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! |
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$$x=\log k\Longrightarrow e^{\log k}-k\log k=k-k\log k$$ 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... |
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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 $x=\ln k$. |
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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. |
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