If x is the side length of the base square and h is the height of the box, then the area of the open box is

    A(x,h) =  + .    (1)


The volume of the box is

    V(x,h) = .        (2)


We need to minimize the surface area (1) under the restriction

     = 62.5  cm^3.     (3)


From (3), express  h =  and substitute it into (1). You will get then

    A(x) =  +  =  + .    (4).


Thus we need to minimize function A(x) of one variable "x" expressed by (4).


For it, take the derivative


    A'(x) = 2x - 


and equate it to 0 (zero).  You will get


    2x -  = 0,

    2x^3 - 250 = 0,

    x^3 = 250/2 = 125,

    x =  = 5  cm.


Then  h =  =  = 2.5 cm.


Thus the optimal dimensions are  x = 5 cm;  h = 2.5 cm.                                     ANSWER

The surface area of the open box is  A =  + 4*x*h =  + 4*5*2.5 = 25 + 50 = 75 cm^2.    ANSWER


Partial CHECK.  The volume =  =  = 62.5 cm^3.    ! Correct !