Solving for profit function $\pi (w,p)$ given the output of production function $f(z) = \sqrt{2z_1 + 3z_2}$ Solving for profit function $\pi (w,p)$ given the output production function $f(z) = \sqrt{2z_1 + 3z_2}$.
I approached this problem by trying to solve the $p\nabla f(z) = w$. This is derived from setting up the Lagrangian for the Profit Maximization Problem,
\begin{align*}
\text{maximize } &amp;pf(z)-w^Tz\\\
\Rightarrow \mathcal{L}(z) &amp;= pf(z) -w^Tz
\end{align*}
Then taking the partial of the Lagrange to zero,
\begin{align*}
\frac{\partial \mathcal{L}}{\partial z} = 0 = p\nabla f(z) - w\\
\Rightarrow p\nabla f(z) = w.
\end{align*}
The issue is, I thought that I could solve for an optimal $z^*$, but that does not seem possible, but I know that a solution exists.
To showing this issue simply, let $q=f(z)$, then the gradient is:
\begin{align}
\nabla f(z) = \begin{bmatrix}
1/q\\
3/2q
\end{bmatrix}
\end{align}
So solving our equation $p\nabla f(z) = w$, should let us solve for $z_1,z_2$, but as you can see,
\begin{align*}
\begin{bmatrix}
1/q\\
3/2q
\end{bmatrix} = \begin{bmatrix}
w_1/p\\
w_2/p
\end{bmatrix}\\
\Rightarrow \begin{bmatrix}
q\\
q
\end{bmatrix} = \begin{bmatrix}
p/w_1\\
3p/2w_2
\end{bmatrix}\\
\Rightarrow \begin{bmatrix}
\sqrt{2z_1 + 3z_2}\\
\sqrt{2z_1 + 3z_2}
\end{bmatrix} = \begin{bmatrix}
p/w_1\\
3p/2w_2
\end{bmatrix}
\end{align*}
This shows I cannot isolate $z_1$ or $z_2$. Without an optimal $z^*=&lt;z_1^*,z_2^*&gt;$, I cannot find my profit function $\pi(w,p) = pf(z^*) - w^Tz^*$.
EDIT
My guess is that actually, for some output $q$, my profit function is what I already solved for, $\pi(w,p)=\max{\{pw_1, 3p/2w_2\}}$
Can anyone confirm this?
The production function has a particular feature: the inputs are perfectly substitutable. One unit of input 1 can be substituted by 2/3 of input 2 to produce the same quantity of output.
Intuitively, a producer would optimally use only one output to produce.
Suppose the production plan is $(z_1,z_2)$. By choosing $(z_1-1,z_2+2/3)$, the firm produces the same quantity but profits are increased by $w_1-2/3w_2$. We can conclude: the producer would buy only input 1 ($z_2=0$) if $w_1&lt;2w_2/3$, and they would buy only input 2 ($z_1=0$) if $w_1&gt;2w_2/3$. From that point, you can solve the maximisation problem by distinguishing these two cases.
The question can also be solved as a constrained optimisation problem if you add the constraints $z_1\geq 0$ and $z_2\geq 0$ to the Lagrangian. If $\lambda_1$ and $\lambda_2$ denote the Lagrangian parameters of these two constraints, you obtain the two cases aforementioned when i) $\lambda_1=0$ and ii) $\lambda_2=0$.