<h1>Hint</h1>

<p>For profit maximization, either $x_1$ or $x_2$ (but not both) must be zero. If not, say $x_1^*&gt;x_2^*&gt;0$ at the optimum, then one could increase profit by lowering cost by reducing $x_2^*$ without affecting output and thus revenue. </p>

<p>Let $z=\max\{x_1,8x_2\}$. The profit function can be written as 
\begin{equation}
p[3(x_3)^{1/3}(z)^{1/3}]-w_3x_3-c_zz,\tag1
\end{equation}
where 
\begin{equation}
c_z=
\begin{cases}
w_1&amp;\text{if }x_1&gt;8x_2\\
w_2/8&amp;\text{if }x_1&lt;8x_2\\
\min\{w_1,w_2/8\}&amp;\text{if }x_1=8x_2
\end{cases}
\end{equation}</p>

<p>Solve $(1)$ for the condition determining the optimal level of $z$, and compare the costs of achieving this level using either $x_1$ or $x_2$. Use the one that entails the lower cost.</p>

CC BY-SA 4.0
2018-09-01T22:49:59.303Z
I like this response so much, it is extremely clear, thank you! However, wouldn't it be the other way around? You'd choose z based on the comparison of w1 against w2/8, or am I in the wrong? My point is: you decide product after evaluating prices, I'm not sure how you'd choose z and thus the w to use if you don't know wether x1 or 8x2 is greater. I hope you can help me once more. Thank you!
mudcake
CC BY-SA 4.0
2018-09-02T01:36:58.763Z
@mudcake: You're welcome. And yes, ultimately, whether $x_1$ or $x_2$ gets chosen depends on the comparison between $w_1$ and $w_2/8$. Note that $z$ is just a placeholder in the maximization. Perhaps put in another way, you'd have compare $$\pi_1=p[3(x_3)^{1/3}(x_1)^{1/3}]-w_3x_3-w_1x_1$$ with $$\pi_2=p[3(x_3)^{1/3}(8x_2)^{1/3}]-w_3x_3-w_2x_2$$ and see which input leads to a higher profit. The above two profit maximization problems should be straightforward to solve and compare.
Herr K.
CC BY-SA 4.0
2018-09-01T19:22:48.453Z
2018-09-01T19:22:48.453Z
Herr K.
<p>Let us first solve the cost minimization problem to derive the cost function and then use that to solve the profit maximization problem:
<span class="math-container">$$\begin{aligned}  \min_{x_1,x_2,x_3} \quad &amp; w_1x_1+w_2x_2+w_3x_3\\ \textrm{s.t.} \quad &amp; 3(x_3)^{1/3}[\max(x_1,8x_2)]^{1/3}\geq y\\ \\ \min_{x_3}&amp; {\begin{aligned} \begin{bmatrix} \min_{x_1,x_2} \quad &amp; w_1x_1+w_2x_2+w_3x_3 \\ \textrm{s.t.} \quad &amp; \max(x_1,8x_2)\geq \frac{y^3}{27x_3} \end{bmatrix} \end{aligned}} \\ \\ \text{gives} \; &amp;(x_1,x_2)^d (w_1,w_2,y,x_3)\in \left\{\begin{matrix} \left \{ (\frac{y^3}{27x_3},0),(0,\frac{y^3}{216x_3}) \right \} &amp; \text{if} \; \frac{w_1}{w_2}=\frac{1}{8}\\  \left \{ (\frac{y^3}{27x_3},0) \right \} &amp; \text{if} \; \frac{w_1}{w_2}&lt;\frac{1}{8} \\  \left \{ (0,\frac{y^3}{216x_3}) \right \} &amp; \text{if} \; \frac{w_1}{w_2}&gt; \frac{1}{8} \end{matrix}\right.\\ \text{for}\; \frac{w_1}{w_2} \leq \frac{1}{8}: \; &amp;\min_{x_3 &gt;0 } \; w_3x_3+w_1\frac{y^3}{27x_3} \\ \text{for}\; \frac{w_1}{w_2}&gt; \frac{1}{8}: \; &amp;\min_{x_3 &gt;0 } \; w_3x_3+w_2\frac{y^3}{216x_3} \\ \therefore \; &amp; \boxed{(x_1,x_2,x_3)^d(w_1,w_2,w_3,y)= \left\{\begin{matrix} (\sqrt{\frac{w_3y^3}{27w_1}},0,\sqrt{\frac{w_1y^3}{27w_3}})&amp; \; \text{if} \; \frac{w_1}{w_2}\leq \frac{1}{8}  \\  (0,\sqrt{\frac{w_3y^3}{216w_2}} ,\sqrt{\frac{w_2y^3}{216w_3}})&amp; \; \text{if} \; \frac{w_1}{w_2} &gt; \frac{1}{8}  &amp;  \end{matrix}\right.} \end{aligned}$$</span></p>
<p>using the input demands, we can determine the cost function: <span class="math-container">$$\boxed{C(w_1,w_2,w_3,y)=\left\{\begin{matrix} y^{3/2}(\frac{w_1w_3}{27})^{1/2} &amp; \text{if} \; \frac{w_1}{w_2} \leq \frac{1}{8}\ \\ y^{3/2}(\frac{w_2w_3}{27})^{1/2} &amp; \text{if} \; \frac{w_1}{w_2}&gt;\frac{1}{8}\end{matrix}\right.}$$</span></p>
<p>Since the technology is of DRS nature and assuming that the firm is a price taker, we can obtain the supply function by using the FOC of the profit maximization problem of a competitive firm:
<span class="math-container">$$\begin{aligned}
&amp; p=C'(y)=\left\{\begin{matrix} \frac{3}{2}(\frac{w_1w_3}{27})^{1/2}y^{1/2} &amp; \text{if} \; \frac{w_1}{w_2} \leq \frac{1}{8}\ \\ \frac{3}{2}(\frac{w_2w_3}{27})^{1/2}y^{1/2} &amp; \text{if} \; \frac{w_1}{w_2}&gt;\frac{1}{8}\end{matrix}\right. \\ \\
\text{supply function is} \; &amp; \boxed{y^s(p,w_1,w_2,w_3)=\left\{\begin{matrix} \frac{12}{w_1w_3}p^2 &amp; \text{if} \; \frac{w_1}{w_2} \leq \frac{1}{8}\ \\ \frac{12}{w_2w_3}p^2 &amp; \text{if} \; \frac{w_1}{w_2}&gt;\frac{1}{8}\end{matrix}\right.}
\end{aligned}$$</span></p>

CC BY-SA 4.0
2023-04-19T14:11:07.270Z
2023-04-19T14:48:49.497Z
2023-04-19T14:48:49.497Z
mynameparv
<p>I've been presented with the following problem:</p>

<p>$$y=3(x_3)^{\frac13}(\max\{x_1,8x_2\})^{\frac13}$$</p>

<p>And the objective is to both maximize profit and minimize cost. First of all, if the problems are dual, does that mean the result will be the same in variables such as demands?</p>

<p><strong>STILL, MY BIGGEST ISSUE IS THIS:</strong> When you get rid of the <em>max</em>, this becomes a Cobb Douglas piece of cake. But I just don't understand how to do that. So far, all I've got is that this function allows corner solutions, so it's not solved just as a Leontieff function. How would you go about choosing each good? I'm also sure it has to do with input prices $w_1$ and $w_2$</p>

<production-function><cost><profit-maximization><cost-functions>
Decision over "max" production function:
mudcake
# Decision over "max" production function:

(Asked by: username_0 on 2018-09-01)

I've been presented with the following problem:

$$y=3(x\_3)^{\frac13}(\max\{x\_1,8x\_2\})^{\frac13}$$

And the objective is to both maximize profit and minimize cost. First of all, if the problems are dual, does that mean the result will be the same in variables such as demands?

**STILL, MY BIGGEST ISSUE IS THIS:** When you get rid of the *max*, this becomes a Cobb Douglas piece of cake. But I just don't understand how to do that. So far, all I've got is that this function allows corner solutions, so it's not solved just as a Leontieff function. How would you go about choosing each good? I'm also sure it has to do with input prices $w\_1$ and $w\_2$

**Tags:** production-function, cost, profit-maximization, cost-functions


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## 2 answers

### Answer by username_1 on 2018-09-01. Score: 4

# Hint

For profit maximization, either $x\_1$ or $x\_2$ (but not both) must be zero. If not, say $x\_1^\*>x\_2^\*>0$ at the optimum, then one could increase profit by lowering cost by reducing $x\_2^\*$ without affecting output and thus revenue.

Let $z=\max\{x\_1,8x\_2\}$. The profit function can be written as
\begin{equation}
p[3(x\_3)^{1/3}(z)^{1/3}]-w\_3x\_3-c\_zz,\tag1
\end{equation}
where
\begin{equation}
c\_z=
\begin{cases}
w\_1&\text{if }x\_1>8x\_2\\
w\_2/8&\text{if }x\_1<8x\_2\\
\min\{w\_1,w\_2/8\}&\text{if }x\_1=8x\_2
\end{cases}
\end{equation}

Solve $(1)$ for the condition determining the optimal level of $z$, and compare the costs of achieving this level using either $x\_1$ or $x\_2$. Use the one that entails the lower cost.

> **Comments:**
> **username_0**: I like this response so much, it is extremely clear, thank you! However, wouldn't it be the other way around? You'd choose z based on the comparison of w1 against w2/8, or am I in the wrong? My point is: you decide product after evaluating prices, I'm not sure how you'd choose z and thus the w to use if you don't know wether x1 or 8x2 is greater. I hope you can help me once more. Thank you!  
> **username_1**: @username_0: You're welcome. And yes, ultimately, whether $x_1$ or $x_2$ gets chosen depends on the comparison between $w_1$ and $w_2/8$. Note that $z$ is just a placeholder in the maximization. Perhaps put in another way, you'd have compare $$\pi_1=p[3(x_3)^{1/3}(x_1)^{1/3}]-w_3x_3-w_1x_1$$ with $$\pi_2=p[3(x_3)^{1/3}(8x_2)^{1/3}]-w_3x_3-w_2x_2$$ and see which input leads to a higher profit. The above two profit maximization problems should be straightforward to solve and compare.  

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### Answer by username_2 on 2023-04-19. Score: 0

Let us first solve the cost minimization problem to derive the cost function and then use that to solve the profit maximization problem:
$$\begin{aligned}  \min\_{x\_1,x\_2,x\_3} \quad & w\_1x\_1+w\_2x\_2+w\_3x\_3\\ \textrm{s.t.} \quad & 3(x\_3)^{1/3}[\max(x\_1,8x\_2)]^{1/3}\geq y\\ \\ \min\_{x\_3}& {\begin{aligned} \begin{bmatrix} \min\_{x\_1,x\_2} \quad & w\_1x\_1+w\_2x\_2+w\_3x\_3 \\ \textrm{s.t.} \quad & \max(x\_1,8x\_2)\geq \frac{y^3}{27x\_3} \end{bmatrix} \end{aligned}} \\ \\ \text{gives} \; &(x\_1,x\_2)^d (w\_1,w\_2,y,x\_3)\in \left\{\begin{matrix} \left \{ (\frac{y^3}{27x\_3},0),(0,\frac{y^3}{216x\_3}) \right \} & \text{if} \; \frac{w\_1}{w\_2}=\frac{1}{8}\\  \left \{ (\frac{y^3}{27x\_3},0) \right \} & \text{if} \; \frac{w\_1}{w\_2}<\frac{1}{8} \\  \left \{ (0,\frac{y^3}{216x\_3}) \right \} & \text{if} \; \frac{w\_1}{w\_2}> \frac{1}{8} \end{matrix}\right.\\ \text{for}\; \frac{w\_1}{w\_2} \leq \frac{1}{8}: \; &\min\_{x\_3 >0 } \; w\_3x\_3+w\_1\frac{y^3}{27x\_3} \\ \text{for}\; \frac{w\_1}{w\_2}> \frac{1}{8}: \; &\min\_{x\_3 >0 } \; w\_3x\_3+w\_2\frac{y^3}{216x\_3} \\ \therefore \; & \boxed{(x\_1,x\_2,x\_3)^d(w\_1,w\_2,w\_3,y)= \left\{\begin{matrix} (\sqrt{\frac{w\_3y^3}{27w\_1}},0,\sqrt{\frac{w\_1y^3}{27w\_3}})& \; \text{if} \; \frac{w\_1}{w\_2}\leq \frac{1}{8}  \\  (0,\sqrt{\frac{w\_3y^3}{216w\_2}} ,\sqrt{\frac{w\_2y^3}{216w\_3}})& \; \text{if} \; \frac{w\_1}{w\_2} > \frac{1}{8}  &  \end{matrix}\right.} \end{aligned}$$

using the input demands, we can determine the cost function: $$\boxed{C(w\_1,w\_2,w\_3,y)=\left\{\begin{matrix} y^{3/2}(\frac{w\_1w\_3}{27})^{1/2} & \text{if} \; \frac{w\_1}{w\_2} \leq \frac{1}{8}\ \\ y^{3/2}(\frac{w\_2w\_3}{27})^{1/2} & \text{if} \; \frac{w\_1}{w\_2}>\frac{1}{8}\end{matrix}\right.}$$

Since the technology is of DRS nature and assuming that the firm is a price taker, we can obtain the supply function by using the FOC of the profit maximization problem of a competitive firm:
$$\begin{aligned}
& p=C'(y)=\left\{\begin{matrix} \frac{3}{2}(\frac{w\_1w\_3}{27})^{1/2}y^{1/2} & \text{if} \; \frac{w\_1}{w\_2} \leq \frac{1}{8}\ \\ \frac{3}{2}(\frac{w\_2w\_3}{27})^{1/2}y^{1/2} & \text{if} \; \frac{w\_1}{w\_2}>\frac{1}{8}\end{matrix}\right. \\ \\
\text{supply function is} \; & \boxed{y^s(p,w\_1,w\_2,w\_3)=\left\{\begin{matrix} \frac{12}{w\_1w\_3}p^2 & \text{if} \; \frac{w\_1}{w\_2} \leq \frac{1}{8}\ \\ \frac{12}{w\_2w\_3}p^2 & \text{if} \; \frac{w\_1}{w\_2}>\frac{1}{8}\end{matrix}\right.}
\end{aligned}$$


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