Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the problem
$$ max \ f(x,y)=\left(\frac12\right)x-y $$

$$subject \ to \ x +e^{-x} \leqslant y $$ $$ x\geqslant0 $$

share|improve this question

closed as unclear what you're asking by 900 sit-ups a day, mookid, Michael Albanese, anorton, hardmath Jul 16 at 1:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

2  
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of the imperatives "Prove", "Show" etc. to be rude when asking for help; please consider rewriting your post. –  Dennis Gulko Oct 9 '12 at 21:22

1 Answer 1

Well, assuming what is given, $f(x,y) \le \frac12x-x-e^{-x}$, so for a fixed $x$, the choice $y:=x+e^{-x}$ is the best in maximizing $f$.

Then, you are left with only one variable, and should maximize this $g(x):=-\frac12x-e^{-x}$ whence $x\ge 0$. Possible maximum place is at the edge ($x=0$) or when the tangent of $g$ is horizontal, i.e. where $g'=0$.

share|improve this answer
    
Hi Berci - thank you for your answer. Could you write down the Lagrangian function and necessary Kuhn-Tucker conditions for this problem? –  Elvis Oct 9 '12 at 21:41
    
errhh.. I don't really know what these are. –  Berci Oct 9 '12 at 22:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.