Google Answers Logo
View Question
Q: Minimizing Question ( Answered 5 out of 5 stars,   0 Comments )
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.
Subject: Re: Minimizing Question
Answered By: dannidin-ga on 08 Dec 2002 00:31 PST
Rated:5 out of 5 stars
Hi roadapples-ga,

I use the signs "<=" and ">=" to mean "less/greater than or equal to",

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

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


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.
roadapples-ga rated this answer:5 out of 5 stars and gave an additional tip of: $2.00

There are no comments at this time.

Important Disclaimer: Answers and comments provided on Google Answers are general information, and are not intended to substitute for informed professional medical, psychiatric, psychological, tax, legal, investment, accounting, or other professional advice. Google does not endorse, and expressly disclaims liability for any product, manufacturer, distributor, service or service provider mentioned or any opinion expressed in answers or comments. Please read carefully the Google Answers Terms of Service.

If you feel that you have found inappropriate content, please let us know by emailing us at with the question ID listed above. Thank you.
Search Google Answers for
Google Answers  

Google Home - Answers FAQ - Terms of Service - Privacy Policy