# How to solve this optimization problem

 0 I have an optimization problem in the form max $$(a-\bar{a})(b-\bar{b})$$ subject to $$a+b=1$$ Here $$\bar{a}$$ and $$\bar{b}$$ are known values and both of them are >0. Let $$a_{opt}$$ and $$b_{opt}$$ are the solutions or optimal values then $$a_{opt}-\bar{a}>0$$ and $$b_{opt}-\bar{b}>0$$ How to find this solution for this optimization problem? asked 09 Feb '16, 04:13 dimitrios 11●3 accept rate: 0% 1 How big could be those $$a$$ and $$b$$? (09 Feb '16, 05:48) Slavko Substitute $$b=1-a$$ into your objective function, yielding a quadratic polynomial in $$a$$. To find $$a_\text{opt}$$ and $$b_\text{opt}$$, use either single-variable calculus or a well-known fact about the location of a parabola's vertex with respect to its intercepts. (09 Feb '16, 10:53) Rob Pratt @Rob Pratt, Thank you very much for the hints, i.e., the solutions! (09 Feb '16, 20:34) dimitrios

One Answer:
 1 As Rob suggests, substitute to get a function of one variable (a). Then first derivative to find turning point gives optimal a (second derivative will show it's a maximum). Substitute back to get b_opt. You should find a = (1 + a¯ - b¯)/2, b = (1 - a¯ + b¯)/2 Also, your last set of constraints on the optimal values constrains the sum of a¯ + b¯ < 1 answered 09 Feb '16, 11:55 Grallator 21●2 accept rate: 0% @Grallator, thanks for the answer! (09 Feb '16, 20:35) dimitrios
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Asked: 09 Feb '16, 04:13

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Last updated: 09 Feb '16, 20:35

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