<p>Given the production function <span class="math-container">$\sqrt{2z_1+3z_2}$</span>, cost function can be obtained by minimizing cost:
<span class="math-container">\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*}</span>
Solving it we get conditional input demand as follows:
<span class="math-container">\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*}</span>
Cost function is therefore, <span class="math-container">$C(w_1,w_2, q)=\left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]q^2$</span>.</p>
<p>Now to obtain the profit function we solve the following problem
<span class="math-container">\begin{eqnarray*} \max_{q} \ \ pq- \left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]q^2\end{eqnarray*}</span>
and we get the supply function as:
<span class="math-container">$q(p, w_1, w_2)= \frac{p}{2\left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]}$</span>
and the optimal profit is
<span class="math-container">$\pi (p, w_1, w_2)= \frac{p^2}{4\left[\min\left(\frac{w_1}{2},\frac{w_2}{3}\right)\right]}$</span></p>

CC BY-SA 4.0
2021-02-12T12:10:09.590Z
2021-02-12T12:10:09.590Z
Amit
<p>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.</p>
<p>Intuitively, a producer would optimally use only one output to produce.
Suppose the production plan is <span class="math-container">$(z_1,z_2)$</span>. By choosing <span class="math-container">$(z_1-1,z_2+2/3)$</span>, the firm produces the same quantity but profits are increased by <span class="math-container">$w_1-2/3w_2$</span>. We can conclude: the producer would buy only input 1 (<span class="math-container">$z_2=0$</span>) if <span class="math-container">$w_1&lt;2w_2/3$</span>, and they would buy only input 2 (<span class="math-container">$z_1=0$</span>) if <span class="math-container">$w_1&gt;2w_2/3$</span>. From that point, you can solve the maximisation problem by distinguishing these two cases.</p>
<p>The question can also be solved as a constrained optimisation problem if you add the constraints <span class="math-container">$z_1\geq 0$</span> and <span class="math-container">$z_2\geq 0$</span> to the Lagrangian. If <span class="math-container">$\lambda_1$</span> and <span class="math-container">$\lambda_2$</span> denote the Lagrangian parameters of these two constraints, you obtain the two cases aforementioned when i) <span class="math-container">$\lambda_1=0$</span> and ii) <span class="math-container">$\lambda_2=0$</span>.</p>

CC BY-SA 4.0
2020-10-13T13:30:14.810Z
2020-10-13T13:30:14.810Z
GuiWil
<p>Solving for profit function <span class="math-container">$\pi (w,p)$</span> given the output production function <span class="math-container">$f(z) = \sqrt{2z_1 + 3z_2}$</span>.</p>
<p>I approached this problem by trying to solve the <span class="math-container">$p\nabla f(z) = w$</span>. This is derived from setting up the Lagrangian for the Profit Maximization Problem,
<span class="math-container">\begin{align*}
\text{maximize } &amp;pf(z)-w^Tz\\\
\Rightarrow \mathcal{L}(z) &amp;= pf(z) -w^Tz
\end{align*}</span>
Then taking the partial of the Lagrange to zero,
<span class="math-container">\begin{align*}
\frac{\partial \mathcal{L}}{\partial z} = 0 = p\nabla f(z) - w\\
\Rightarrow p\nabla f(z) = w.
\end{align*}</span>
The issue is, I thought that I could solve for an optimal <span class="math-container">$z^*$</span>, but that does not seem possible, but I know that a solution exists.</p>
<p>To showing this issue simply, let <span class="math-container">$q=f(z)$</span>, then the gradient is:
<span class="math-container">\begin{align}
\nabla f(z) = \begin{bmatrix}
1/q\\
3/2q
\end{bmatrix}
\end{align}</span>
So solving our equation <span class="math-container">$p\nabla f(z) = w$</span>, should let us solve for <span class="math-container">$z_1,z_2$</span>, but as you can see,
<span class="math-container">\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*}</span>
This shows I cannot isolate <span class="math-container">$z_1$</span> or <span class="math-container">$z_2$</span>. Without an optimal <span class="math-container">$z^*=&lt;z_1^*,z_2^*&gt;$</span>, I cannot find my profit function <span class="math-container">$\pi(w,p) = pf(z^*) - w^Tz^*$</span>.</p>
<p><strong>EDIT</strong>
My guess is that actually, for some output <span class="math-container">$q$</span>, my profit function is what I already solved for, <span class="math-container">$\pi(w,p)=\max{\{pw_1, 3p/2w_2\}}$</span></p>
<p>Can anyone confirm this?</p>

CC BY-SA 4.0
2020-10-11T00:59:15.493Z
I am guessing it involves inequalities?
Tsangares
<supply-and-demand><profit-maximization>
Solving for profit function $\pi (w,p)$ given the output of production function $f(z) = \sqrt{2z_1 + 3z_2}$
Tsangares
# Solving for profit function $\pi (w,p)$ given the output of production function $f(z) = \sqrt{2z_1 + 3z_2}$

(Asked by: username_0 on 2020-10-10)

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?

**Tags:** supply-and-demand, profit-maximization

> **Comments:**
> **username_0**: I am guessing it involves inequalities?  

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## 2 answers

### Answer by username_2 on 2020-10-13. Score: 5 | 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 $z\_1\geq 0$ 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) $\lambda\_2=0$.


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### Answer by username_1 on 2021-02-12. Score: 5

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
$\pi (p, w\_1, w\_2)= \frac{p^2}{4\left[\min\left(\frac{w\_1}{2},\frac{w\_2}{3}\right)\right]}$


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