Section 3.5 #163: The general profit function is the difference between the revenue and cost funcitons:
$$P(x) = R(x) - C(x).$$
Suppose the price-demand function and cost functions for the produciton of cordless drills is given respectively by
$$p =143-0.03x$$
and
$$C(x) = 75000+65x,$$
where $x$ is the number of cordless drills that are sold at a price of $p$ dollars per drill (i.e. $p$ has units $\dfrac{\text{dollars}}{\text{drill}}$) and $C(x)$ is the cost of producing $x$ cordless drills.
a. Find the marginal cost function.
b. Find the revenue and marginal revenue function.
c. Find $R'(1000)$ and $R'(4000)$. Interpret the results.
d. Find the profit and marginal profit functions.
e. Find $P'(1000)$ and $P'(4000)$. Interpet the results.
Solution a.: By definition (see blue box on pg. 273), the marginal cost is given by the derivative of the cost function. Therefore the marginal cost funciton is
$$C'(x) = \dfrac{\mathrm{d}}{\mathrm{d}x} \Big[ 75000+65x \Big] = 65 \dfrac{\mathrm{dollars}}{\mathrm{drill}}.$$
Solution b.: To find the revenue function, we need to use the equation price-demand function in a clever way. Notice that the units of $p$ are $\dfrac{\mathrm{dollars}}{\mathrm{drill}}$. Since $x$ is measured as the number of drills, then multiplying $p$ by $x$ has the following effect:
$$px = (143-0.03x)x = 143x - 0.03x^2 \mathrm{dollars}.$$
In other words, the revenue function is
$$R(x) = px = 143x - 0.03x^2 \mathrm{dollars}.$$
Again by pg. 273, the "marginal revenue" is simply the derivative of $R(x)$, so compute it:
$$R'(x) = 143 - 0.06x \dfrac{\mathrm{dollars}}{\mathrm{drill}}.$$
Solution c.: Calculate
$$R'(1000) = 143 - 0.06(1000) = 143 - 60 = 83 \dfrac{\mathrm{dollars}}{\mathrm{drill}}$$
and
$$R'(4000) = 143 - 0.06(4000) = 143 - 240 = -97 \dfrac{\mathrm{dollars}}{\mathrm{drill}}.$$
The interpretation of this is that we gain $83$ dollars on "the $1001$st" drill produced but we lose approximately $97$ dollars on "the $4001$st" drill produced.
Solution d.: Since $R(x)=143x-0.03x^2$ and $C(x)=75000+65x$, then the definition of the profit function tells us that
$$\begin{array}{ll}
P(x) &= R(x)-C(x) \\
&= (143x-0.03x^2) - (75000+65x) \\
&= 143x-0.03x^2-75000-65x \\
&= -0.03x^2 + 78x - 75000.
\end{array}$$
According to pg. 273, the "marginal profit" is defined as the derivative of $P$, i.e. the marginal profit is
$$P'(x) = -0.06x + 78.$$
Solution e.: Calculate
$$P'(1000) = -0.06(1000)+78 = -60 + 78 = 18 \dfrac{\text{dollars}}{\text{drill}}$$
and
$$P'(4000) = -0.06(4000)+78 = -240+78 = -162 \dfrac{\text{dollars}}{\text{drill}}.$$
To interpret this: we make a profit of $18$ dollars on "the $1001$st" drill produced, but we lose 162 dollars on "the 4001st" drill produced.
note: in reality, if our profit function was this $P(x)$, then the business essentially will always lose money -- plot this function to see the profit curve: see the graph here
