Help please!! If 12ft^2 of material is available to make a box with a square base and open top, find the largest possible volume for the box.

The sufrace area, S, of the box will  = b^2 + 4bh   where b is the side of the base and h is the height  and this will =  12ft^2

The volume, V  = b^2 *h

So ....  we have the two equations

12 = b^2 + 4bh

V  = b^2* h

We can solve for the height in the first equation and sub this into the second

12- b^2  = 4bh

h = [12  - b^2]  / 4b

So....the volume is given by

V(b) = b^2 [12  - b^2]  / 4b   =  b[12 - b^2]/4  =  3b - (1/4)b^3

Taking the derivative of this we have

V ' (b)  = 3 - (3/4)b^2     set this to  0

3 - (3/4)b^2  = 0

3 = (3/4)b^2      divide through by 3

1 = (1/4)b^2      multiply both sides by 4

4 = b^2     take the positive root

2  feet = b

And h = [12  - 2^2]  / 4(2)  =  [8] / [8]  = 1 ft

So  a box with a square base of 2 ft on each side and 1 ft high will produce the max volume = b^2*h  =  (2)^2* (1)   =  4ft^3

Here's a graph of the volume where h is in terms of b :  https://www.desmos.com/calculator/27nqcjdoy7

Note that the max volume occurs when b = 2 ft   [ and, by implication, h = 1 ft ]