For what value is the local minimum the largest? 
If  $f(x)=e^x-kx$ for $k>0$, find the values of $k$ for which the local minimum at $x=\ln(k)$ is the largest. 

I found the derivative, which is $e^{x} - k$, and when I set that to $0$ I got $-e^{x}=k$. I'm not really sure if this is useful information and if it is, I am not sure how to use it to answer the questions, so I would appreciate any tips on where to go next! Thanks!
 A: The local minimum, we are told, is at $x=\ln k$. We can also find that out for ourselves by differentiating.
Thus the (local) minimum value is $e^{\ln k}-k\ln k$, or more simply $k-k\ln k$. We want to maximize this function of $k$. So  your task is to find the maximum value of $g(k)$, where $g(k)=k-k\ln k$. Use standard max/min tools. 
Remark: You were finding the local min of $e^x-kx$, differentiated, got $e^x-k$. When you set this equal to $0$, you should get $e^x=k$, which yields $x=\ln k$. 
A: Well, you know that the local minimum occurs at $x = \ln(k)$. So replace in your equation:
$$f(\ln(k)) = k - k\ln(k) = k(1-\ln(k)) = -k (\ln(k) - \ln(e)) = -k \displaystyle \ln \left (\frac{k}{e} \right ) = \ln \left (\frac{3}{k} \right)^k$$
Now, let $g(k) = \ln \left (\frac{3}{k} \right)^k $ and find, using differentiation, for which $k$ does this take its maximum value. That $k$ will be the desired one.
A: $$x=\log k\Longrightarrow e^{\log k}-k\log k=k-k\log k$$
Take now the function 
$$g(x):=x-x\log x\;\;,\;\;x>0\Longrightarrow g'(x)=1-\log x-1=-\log x=0\Longrightarrow x=1$$
and since $\,g''(x)=-\frac{1}{x}<0\,\,\;\;\;\forall\,\,x>0\,\;$ , the point $\,x= 1\,$ gives a maximum...
