Chapter 55, Problem 1CYU

a)

To determine

MC, VC, AFC, AVC, ATC when output rises from 0 to 6

Explanation of Solution

MC, VC, AFC, AVC, ATC when output rises from 0 to 6 would be:

Quantity of pies	Marginal cost of pie	Variable cost	Average fixed cost of pie	Average variable cost of pie	Average total cost of pie
0		$0
1	$1.00	1	9	1	10
2	1.50	2.50	4.50	1.25	5.75
3	2.25	4.75	3	1.58	4.58
4	3.38	8.13	2.25	2.03	4.28
5	5.06	13.19	1.80	2.64	4.44
6	7.59	20.78	1.50	3.46	4.96

Marginal cost of the pie would be calculated by multiplying the previous cost with 1.5

Variable cost is calculated by summing up the marginal cost

Average fixed cost (AFC) of pie would be calculated by $\frac{\text{9}}{\text{Q}}$

Average variable cost (AVC) of pie would be calculated as variable cost/Q

And, Average total cost = AFC+AVC

Then,

For second pie:

  $\begin{array}{l}\text{MC:}\hfill \\ \text{= }\left(\text{1×1}\text{.5}\right)\hfill \\ \text{= 1}\text{.50}\hfill \\ \text{VC:}\hfill \\ \text{= }\left(\text{1+1}\text{.50}\right)\hfill \\ \text{= 2}\text{.50}\hfill \\ \text{AFC:}\hfill \\ \text{= 9/2}\hfill \\ \text{= 4}\text{.5}\hfill \\ \text{AVC:}\hfill \\ \text{= 2}\text{.5/2}\hfill \\ \text{= 1}\text{.25}\hfill \\ \text{ATC:}\hfill \\ \text{= 4}\text{.5+1}\text{.25}\hfill \\ \text{= 5}\text{.75}\hfill \\ \hfill \end{array}$

For third pie:

MC:

  $\begin{array}{l}=\left(1.5\times 1.5\right)\hfill \\ =2.25\hfill \end{array}$

VC:

  $\begin{array}{l}=\left(1+1.50+2.25\right)\hfill \\ =4.75\hfill \end{array}$

AFC:

  $\begin{array}{l}=\frac{9}{3}\\ =3\end{array}$

AVC:

  $\begin{array}{l}=4.75/3\hfill \\ =1.58\hfill \end{array}$

ATC:

  $\begin{array}{l}=1.58+3\hfill \\ =4.58\hfill \end{array}$

For the fourth pie:

  $\begin{array}{l}\text{MC:}\hfill \\ \text{= }\left(\text{2}\text{.25×1}\text{.5}\right)\hfill \\ \text{= 3}\text{.38}\hfill \\ \text{VC:}\hfill \\ \text{= }\left(\text{4}\text{.75+3}\text{.38}\right)\hfill \\ \text{= 8}\text{.13}\hfill \\ \text{AFC:}\hfill \\ =\frac{9}{4}\hfill \\ \text{= 2}\text{.25}\hfill \\ \text{AVC:}\hfill \\ \text{= 8}\text{.13/4}\hfill \\ \text{= 2}\text{.03}\hfill \\ \text{ATC:}\hfill \\ \text{= 2}\text{.25+2}\text{.03}\hfill \\ \text{= 4}\text{.28}\hfill \end{array}$

For the fifth pie:

MC:

  $\begin{array}{l}=\left(3.38\times 1.5\right)\hfill \\ =5.06\hfill \end{array}$

VC:

  $\begin{array}{l}=\left(8.13+5.06\right)\hfill \\ =13.19\hfill \end{array}$

AFC:

  $\begin{array}{l}=9/5\hfill \\ =1.80\hfill \end{array}$

AVC:

  $\begin{array}{l}=\frac{13.19}{5}\\ =2.64\end{array}$

ATC:

  $\begin{array}{l}=2.64+1.80\hfill \\ =4.44\hfill \end{array}$

For the sixth pie:

  $\begin{array}{l}\text{MC:}\hfill \\ \text{= }\left(\text{5}\text{.06×1}\text{.5}\right)\hfill \\ \text{= 7}\text{.59}\hfill \\ \text{VC:}\hfill \\ \text{= }\left(\text{13}\text{.19+7}\text{.59}\right)\hfill \\ \text{= 20}\text{.78}\hfill \\ \text{AFC:}\hfill \\ \text{= 9/6}\hfill \\ \text{= 1}\text{.50}\hfill \\ \text{AVC:}\hfill \\ \text{= }\frac{\text{20}\text{.78}}{6}\hfill \\ \text{= 3}\text{.46}\hfill \\ \text{ATC:}\hfill \\ \text{= 3}\text{.46+1}\text{.50}\hfill \\ \text{= 4}\text{.96}\hfill \end{array}$

Economics Concept Introduction

Introduction: Marginal cost is calculated by subtracting the new total cost from the previous one and the average total cost is determined by dividing the total cost by quantity.

b)

To determine

The range of pies for which the spreading effect dominates and the range for which the diminishing effect dominates.

Explanation of Solution

The decrease in average fixed cost causes the rise in average variable cost for pies as the spreading effect dominates the diminishing returns effect at the time when the average total cost is decreasing.

There is an increase in AVC for pies 1 to 4 due to a decrease in AFC. The diminishing returns effect dominates at the time when ATC is rising due to a decrease in AFC for pies 5 and 6.

Economics Concept Introduction

Introduction: Marginal cost is calculated by subtracting the new total cost from the previous one and the average total cost is determined by dividing the total cost by quantity.

c)

To determine

A's a minimum cost of output

Explanation of Solution

 A's the minimum cost of output would be 4 pies because this generates the minimum ATC, which is $4.28. The marginal cost of a pie is less than the average total cost that is already produced at the time when output is less than 4. Moreover, when output is more than 4, the marginal cost of a pie is higher than the ATC.

Therefore, making an additional pie decreases the ATC. For example, the marginal cost of pie 3 is $2.25, and the ATC for 1 and 2 is $5.75 which shows that making of third pie lowers the ATC which is equal to $4.58.

Economics Concept Introduction

Introduction: Marginal cost is calculated by subtracting the new total cost from the previous one and the average total cost is determined by dividing the total cost by quantity.