Hey Kohoutek.

For concavity you basically have to show that if you take any two points in the set of the function (call them A and B), then you have to show that all the points in-between them (on a straight line path) also lie in the function.

Something is convex if f(a + (b-a)t) where to goes from 0 to 1 and a and b are in the domain.

So to start off with, pick two general points corresponding to f(a) and f(b) where a = -65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1 and b = -65,9569 + 5,08x2 - 0,83y2 - 5,27z2 + 6,65w2. Now if you can show that f(a + (b-a)t) is always in the possible function values. If you show that it fails then its concave.

So basically you have f(a + (b-a)t) = f(-65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1 + (-65,9569 + 5,08x2 - 0,83y2 - 5,27z2 + 6,65w2 - (-65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1))t)

= f(-65,9569 + 5,08x1 - 0,83y1 - 5,27z + 6,65w1 + (5,08(x2-x1)t - 0,83(y2-y1)t - 5,27(z2-z1)t + 6,65(w2-w1)t) where t is in between 0 and 1.

So for all values of t, do these values lie in the set of potential function values? If they do your function is convex.

Basically the question now becomes what the constraints are on your x,y,z,w parameters and in some cases you will get convexity and in others you won't.