How to find the max of the value? I have $\frac{(a-1)}{4}+\frac{b}{2}=x$ and $0\leq a \leq 1$, $0\leq b \leq 1$. How can I find the values for a, b such that x is maximized? Thanks.
EDIT: Sorry, I forgot to clarify, I am working with events (and related pay-offs), and I have probabilities a, b, and (1-a-b) for three of the events, so it's a bit trickier (they are related because all together can't give more than one, so for example both a and b can't be .6, as that would add up to 1.2 when max probability is 1).
 A: In this case, since $a$ and $b$ are independent, and they don't interact in the value of the function, all you need to do is find the value of $a$ that maximizes $\frac{a-1}{4}$ and find the value of $b$ that maximizes $\frac{b}{2}$. They are both rather trivial to do. If it helps, the maximum value of $x$ is $\frac{1}{2}$.
In view of the edit, the actual problem is to maximize
$$\frac{a-1}{4} + \frac{b}{2}$$
subject to the constraints
$$0\leq a\leq 1,\qquad 0\leq b\leq 1,\qquad 0\leq a+b\leq 1.$$
That means that you are trying to maximize this function on on right triangle with vertices on $(0,0)$, $(1,0)$, and $(0,1)$, instead of on the unit square.
It is plain that the maximum will occur on the boundary, because moving further away from $(0,0)$ will never decrease the summands.
So it either occurs on the line $a=0$, $0\leq b\leq 1$ (the maximum along this line is $\frac{1}{4}$, obtained at $(0,1)$); or on the line $b=0$, $0\leq a\leq 1$ (the maximum along this line is $0$, obtained at $(1,0)$); or along the line $a+b = 1$. On this line, the function equals $\frac{b}{4}$, so the maximum occurs when $b=1$, $a=0$, same as before.
So the maximum value for $x$ is $\frac{1}{4}$, when $a=0$ and $b=1$. 
