Explanation:
Let x, be the length of the square box and S be the surface of the box,
Let y, be the height of the square box and V for the volume.
The top of the box is open and the material of the box is equal to the surface area of box.
then,
Putting the value of S (surface area).
subtitutting the following equation in volume at the place of y.
Then,
Differentiate the volume to 0.
Then,
∴
and
Then, we have to find maximum volume by putting the value of x.