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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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