Question

7.    Given that TC=2000+100Q-10Q²+Q³ where Q is level of output and maximum level of MP is 15 units, answer the following question based on the given information.

A.   Determine the TFC and TVC functions

B.   Find the AVC,AFC,AC and MC functions

C.   Calculate the level of output at which AVC reaches its minimum

D.   Calculate the level of output at which AC reaches minimum

E.    Draw the relationship between production and cost curves

F.    What relationship do you observe between production and cost curves? Why is this so?

G.   At 5 units AVC,APL is 10,what is the maximum APL? 

Answer 1

Given

$TC=2000+100Q-10Q^2+Q^3$

The maximum level of MP = 15 units 

a)TFC:- refers to the costs of the assets that do not change with the change in the level of output.

$TC=2000+100Q-10Q^2+Q^3$

here fixed cost is 2000 as it does not change with the change in the level of Output.

TFC = 2000

TVC:- refers to the costs of the assets that change with the change in the level of output.

$TC=2000+100Q-10Q^2+Q^3$

here variable cost is $100Q-10Q^2+Q^3$  as it changes with the change in the level of Output.

$TVC = 100Q-10Q^2+Q^3$

b)

AC:- Divide the total cost with Q

$AC = \frac{TC}{Q}\\ AC = \frac{2000+100Q-10Q2+Q3}{Q}\\ AC =\frac{ 2000}{Q+100-10Q+Q^2}$

AVC:- Divide the TVC with Q

$AVC =\frac{TVC}{Q}\\ AVC = \frac{100Q-10Q^2+Q^3}{Q}\\ AVC = 100-10Q+Q^2$

AFC:- Divide the Fixed Cost with respect to Q

$AFC = \frac{TFC}{Q}\\ AFC = \frac{2000}{Q}$

MC:- refers to the change in TC that comes from making or producing one additional unit. 

$MC =\frac{ dTC}{dQ}\\ MC = 100 - 20Q + 3Q^2$

c)

Output at which AVC minimum is calculated as 

• Take the derivative of AVC
• Put derivative of AVC = 0

$AVC = 100-10Q+Q^2\\ \frac{dAVC}{ dQ} = −10 + 2Q$

now put the $\frac{dAVC}{ dQ} =0,$  we get

$−10 + 2Q = 0\\ 2Q = 10\\ Q = \frac{10}{2}\\ Q = 5$

AVC is minimum at Q = 5

d)

Output at which AC minimum is calculated as 

MC = AC when AC is Minimum

$100 − 20Q + 3Q ^2 = \frac{2000} {Q} +100−10Q+Q ^2\\ 100 − 20Q + 3Q ^2 − 100+ 10Q−Q ^2 = \frac{2000}{ Q}\\ − 10Q + 2Q^ 2 =\frac{ 2000}{ Q}\\− 10Q ^2 + 2Q ^3 = 2000 \\ 2Q ^3 − 10Q ^2 −2000 = 0\\ Q= 11.974 Approx$

e)

f)

• when MC is Minimum MP is Maximum
• MC cuts AC at its lowest point where MP cuts AP at its Maximum Point but the output is the same.
• there is a point where MC = AC = AP = MP.
• After all, these are equal MP and AP start falling whereas MC and AVC start Increasing

g)

$AVC =\frac{APL}{Q}\\APL=10\times 5=50$