Initially I had trouble generalizing the multivector Lagrangian to include both the electric and magnetic sources without using two independent potentials. However, this can be done, provided one is careful enough. Recall that we found that a useful formulation for the field in terms of two potentials is
\label{eqn:maxwellLagrangian:2050}
F = F_{\mathrm{e}} + I F_{\mathrm{m}},

where
\label{eqn:maxwellLagrangian:2070}
\begin{aligned}
F_{\mathrm{e}} = \grad \wedge A \\
\end{aligned}

and where $$A, K$$ are arbitrary four-vector potentials.
Use of two potentials allowed us to decouple Maxwell’s equations into two separate gradient equations. We don’t want to do that now, but let’s see how we can combine the two fields into a single multivector potential. Letting the gradient act bidirectionally, and introducing a dummy grade-two selection into the mix, we have
\label{eqn:maxwellLagrangian:2090}
\begin{aligned}
F
&= \rgrad \wedge A + I \lr{ \rgrad \wedge K } \\
&= – A \wedge \lgrad – I \lr{ K \wedge \lgrad } \\
\end{aligned}

Now, we call
\label{eqn:maxwellLagrangian:2110}
N = A + I K,

(a 1,3 multivector), the multivector potential, and write the electromagnetic field not in terms of curls explicitly, but using a grade-2 selection filter
\label{eqn:maxwellLagrangian:2130}

We can now form the following multivector Lagrangian
\label{eqn:maxwellLagrangian:2150}
\LL = \inv{2} F^2 – \gpgrade{ N \lr{ J – I M } }{0,4},

and vary the action to (eventually) find our multivector Maxwell’s equation, without ever resorting to coordinates. We have
\label{eqn:maxwellLagrangian:2170}
\begin{aligned}
\delta S
&= \int d^4 x \inv{2} \lr{ \lr{ \delta F } F + F \lr{ \delta F } } – \gpgrade{ \delta N \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ \lr{ \delta F } F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ -\gpgradetwo{ \lr{ \delta N} \lgrad } F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ -\gpgradetwo{ \lr{ \delta N} \lrgrad } F +\gpgradetwo{ \lr{ \delta N} \rgrad } F – \lr{ \delta N } \lr{ J – I M } }{0,4}.
\end{aligned}

The $$\lrgrad$$ term can be evaluated using the fundamential theorem of GC, and will be zero, as $$\delta N = 0$$ on the boundary. Let’s look at the next integrand term a bit more carefully
\label{eqn:maxwellLagrangian:2190}
\begin{aligned}
&=
\gpgrade{ \gpgradetwo{ \lr{ \lr{ \delta A } + I \lr{ \delta K } } \rgrad } F }{0,4} \\
&=
\gpgrade{ \lr{ \lr{\delta A} \wedge \rgrad + I \lr{ \lr{ \delta K } \wedge \rgrad }} F }{0,4} \\
&=
\gpgrade{ \lr{\delta A} \rgrad F – \lr{ \lr{\delta A} \cdot \rgrad} F + I \lr{ \delta K } \rgrad F – I \lr{ \lr{ \delta K } \cdot \rgrad} F }{0,4} \\
&=
&=
\gpgrade{ \lr{ \lr{\delta A} + I \lr{ \delta K} } \rgrad F }{0,4} \\
&=
\end{aligned}

so
\label{eqn:maxwellLagrangian:2210}
\begin{aligned}
\delta S
&= \int d^4 x \gpgrade{ \lr{ \delta N} \rgrad F – \lr{ \delta N } \lr{ J – I M } }{0,4} \\
&= \int d^4 x \gpgrade{ \lr{ \delta N} \lr{ \rgrad F – \lr{ J – I M } } }{0,4}.
\end{aligned}

for this to be zero for all variations $$\delta N$$ of the 1,3-multivector potential $$N$$, we must have
\label{eqn:maxwellLagrangian:2230}
\grad F = J – I M.

This is Maxwell’s equation, as desired, including both electric and (if desired) magnetic sources.