there is none

Axel Vogt — Mon, 11 Jan 2010 01:40:00 Z

There is no minimum. Even if I suppose it is homework:

The last terms are linear, where x,y,z live in a cube, thus this gives
something bounded (IIRC Analysis correctly).

So your 1st term matters. Transform to positive x,y,z:

subs(z=-z,%); subs(y=y+1, %);

a*b^x*c^(y+1)*d^(-z)-x-2*y-2+3*z

now with 0 < z < 1, 0 < y < 2

This suggests you want b,c,d to be positive to avoid imaginary values
(may be you forgot to say that).

Now choose: subs(x=1/2, y=1, z=1/2, %);

This gives a*somePositiveConstant - anotherConstant

So for a ---> - infinity ...

Or similar ...

MaplePrimes - answers and comments on Question, solve the nonlinear programming problem

there is none