Step 1 The average cost function is given as A(q) = 125/q+ q2/16 -4, where q is the firm's output. 1. What part of the decreasing range is practically feasible? 2. Write an equation for the total cost function. 3. Hence calculate the level of output for which total costs are minimum.

$\begin{array}{l}\left(1\right)\\ \mathrm{Consider} \mathrm{the} \mathrm{average} \mathrm{cost} \mathrm{function} \mathrm{is}\\ A\left(q\right)=\frac{125}{q}+\frac{q^2}{16}-4,\\ \mathrm{where} q \mathrm{is} \mathrm{the} \mathrm{firm}\text{'}\mathrm{s} \mathrm{output}.\\ \mathrm{Now},\mathrm{differentiate} \mathrm{the} \mathrm{function} A \mathrm{w}.\mathrm{r}.\mathrm{t} q,\mathrm{then}\\ A\text{'}\left(q\right)=-\frac{125}{q^2}+\frac{2q}{16}\\ \mathrm{Put} A\text{'}\left(q\right)=0\\ -\frac{125}{q^2}+\frac{2q}{16}=0\\ -\frac{125}{q^2}+\frac{q}{8}=0\\ \frac{q^3-1000}{8q^2}=0\\ \Rightarrow q=10\\ \begin{array}{cccc}\mathrm{No}-& \mathrm{Domain}& \mathrm{Behaviour} \mathrm{of} A\text{'}\left(q\right)& \mathrm{Nature}\\ 1& \mathrm{For} q>10& A\text{'}\left(q\right)>0& \mathrm{Increasing}\\ 2& \mathrm{for} q<10& A\text{'}\left(q\right)<0& \mathrm{Decreasing}\end{array}\\ \mathrm{It} \mathrm{shows} \mathrm{that} \mathrm{cost} \mathrm{function} \mathrm{is} \mathrm{decreaing} \mathrm{in} \mathrm{the} \mathrm{range} \mathrm{of} \mathrm{output} q\in \left(0,10\right).\end{array}$