1. ## Business Calculus - Need help! :(

I've no idea is it the right section to post this,
But I had some problems dealing with this problem.

Total costs

Market price

Tax (per unit produced)

Find:
1. Quantity to produce to maximise profits
2. What should t be, so as to maximise tax revenue?

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

What I've done:

I derived the profit function as:

Then, I differentiate the profit function:

Then I'm stuck cos' I can't find x.

2. Originally Posted by geekie
I've no idea is it the right section to post this,
But I had some problems dealing with this problem.

Total costs

Market price

Tax (per unit produced)

Find:
1. Quantity to produce to maximise profits
2. What should t be, so as to maximise tax revenue?

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

What I've done:

I derived the profit function as:

Then, I differentiate the profit function:

Then I'm stuck cos' I can't find x.

Conditional denotations:

$\displaystyle {\text{TC}}\left( x \right)$ is function of total costs
$\displaystyle {\text{P}}\left( x \right)$ is function of market price
$\displaystyle {\text{TR}}\left( x \right)$ is function of revenue
$\displaystyle {\text{T}}\left( x \right)$ is function of tax
$\displaystyle {\text{Pr}}\left( x \right)$ is function of profit

$\displaystyle \begin{gathered}{\text{TC}}\left( x \right) = \frac{7}{8}x^2 + 5x + 1 \hfill \\{\text{P}}\left( x \right) = 15 - \frac{3}{8}x{\text{ }} \Rightarrow {\text{ TR}}\left( x \right) = x \cdot {\text{P}}\left( x \right) = x\left( {15x - \frac{3}{8}x^2 } \right) = 15x - \frac{3}{8}x^2 \hfill \\{\text{T}}\left( x \right) = tx \hfill \\ \end{gathered}$

$\displaystyle \begin{gathered}{\text{Pr}}\left( x \right) = {\text{TR}}\left( x \right) - {\text{TC}}\left( x \right) - {\text{T}}\left( x \right) = \hfill \\= 15x - \frac{3}{8}x^2 - \left( {\frac{7}{8}x^2 + 5x + 1} \right) - tx = 10x - \frac{5}{4}x^2 - tx - 1 \hfill \\ \end{gathered}$

$\displaystyle \begin{gathered}{\text{Pr}}\left( x \right) \to \max {\text{ if }}\frac{d}{{dx}}{\text{Pr}}\left( x \right) \to 0 \hfill \\\frac{d}{{dx}}{\text{Pr}}\left( x \right) = \frac{d}{{dx}}\left( {10x - \frac{5}{4}x^2 - tx - 1} \right) = 10 - \frac{5}{2}x - t \hfill \\ \end{gathered}$

$\displaystyle 10 - \frac{5}{2}x - t = 0 \Leftrightarrow \frac{5}{2}x = 10 - t \Leftrightarrow 5x = 20 - 2t \Leftrightarrow \left[ \begin{gathered}x = 4 - \frac{2}{5}t \hfill \\t = 10 - \frac{5}{2}x \hfill \\ \end{gathered} \right.$