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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).

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    If you stare at $a+b+1-a-b=1$ for$a$long time, wouldn't you agree that it's a tautology and thus helps your problem not even a smidgen?2011-04-07

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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 $b=1$.

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    I forgot to mention constraints on a, b, and 1-a-b. They are three probabilities that are supposed to add up to one, and no two can add up to more than one, since we are dealing with probabilities. Is this solvable then?2011-04-07