Given,

• Side of the square piece is 12 cms.

• the volume of the formed box is maximum.

Let us consider,

• ‘x’ be the length and breadth of the piece cut from each vertex of the piece.

• Side of the box now will be (12-2x)

• The height of the new formed box will also be ‘x’.

Let the volume of the newly formed box is :

V = (12-2x)² × (x)

V = (144 + 4x² – 48x) x

V = 4x³ -48x² +144x ------ (1)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (1) with respect to x:

-------- (2)

[Since ]

To find the critical point, we need to equate equation (2) to zero.

x² – 8x +12 = 0

x = 6 or x =2

x= 2

[as x = 6 is not a possibility, because 12-2x = 12-12= 0]

Now to check if this critical point will determine the maximum area of the box, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

Now let us find the value of

As , so the function A is maximum at x = 2

Now substituting x = 2 in 12 – 2x, the side of the considered box:

Side = 12-2x = 12 - 2(2) = 12-4= 8cms

Therefore side of wanted box is 8cms and height of the box is 2cms.

Now, the volume of the box is

V = (8)² × 2 = 64 × 2 = 128cm³

Hence maximum volume of the box formed by cutting the given 12cms sheet is 128cm³ with 8cms side and 2cms height.