =
4a.

As the second derivative is positive at x = a/2, we
have a minimum value here.

The volume of the box is maximum
at x = a/6 and the volume is equal to a^2*a/6 + 4(a/6)^3 –
4a*(a/6)^2

=> a^3 / 6 + a^3(4/6^3) –
(4/36)*a^3

=>
2a/27

The maximum volume of the box is
2a/27.