# 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 } &pf(z)-w^Tz\\\ \Rightarrow \mathcal{L}(z) &= 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 $$z^*=$$, 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?

• I am guessing it involves inequalities? Oct 11 '20 at 0:59

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 $$w_1<2w_2/3$$, and they would buy only input 2 ($$z_1=0$$) if $$w_1>2w_2/3$$. 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) $$\lambda_2=0$$.
Given the production function $$\sqrt{2z_1+3z_2}$$, cost function can be obtained by minimizing cost: $$\begin{eqnarray*} \min_{z_1, z_2} \ \ w_1z_1+w_2z_2 \\ \text{s.t.} \sqrt{2z_1+3z_2} \geq q\end{eqnarray*}$$ Solving it we get conditional input demand as follows: $$\begin{eqnarray*} (z_1, z_2) = \begin{cases} \left(\frac{q^2}{2}, 0\right) \ \text{if } \frac{w_1}{w_2} \leq \frac{2}{3} \\ \left(0,\frac{q^2}{3}\right) \ \text{if } \frac{w_1}{w_2} \geq \frac{2}{3} \end{cases} \end{eqnarray*}$$ Cost function is therefore, $$C(w_1,w_2, q)=\left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]q^2$$.
Now to obtain the profit function we solve the following problem $$\begin{eqnarray*} \max_{q} \ \ pq- \left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]q^2\end{eqnarray*}$$ and we get the supply function as: $$q(p, w_1, w_2)= \frac{p}{2\left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]}$$ and the optimal profit is $$\pi (p, w_1, w_2)= \frac{p^2}{4\left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]}$$