# Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 28

The equation $c(x) = 339 + 25x - 0.09x^2 + 0.0004 x^3$ represents the cost function for production of a commodity.

a.) Find and interpret $c'(100)$

b.) Compare $c'(100)$ with the cost of producing the 101st item.

\begin{aligned} \text{a.) } c'(x) &= 25 - 0.18 x + 0.0012 x^2\\ \\ c'(100) &= 25 - 0.18(100) + 0.0012(100)^2\\ \\ c'(100) &= 19 \end{aligned}

$c'(100)$ means that the cost per 100 units of production is changing at a rate of 19 $\displaystyle \frac{\text{unit cost}}{\text{unit production}}$

\begin{aligned} \text{b.) } c(x) &= 339 + 25x - 0.09x^2 + 0.0004x^3\\ \\ c'(101) &= c(101) - c(100)\\ \\ &= 339 + 25(101) - 0.09 (101)^2 + 0.0004(101)^2 - \left[ 339 + 25 (100) - 0.09 (100)^2 + 0.0004 (100)^2\right]\\ \\ c'(101) & = 19.0304 \end{aligned}

It means that the cost increases by 0.0304 as the number of unit production increases by 1 from 100.

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