Question

1.   By using the knowledge of relationships among the various costs of production concepts fill the blank space of the following table

Quantity TC TFC TVC ATC AVC MC

0 125

  10 5

20 105

30 110

40 255

50 3

60 3

Answer 1

i)TVC=TC-TFC$\therefore$ (125-125=0,(5-125)=-120,(105-125)=-20,(110-125)=-15,(255-125)=130, (3-125)=-122,(3-125)=-122

so TVC values are; 0,-120,-20,-15,130,-122,-122

ii)ATC=$\frac {TC}{OUTPUT}$ $\therefore$ ($\frac{125}{0}$ )=0 ,$(\frac{5}{10})$ =0.5 ,$\frac {105}{20}$ =5.25, $\frac{110}{30}$ =3.66, $\frac{255}{40}$ =6.375, $\frac{3}{50}$ =0.06, $\frac{3}{60}$=0.05

thus TVC values are; 0 , 0.5, 5.25, 3.66, 6.75, 0.06, 0.05

iii)AVC=$\frac{TVC}{OUTPUT}$ =$\frac{0}{0}$ 0, $-\frac{120}{10}$ =-12, $-\frac{20}{20}$=-1 ,$-\frac{15}{30}$ =-0.5, $\frac{130}{40}$ =3.25, $-\frac{122}{50}$ =-2.44, $-\frac{122}{60}$ =-2.033

values of AVC are;0, -12, -1, -0.5, 3.25, -2.44, -2.0333

iv) MC=$\frac{change in TC}{change in output(Q)}$ =($\frac{5-125}{20-10})$ =-12, $(\frac{105-5}{30-20})$ =10, $(\frac{110-105}{40-30})$=0.5, $(\frac{255-105}{50-40})$ =14.5, $(\frac{3-255}{60-50})$ =-12.2

the first and the last MC are =0

the MC values are; 0, -12, 10, 0.5, 14.5,-12.2, 0