The volume equation is  


    x^2*h = 108.     (1)


The surface area expression for the open box is  


    A(x,h) = x^2 + 4xh.     (2)


So, you need to find dimensions which minimize the function A(x,h) (2)  under the condition (1).


To solve the problem, express h =  from (1) and substitute it into (2), making A function of only one variable x:


    A(x) = x^2 + 4x* = x^2 + .    (3)


Now you have this function A(x) of one variable x, and you should find its minimum.


Differentiate; equate the derivative to zero


    A'(x) = 2x -  = 0

and get


    2x^3 - 432 = 0  ====>  x^3 = 432/2 = 216  ====>  x =  = 6.


Answer.  x= 6;  h =  =  =  = 3.

If the box is flattened it will look like this:
A square base whose dimensions are x inches by x inches, and
4 equal flaps each of whose dimensions are x inches by h inches:



The formula for the volume is:

 with V = 108 in³, l = length = x, and width = w = x, and height = x





The formula for the surface area, the square base whose dimensions are x
inches by x inches, plus 4 equal flaps each of whose dimensions are x inches
by h inches is:

 and we substitute  for h





Set that = 0

Divide through by 2:

Rewrite x-2 as x² in the denominator

Multiply through by LCD = x²


 in

Then use:


 




Edwin