I'll answer in reverse order:

Multiplying that out,



We find the values of x at the endpoints 200 and 810:

 and 

 and 

Using synthetic division to factor them both, we get:

 and 

Those have 5 and 7 respectively as their only real solutions

The volume (2x-5)*(2x-5)*(x+3) is steadily increasing as the volume increases
from 200 cm3 to 810 cm3.

So the interval for x corresponding to  is .

Area of the square base = base2 = (2x-5)2.

We build that up from this inequality:



We multiply all three sides by 2:





We subtract 5 from all three sides:





We square all three sides (which keeps the same inequality since they are > 1).





Edwin