Chapter 3 - Derivatives - 3.6 Derivatives as Rates of Change - 3.6 Exercises - Page 182: 23

\[\begin{align} & \text{a}\text{. }\overline{C}\left( x \right)=-0.01x+40+\frac{100}{x},\text{ }C'\left( x \right)=-0.02x+40 \\ & \text{b}\text{.}\overline{C}\left( 1000 \right)=\$30.10/\text{item},\text{ }C'\left( 1000 \right)=\$20/\text{item } \\ & \text{c}\text{.The average cost per item when 1000 items are produced is } \\ & \$30.10/\text{item}\text{.} \\ & \text{The cost of producing 1001st item is 20. }\!\\\!\!\text{ } \\ \end{align}\]

\[\begin{align} & \text{Let }C\left( x \right)=-0.01{{x}^{2}}+40x+100,\text{for }0\le x\le 1500,\text{ }a=1000 \\ & \\ & \text{a}\text{. The average cost function is given by: }\overline{C}\left( x \right)=\frac{C\left( x \right)}{x},\text{ then} \\ & \overline{C}\left( x \right)=\frac{C\left( x \right)}{x}=\frac{-0.01{{x}^{2}}+40x+100}{x} \\ & \text{Simplifying} \\ & \overline{C}\left( x \right)=-0.01x+40+\frac{100}{x} \\ & \text{The marginal cost function is }C'\left( x \right) \\ & C'\left( x \right)=\frac{d}{dx}\left[ -0.01{{x}^{2}}+40x+100 \right] \\ & C'\left( x \right)=-0.02x+40 \\ & \\ & \text{b}\text{. The average and marginal cost when }x=a=1000\text{ is} \\ & \overline{C}\left( 1000 \right)=-0.01\left( 1000 \right)+40+\frac{100}{1000} \\ & \overline{C}\left( 1000 \right)=30.1 \\ & or \\ & \overline{C}\left( 1000 \right)=\$30.10/\text{item} \\ & \\ & and \\ & \\ & C'\left( 1000 \right)=-0.02\left( 1000 \right)+40 \\ & C'\left( 1000 \right)=20 \\ & C'\left( 1000 \right)=\$20/\text{item} \\ & \\ & \text{c}\text{. From the result of the part b, the average cost per item when} \\ & \text{1000 items are produced is }\$30.10/\text{item}\text{.} \\ & \text{The cost of producing 1001st item is 20. }\!\\\!\!\text{ } \\ \end{align}\]