= Solution
{c}
Use upward <generalized coordinates>. Let $h$ be the displacement of the lower <pulley> and $z$ that of the $4m$ <mass>. The two inextensible <mechanical strings> impose
$$
x+y-2h=0,\qquad z+h=0,
\qquad h=\frac{x+y}{2},\qquad z=-\frac{x+y}{2}.
$$
For ideal massless <pulleys> and <mechanical strings>, only the three <masses> contribute <kinetic energy>. Hence
$$
\begin{aligned}
K&=\frac12(4m)\left(\frac{\dot x+\dot y}{2}\right)^2
+\frac12(5m)\dot x^2+\frac12(3m)\dot y^2\\
&=m(3\dot x^2+\dot x\dot y+2\dot y^2).
\end{aligned}
$$
Up to a constant, their <gravitational potential energy> is
$$
U=4mgz+5mgx+3mgy=mg(3x+y).
$$
The <Lagrangian> of this <compound Atwood machine> is therefore
$$
\boxed{\mathcal L=K-U=m(3\dot x^2+\dot x\dot y+2\dot y^2)-mg(3x+y).}
$$
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