<ol>
<li>The pre-tax profit is given by <span class="math-container">$TR-TC$</span>. Since the firm has to pay profit times <span class="math-container">$c$</span> in taxes, the net profit equation is given by <span class="math-container">$(TR-TC)(1-c)$</span>. Since this is what the firm actually keeps, this is what the firm maximizes.</li>
<li>The choice variable for the firm is <span class="math-container">$q$</span>. So, the firm wants to choose <span class="math-container">$q$</span> that maximizes <span class="math-container">$\pi$</span>. This is done by taking the derivative of <span class="math-container">$\pi$</span> with respect to <span class="math-container">$q$</span> and setting that equal to zero. If this step is confusing, do an internet search for something like &quot;calculus maximization&quot;.</li>
<li>The author says that <span class="math-container">$(1-c) \ne 0$</span>. This is important so that we can solve for <span class="math-container">$q$</span>. Divide both sides of <span class="math-container">$(360-5q)(1-c)=0$</span> by <span class="math-container">$(1-c)$</span> to get <span class="math-container">$360-5q=0 \implies q=72$</span>. You may want to review the mechanics of solving algebraic equations so that you can focus on the concepts in your class.</li>
</ol>

CC BY-SA 4.0
2023-04-14T18:24:55.317Z
In step 2 why doesn't the author expand the profit function with (1-c), then differentiate from then on?
sffffii
CC BY-SA 4.0
2023-04-14T18:47:27.557Z
Why would the author want to complicate the algebra unnecessarily? This shows that the optimal $q$ is independent of $c$ for $c \in [0,1)$.
A. Miller
CC BY-SA 4.0
2023-04-14T19:32:27.693Z
I don't understand how you can differentiate a function like that though? When there are brackets with one of them containing the variable in question.
sffffii
CC BY-SA 4.0
2023-04-14T22:21:13.470Z
@sffffii You would presumable accept that $\frac{d}{dx}k\,f(x)= k f'(x)$.  Here $k=1-c$, a constant that does not depend on $x$
Henry
CC BY-SA 4.0
2023-04-15T11:42:55.770Z
@Henry oh I see, would you get a similar answer if you decided to expand with 1-c initially ?
sffffii
CC BY-SA 4.0
2023-04-15T11:59:04.350Z
As in $\frac{d}{dx}(1-c)f(x)= \frac{d}{dx}f(x)-\frac{d}{dx}cf(x)=f'(x)-cf'(x)=(1-c)f'(x)$ ?
Henry
CC BY-SA 4.0
2023-04-15T17:43:43.950Z
Yeah I guess so, so if it were 360 - 360c - 5q -5qc, and you took the derivative of that instead - you'd end up at the same answer?
sffffii
CC BY-SA 4.0
2023-04-14T17:39:46.403Z
2023-04-14T17:39:46.403Z
A. Miller
<p>Profit is made up of two functions, total revenue <span class="math-container">$(TR = f(q) = 360 - 2.1q^2)$</span> and total cost <span class="math-container">$(TC = 50 + 0.4q^2)$</span>. In this case, <span class="math-container">$1-c$</span> is the proportion of profit allocated to taxation. In the first equation, the profit equation was simply rearranged to facilitate algebraic manipulations. The following two equations are the first and second order conditions, respectively. Next, the term <span class="math-container">$(1-c)$</span> disappears because the example assumes a <span class="math-container">$100\%$</span> tax on profit, that is, <span class="math-container">$c = 1$</span>, which makes <span class="math-container">$1-c = 0$</span>, resulting in an optimal production of 72 units. In short, since <span class="math-container">$0\leq c \leq 1$</span>, then <span class="math-container">$1-c$</span> only disappears when <span class="math-container">$c=1$</span>.</p>

CC BY-SA 4.0
2024-01-21T08:49:38.620Z
"$1-c$ only disappears when $c=1$." this is false, it disappears due to algebraic manipulation, so (-1).
Giskard
CC BY-SA 4.0
2024-01-10T10:28:18.267Z
2024-01-10T10:28:18.267Z
Right_4ever
<p><a href="https://i.sstatic.net/R30KH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/R30KH.png" alt="enter image description here" /></a></p>
<p>I don't really understand this. Why didn't the author multiply out 1-c then find the first order condition? And why does the 1-c term just disappear?</p>

<mathematical-economics>
Profit tax analysis
sffffii
# Profit tax analysis

(Asked by: username_0 on 2023-04-14)

[![enter image description here](https://i.sstatic.net/R30KH.png)](https://i.sstatic.net/R30KH.png)

I don't really understand this. Why didn't the author multiply out 1-c then find the first order condition? And why does the 1-c term just disappear?

**Tags:** mathematical-economics

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## 2 answers

### Answer by username_1 on 2023-04-14. Score: 1

1. The pre-tax profit is given by $TR-TC$. Since the firm has to pay profit times $c$ in taxes, the net profit equation is given by $(TR-TC)(1-c)$. Since this is what the firm actually keeps, this is what the firm maximizes.
2. The choice variable for the firm is $q$. So, the firm wants to choose $q$ that maximizes $\pi$. This is done by taking the derivative of $\pi$ with respect to $q$ and setting that equal to zero. If this step is confusing, do an internet search for something like "calculus maximization".
3. The author says that $(1-c) \ne 0$. This is important so that we can solve for $q$. Divide both sides of $(360-5q)(1-c)=0$ by $(1-c)$ to get $360-5q=0 \implies q=72$. You may want to review the mechanics of solving algebraic equations so that you can focus on the concepts in your class.

> **Comments:**
> **username_0**: In step 2 why doesn't the author expand the profit function with (1-c), then differentiate from then on?  
> **username_1**: Why would the author want to complicate the algebra unnecessarily? This shows that the optimal $q$ is independent of $c$ for $c \in [0,1)$.  
> **username_0**: I don't understand how you can differentiate a function like that though? When there are brackets with one of them containing the variable in question.  
> **username_2**: @sffffii You would presumable accept that $\frac{d}{dx}k\,f(x)= k f'(x)$.  Here $k=1-c$, a constant that does not depend on $x$  
> **username_0**: @Henry oh I see, would you get a similar answer if you decided to expand with 1-c initially ?  
> **username_2**: As in $\frac{d}{dx}(1-c)f(x)= \frac{d}{dx}f(x)-\frac{d}{dx}cf(x)=f'(x)-cf'(x)=(1-c)f'(x)$ ?  
> **username_0**: Yeah I guess so, so if it were 360 - 360c - 5q -5qc, and you took the derivative of that instead - you'd end up at the same answer?  

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### Answer by username_3 on 2024-01-10. Score: -1

Profit is made up of two functions, total revenue $(TR = f(q) = 360 - 2.1q^2)$ and total cost $(TC = 50 + 0.4q^2)$. In this case, $1-c$ is the proportion of profit allocated to taxation. In the first equation, the profit equation was simply rearranged to facilitate algebraic manipulations. The following two equations are the first and second order conditions, respectively. Next, the term $(1-c)$ disappears because the example assumes a $100\%$ tax on profit, that is, $c = 1$, which makes $1-c = 0$, resulting in an optimal production of 72 units. In short, since $0\leq c \leq 1$, then $1-c$ only disappears when $c=1$.

> **Comments:**
> **username_4**: "$1-c$ only disappears when $c=1$." this is false, it disappears due to algebraic manipulation, so (-1).  

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