= Solution
The inverse <change of variables> is $x=u$, $y=(v-u)/a$, with absolute <Jacobian determinant> $1/a$. Because the original <random variables> are independent, their joint density is $f(x)g(y)$. Thus the transformed <joint density> is
$$
\boxed{f_{U,V}(u,v)=\begin{cases}
\dfrac1a f(u)g\left(\dfrac{v-u}a\right),&0\leq u\leq v,\\
0,&\text{otherwise}.
\end{cases}}
$$
The support is important: $X,Y\geq0$ implies $V\geq U\geq0$.
For independent <exponential random variables> of rate $\lambda>0$ and $a=1/2$, this becomes $2\lambda^2e^{-2\lambda v+\lambda u}$ on $0\leq u\leq v$. Integrating out $u$ gives the density of the sum:
$$
\boxed{f_{X+Y/2}(v)=2\lambda^2e^{-2\lambda v}\int_0^v e^{\lambda u}\,du
=2\lambda(e^{-\lambda v}-e^{-2\lambda v}),\quad v\geq0.}
$$
It vanishes for $v<0$. On the other hand, the <cumulative distribution function> of the maximum is
$$
P(\max(X,Y)\leq v)=P(X\leq v)P(Y\leq v)=(1-e^{-\lambda v})^2\quad(v\geq0).
$$
Differentiating yields exactly the same density. \b[The maximum and $X+Y/2$ have the same distribution], as described by <maximum of two exponentials as a scaled sum>; this does not assert that they are equal for the same pair of sampled values.
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