Chapter 3 - Derivatives - 3.6 Derivatives as Rates of Change - 3.6 Exercises - Page 182: 24

\[\begin{align} & \text{a}\text{. }\overline{C}\left( x \right)=-0.04x+100+\frac{800}{x},\text{ }C'\left( x \right)=-0.08x+100 \\ & \text{b}\text{.}\overline{C}\left( 500 \right)=\$81.60/\text{item},\text{ }C'\left( 500 \right)=\$60/\text{item } \\ & \text{c}\text{.The average cost per item when 500 items are produced is } \\ & \$81.60/\text{item}\text{.} \\ & \text{The cost of producing 501st item is 60. }\!\\\!\!\text{ } \\ \end{align}\]

\[\begin{align} & \text{Let }C\left( x \right)=-0.04{{x}^{2}}+100x+800,\text{for }0\le x\le 1000,\text{ }a=500 \\ & \\ & \text{a}\text{. The average cost function is given by: }\overline{C}\left( x \right)=\frac{C\left( x \right)}{x},\text{ then} \\ & \overline{C}\left( x \right)=\frac{C\left( x \right)}{x}=\frac{-0.04{{x}^{2}}+100x+800}{x} \\ & \text{Simplifying} \\ & \overline{C}\left( x \right)=-0.04x+100+\frac{800}{x} \\ & \text{The marginal cost function is }C'\left( x \right) \\ & C'\left( x \right)=\frac{d}{dx}\left[ -0.04{{x}^{2}}+100x+800 \right] \\ & C'\left( x \right)=-0.08x+100 \\ & \\ & \text{b}\text{. The average and marginal cost when }x=a=500\text{ is} \\ & \overline{C}\left( 500 \right)=-0.04\left( 500 \right)+100+\frac{800}{500} \\ & \overline{C}\left( 500 \right)=81.6 \\ & or \\ & \overline{C}\left( 500 \right)=\$81.60/\text{item} \\ & \\ & and \\ & \\ & C'\left( 500 \right)=-0.08\left( 500 \right)+100 \\ & C'\left( 500 \right)=60 \\ & C'\left( 500 \right)=\$60/\text{item} \\ & \\ & \text{c}\text{. From the result of the part b, the average cost per item when} \\ & \text{500 items are produced is }\$81.60/\text{item}\text{.} \\ & \text{The cost of producing 501st item is 60. }\!\\\!\!\text{ } \\ \end{align}\]