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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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It's a plane with the domain restricted to a rectangular region. Look at the corners and see which gives the highest value. –  J. M. Apr 7 '11 at 4:32
    
Hint: Small $a$ hurts, big $b$ helps. –  André Nicolas Apr 7 '11 at 4:34
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The constrant you wrote doesn't make sense: a+b+1-a-b is always equal to $1$ and places no constraints on either a nor b. The constraint you are looking for is $0\leq a+b\leq 1$, not what you wrote. –  Arturo Magidin Apr 7 '11 at 4:41
    
@Arturo: I understand what you are saying, but that's how I had to phrase the problem -- I am trying to find the values for a, b, such that pay-offs are maximized, and both player 1 and 2 have 3 options. I know the options for player 2 (they are given to me), and I know the pay off for every combination of the three options, and I need to find such a (for player 1 opt 1) b (for player 1 opt 2) and 1-a-b (player 1 opt 3) such that expected pay-offs are maximized. So those three add to 1 (always, as you said), but also a+b has to be $\leq$ to 1. –  user9237 Apr 7 '11 at 4:43
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@Dantist: Which is exactly what I said; when you stated your constraints by saying "$0\leq a\leq 1$, $0\leq b\leq 1$, $a+b+1-a-b = 1$", the last equation said nothing. The correct way to state the constraints you want is to say "$0\leq a\leq 1$, $0\leq b\leq 1$, and $0\leq a+b\leq 1$". The origin of the problem is irrelevant, the "constraint" $a+b+1-a-b=1$ is vacuous. –  Arturo Magidin Apr 7 '11 at 4:49

1 Answer 1

up vote 1 down vote accepted

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? –  user9237 Apr 7 '11 at 4:38

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