Abridged answers

1) minimum of 9 at (0, 0, 3)
2) 54*pi
3) 6
4) x = y = z = (V_0)^(1/3)
5) df/dx = (\sqrt(y) + 1) / 
           (2*(1+y*x^2+x*y^2 - 2*x^(3/2)*y^(3/2))*sqrt(x))  
   df/dy = (\sqrt(x) - 1) / 
           (2*(1+y*x^2+x*y^2 - 2*x^(3/2)*y^(3/2))*sqrt(y))  
6) x = y = z = S_0 / 12
7) x + y - 6z = -2
8) maximum of 3 at (1,2)
   minimum of -12/5 at (8/5,-2/5)
9) (2/3, 3/2)
10) (24x^2, -18y)
11) (integral from 0 to 2) (integral from 0 to cosx) f(x,y) dy dx
12) 2*pi*ln(3/2)
13) max of 27*4 / 5^(5/2) at (\sqrt{2/5), \sqrt(3/5))
    min of -27*4 / 5^(5/2) at (-\sqrt{2/5), \sqrt(3/5))
14) 0
15) More or less unsolvable...sorry about that. You get a 4th order
polynomial to solve for y (or x).
16) I_x = 1024 / 9
    I_y = 128/6 - 128/14
    I_z = I_x + I_y
17) 1/12