= Solution
Write $\Phi$ for the <standard normal cumulative distribution function>. Since $B_t\leq M_t$, for $y\geq0$ and $x>y$,
$$
\mathbb P(B_t\leq x,M_t\leq y)
=\mathbb P(M_t\leq y)
=2\Phi\!\left(\frac y{\sqrt t}\right)-1.
$$
For $x\leq y$, the <Brownian reflection principle> gives
$$
\begin{aligned}
\mathbb P(B_t\leq x,M_t\leq y)
&=\mathbb P(B_t\leq x)
-\mathbb P(B_t\leq x,M_t>y)\\
&=\Phi\!\left(\frac x{\sqrt t}\right)
-\mathbb P(B_t>2y-x)\\
&=\Phi\!\left(\frac x{\sqrt t}\right)
+\Phi\!\left(\frac{2y-x}{\sqrt t}\right)-1.
\end{aligned}
$$
The <joint distribution function> is therefore
$$
\boxed{
F_{B_t,M_t}(x,y)=
\begin{cases}
0,&y<0,\\
\Phi(x/\sqrt t)+\Phi((2y-x)/\sqrt t)-1,&y\geq0,\ x\leq y,\\
2\Phi(y/\sqrt t)-1,&y\geq0,\ x>y.
\end{cases}}
$$
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