Example 5.2.1. Change in revenue with respect to expense, doable two ways.
We can buy widgets wholesale for $10 a widget. In the. retail market, the demand price of widgets is $20 minus 0.1 times the quantity to be sold. Find the derivative of revenue with respect to expense.
Solution 1. Solution A
The revenue and cost functions for widgets depend on the quantity \(q\text{.}\) The formulas for revenue and cost are:
\begin{align*}
\revenue\amp =q(20-0.1q)=20q-0.1q^2\\
\cost\amp =10q\text{.}
\end{align*}
We can solve the second equation for quantity and substitute back into the first equation. This now gives us the revenue function in terms of cost (\(c\)).
\begin{align*}
\quantity \amp =0.1*c\\
\revenue \amp =2c-0.001 c^2\text{.}
\end{align*}
It is straightforward to take the derivative:
\begin{equation*}
\frac{d\revenue}{d\cost}=2-0.002*\cost\text{.}
\end{equation*}
Note that the derivative is positive for cost between $0 and $1000. This implies that the revenue is rising until the cost is $1000. After we hit a cost of $1000, the derivative becomes negative. This indicates that the revenue will actually decrease.
Solution 2. Solution B
The alternative method is to differentiate the equations for revenue (\(r\)) and cost (\(c\)) with respect to quantity (\(q\)), and find the two derivatives \(\frac{d r}{d q}\) and \(\frac{d c}{d q}\text{,}\) then treat them as fractions. The derivative we want is the quotient of these fractions.
The revenue and cost functions for widgets are the same as above.
\begin{align*}
\revenue\amp =20q-0.1 q^2\\
\cost\amp =10q
\end{align*}
We now differentiate:
\begin{align*}
\frac{d r}{d q} \amp =20-0.2 q\\
\frac{d c}{d q}\amp =10
\end{align*}
We divide these derivatives to get the desired derivative.
\begin{equation*}
\frac{\text{change in revenue}}{\text{change in cost}}: \frac{d r}{d c}=\frac{d r}{d q}/\frac{d c}{d q}=(20-0.2 q)/10\text{.}
\end{equation*}
Substituting \(q =0.1 c\) gives the same solution we had from the first method.
