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 Subject: Minimizing Question Category: Science > Math Asked by: roadapples-ga List Price: \$7.50 Posted: 07 Dec 2002 23:26 PST Expires: 06 Jan 2003 23:26 PST Question ID: 121237
 ```Solve the problem of minimizing a^2 + (b - 1)^2 subject to e^b - a is less than or equal to zero.```
 ```Hi roadapples-ga, I use the signs "<=" and ">=" to mean "less/greater than or equal to", respectively. Define f(a,b) = a^2 + (b-1)^2, and g(b) = e^(2b) + (b-1)^2. Then, under the constraint e^b - a <= 0 (or, equivalently, a >= e^b), we have f(a,b) >= g(b) So let us first minimize g(b) as a function of b. Differentiate: g'(b) = 2e^(2b) + 2(b-1) = 2(e^(2b) + b - 1) So g'(b)=0 if and only if e^(2b)+b-1 = 0, this happens for b=0, and ONLY for b=0, since g''(b) = 4e^(2b)+2 is a strictly positive function, so g' is increasing and can only hit zero once. Furthermore since g''(0) is positive this means that g has a local minimum at 0, but because it's the only critical point it's also a global minimum. So we have shown: g(b) >= g(0) = 2 for all b and f(a,b) >= g(b) >= 2 for all a,b satisfying the constraint. But setting b=0, a=e^b=1 gives f(a,b)=2. This proves that a=1, b=0 is the global minimum for f(a,b) under the constraint. This is the answer. Hope this helps, please ask for clarification if you have any doubts. Regards, dannidin```