Answer to Question #237118 in Economics of Enterprise for KEBEDE AJEME

Question

Consider the following total cost function: TC = 2/3Q3 – 10Q2 + 200Q + 50 A) Identify the FC and VC function? B) Calculate AVC, AFC, ATC, and MC functions C) Determine the level of output at which AVC reaches minimum point and the minimum AVC at that level of output?

Answer 1

Here, cost function is given as:$TC = \frac{2}{3}Q^3 – 10Q^2 + 200Q$

A) Using given information, FC would be 50 as it is a constant in the total cost function that is not dependent on output. If FC is zero, total cost will be equal to variable cost. Therefore, VC would be:

$VC = \frac{2}{3}Q^3 – 10Q^2 + 200Q$

B) Similarly, AVC would be:

$VC = \frac{2}{3}Q^3 – 10Q^2 + 200Q\\AVC=\frac{VC}{Q}= \frac{2}{3}Q^2 – 10Q + 200$

AFC would be:

$FC=50\\AFC=\frac{FC}{Q}=\frac{50}{Q}$

ATC would be:

$TC = \frac{2}{3}Q^3 – 10Q^2 + 200Q\\ATC=\frac{TC}{Q}= \frac{2}{3}Q^2 – 10Q+ 200+\frac{50}{Q}$

MC would be:

$TC = \frac{2}{3}Q^3 – 10Q^2 + 200Q\\\frac{dTC}{dQ}=2Q^2-20Q+200$

C) At the minimum of AVC, derivative of AVC will be zero such that:

$AVC= \frac{2}{3}Q^2 – 10Q + 200\\\frac{dAVC}{dQ}=\frac{4}{3}Q-10\\0=\frac{4}{3}Q-10\\10\times\frac{3}{4}=Q\\7.5=Q$

At quantity of 7.5, minimum AVC would be:

$AVC=23Q^2−10Q+200\\AVC=\frac{2}{3}(7.5)^2−10(7.5)+200\\AVC=37.5−75+200\\AVC=232.5$