Solving for profit function $pi (w,p)$ given the output of production function $f(z) = sqrt{2z_1 + 3z_2}$

Question

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 $pnabla f(z) = w$. 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 = pnabla f(z) - w
Rightarrow pnabla 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 $pnabla f(z) = w$, 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 $z^*=<z_1^*,z_2^*>$, 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?

GuiWil · Accepted Answer

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 $w_1<2w_2/3$, 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 $z_1geq 0$ and $z_2geq 0$ 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) $lambda_2=0$.

Amit · Answer

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 qend{eqnarray*}
Solving it we get conditional input demand as follows:
begin{eqnarray*} (z_1, z_2) = begin{cases} left(frac{q^2}{2}, 0right)  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, $C(w_1,w_2, q)=left[minleft(frac{w_1}{2},frac{w_2}{3}right)right]q^2$.
Now to obtain the profit function we solve the following problem
begin{eqnarray*} max_{q}   pq- left[minleft(frac{w_1}{2},frac{w_2}{3}right)right]q^2end{eqnarray*}
and we get the supply function as:
$q(p, w_1, w_2)= frac{p}{2left[minleft(frac{w_1}{2},frac{w_2}{3}right)right]}$
and the optimal profit is
$pi (p, w_1, w_2)= frac{p^2}{4left[minleft(frac{w_1}{2},frac{w_2}{3}right)right]}$

Solving for profit function $pi (w,p)$ given the output of production function $f(z) = sqrt{2z_1 + 3z_2}$

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