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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