4
From FOC, we know that:
\begin{align}
\nabla_x\pi(\mathbf{x},\mathbf{w})=p\nabla f(\mathbf{x})-\mathbf{w}=\mathbf{0} \tag{1}
\end{align}
This will be true at equilibrium, i.e. for any given $\mathbf{w}$, the input vector $\mathbf{x}$ will adjust so that the above holds.
Now consider $d\pi(\mathbf{x},\mathbf{w})/d w_i$ (and using $(1)$):
\begin{align}
\frac{d\...
2
(Without using differentiation) When $w \leq w'$ it follows that $pf(x) − w · x \geq pf(x) − w' · x$ and so $\pi(p,w) \geq \pi(p,w')$.
EDIT 1. The last inequality (first left as an exercise) can be justified as follow: $w \leq w'$ implies that
$$pf(x) − w · x \geq pf(x) − w' · x$$
for any $x \geq 0$ and admissible. The inequality is in particular true for $x=...
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