Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 54

a.) Prove that if the profit $P(x)$ is a maximum, then the marginal revenue equals the marginal cost.
b.) If $C(x) = 16,000 + 500x - 1.6x^2 = 0.004x^3$ is the cost function and $P(x) = 1700 - 7x$ is the demand function, find the production level that will maximize profit.

The total profit $P(x) = R(x) - C(x)$ where $R(x)$ is the revenue function and $C(x)$ is the cost function
a.) By taking the derivative of $P(x)$ we have,
$P'(x) = R'(x) - C'(x)$
when $P'(x)$ is maximum, then $P'(x) = 0$

$
\begin{equation}
\begin{aligned}
0 &= R'(x) - C'(x) \\
\\
R'(x) &= C'(x)
\end{aligned}
\end{equation}
$


b.) We know that the profit is maximum when $R'(x) = C'(x)$...

$
\begin{equation}
\begin{aligned}
\text{Recall that } R(x) &= xP(x), \text{ so,}\\
\\
R(x) &= 1700x - 7x^2
\end{aligned}
\end{equation}
$

So,
$R'(x) = C'(x)$

$
\begin{equation}
\begin{aligned}
1700 - 14x &= 500 - 3.2x + 0.012x^2\\
\\
0 &= 0.012x^2 + 10.8x - 1200
\end{aligned}
\end{equation}
$


By using Quadratice Formula, we get...
$x = 100$ and $x = -100$

Since $R'"(100) < C''(100)$,
$x = 100$ gives the maximum profit.