这是一份uwa西澳大学MATH1012的成功案例

requires that
$$
\begin{gathered}
p_{i}-\sum_{\nu=1}^{s} b_{v i} w_{i} \leqslant 0 \quad \
\mathrm{p}^{\prime} \leqslant \mathrm{w}^{\prime} \mathrm{B} \quad \mathrm{y} \geqslant 0 \quad i=1, \ldots, n
\end{gathered}
$$
or
where $\mathrm{B}$ is $s \times n$, and $w$ is $s \times 1$ and $\mathrm{p}$ is $n \times 1$. This condition maximises profits $\pi(\mathbf{p}, \mathbf{w})$ at the zero level. If there exists a resource constraint of $\mathrm{r}^{0}$ then
$$
r_{v}^{0} \geqslant \sum_{i=1}^{n} r_{v i}=\sum_{i=1}^{n} b_{v i} y_{i} \quad v=1, \ldots, s
$$
i.e.
$$
\mathbf{r}^{0} \geqslant \mathbf{B y}
$$
where, as before, $\mathbf{B}$ is an $s \times n$ matrix, $\mathbf{r}^{0}$ an $s \times 1$ vector of given resource endowments and $\mathrm{y}$ an $n \times 1$ output vector. Hence, for given factor endowment $\mathrm{r}^{0}$ we can find a non-negative output vector $y \geqslant 0$ which maximises $\pi(p, w)$ at the zero level if
$$
\begin{aligned}
&\mathbf{p}^{\prime} \leqslant \mathbf{w}^{\prime} \mathbf{B} \
&\mathbf{r}^{0} \geqslant \mathbf{B y}
\end{aligned}
$$
and

MATH1012 COURSE NOTES ：

The Cobb-Douglas production function, which for the two-input case takes the form
$$
q=a_{0} x_{1}^{a_{1}} x_{2}^{a_{2}} \quad a_{0}, a_{1} a_{2}>0
$$
or more generally,
$$
q=a_{0} \prod_{i=1}^{s} x_{i}^{a_{i}} \quad a_{i}>0 \forall i
$$
The Constant Elasticity of Substitution production function, which for the two-input case takes the form
$$
q=a_{0}\left[a_{1} x_{1}^{-\rho}+a_{2} x_{2}^{-\rho}\right]-v / \rho \quad a_{0}, a_{1}, a_{2}>0, \nu>0, \rho \geqslant-1
$$
or more generally,
$$
q=a_{0}\left[\sum_{i=1}^{s} a_{i} x_{i}^{-\rho}\right]-v / \rho \quad a_{i}>0 \forall i, v>0, \rho \geqslant-1 .
$$