= Solution
{c}
Let $n=2m\ge2$. Between the two middle <order statistics>, $k=m$ and the slope $A'(\mu)=2m-n$ is zero. To the left $A$ decreases and to the right it increases. Consequently all joint <maximum-likelihood estimates> are
$$
\boxed{\widehat\mu\in[x_{(m)},x_{(m+1)}],\qquad
\widehat\lambda=\frac{n}{A_{\min}},}
$$
where the constant minimum is
$$
\boxed{A_{\min}=\sum_{i=m+1}^{2m}x_{(i)}-\sum_{i=1}^mx_{(i)}>0.}
$$
The strict positivity follows from distinct observations. Thus the location <maximum-likelihood estimate> is not unique, while the rate <maximum-likelihood estimate> is unique and the same for every location in the displayed <interval>. Choosing the arithmetic midpoint of the two middle values is the usual <sample median> convention and gives one member of this entire maximizing family.
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