# 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...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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.

Approved by eNotes Editorial Team