Let us Say that the total length of the rectangular box is AF = 95 cms and the width is 47 cms. Assume that the on cutting a square of length “x” cms along the length leads to the maximum volume of the pizza box.

Effectively we will have the length of the pizza box as BC and the width as GH.

Since 4 squares are cut along a length, the length of BC = (95-4x)/2 and the width GH would be 47-2x (Please refer the picture)

Now we have the rectangular box of length BC, width 47-2x and the height “x”.

Volume of the Box is given by V= length X width X height V= (95-4x)*(47-2x)*x/2

V=4x^3-189x^2+2232.5x

Now we need to derivate this function and equate it to 0 to find the critical points. V’ = 12x^2 -378x +2232.5=0 Therefore on solving for the roots using the quadratic formula we can have x= 7.875cms or x=23.625cms .

So the square of dimension x= 7.875 cms nearly is to be cut to get the maximum volume.

Let us Say that the total length of the rectangular box is AF = 95 cms and the width is 47 cms.

Assume that the on cutting a square of length “x” cms along the length leads to the maximum volume of the pizza box.

Effectively we will have the length of the pizza box as BC and the width as GH.

Since 4 squares are cut along a length, the length of BC = (95-4x)/2 and the

width GH would be 47-2x (Please refer the picture)

Now we have the rectangular box of length BC, width 47-2x and the height “x”.

Volume of the Box is given by

V= length X width X height

V= (95-4x)*(47-2x)*x/2

V=4x^3-189x^2+2232.5x

Now we need to derivate this function and equate it to 0 to find the critical points.

V’ = 12x^2 -378x +2232.5=0

Therefore on solving for the roots using the quadratic formula we can have x= 7.875cms or x=23.625cms .

So the square of dimension x= 7.875 cms nearly is to be cut to get the maximum volume.