## Calculus with Applications (10th Edition)

a. $\frac{8}{35}$ b. $\frac{31}{240}$ c. $\frac{3x+2}{x^2+4x}$ d. $\frac{-3x^2-4x-8}{(x^2+4x)^2}$
We are given $C(x)=\frac{3x+2}{x+4}$ The average cost is defined by: $\bar{C}(x)=\frac{C(x)}{x}=\frac{3x+2}{x+4}.\frac{1}{x}=\frac{3x+2}{x^2+4x}$ a. 10 units then the average cost is: $\bar{C}(x)=\frac{3x+2}{x^2+4x}=\frac{3(10)+2}{10^2+4(10)}=\frac{8}{35}$ b. 20 units then the average cost is: $\bar{C}(x)=\frac{3x+2}{x^2+4x}=\frac{3(20)+2}{20^2+4(20)}=\frac{31}{240}$ c. x units then the average cost is: $\bar{C}(x)=\frac{3x+2}{x^2+4x}$ d. The marginal average cost is given by: $\frac{d}{dx}[\bar{C}(x)]=\frac{3(x^2+4x)-(2x+4)(3x+2)}{(x^2+4x)^2}$ $=\frac{3x^2+12x-6x^2-4x-12x-8}{(x^2+4x)^2}$ $=\frac{-3x^2-4x-8}{(x^2+4x)^2}$