Let x be the size of the square base and h be the height of the pot, in centimeters.


Then the volume of the pot is

     V = .                           (1)


The surface are of this pot, which has no top, is

      A =  = 500 cm^2.            (2)


So, we want to find optimal values of "x" and "h" to maximize the volume (1) at given restriction (2) on surface area.


From (2), we have 

     h =  = .            (3)


Substitute it into (1) to get

     V =  =  - .     (4)


Now we need to find a maximum value for V in formula (4) considering the volume  as a function of "x" only.


For it, take the derivative of  V(x) and equate it to zero

     V'(x) = 125 -  = 0.


It implies

     500 = 3x^2

     x^2 = 

     x =  = 12.91 cm (approximately.


Then

     h = (see formula (3)) =  =  = 6.45 cm.



So, the problem is just solved.

Optimal dimensions are: the square base size of 12.91 cm and the height of 6.45 cm.



The maximum volume is   =  = 1075 cm^3  (approximately).