Part A:

A house maximizes net income when it equates MRPL = ( MR ) * ( MPL ) = MCL MPL= dQ/dL =1 – L/400
Therefore ( 40 ) * ( 1-L/400 ) = 20. The solution is L = 200.
In bend. Q = 200 – ( 2002/800 ) . The solution is Q = 150.
The houses profit is= PQ – ( MC ) L= ( $ 40 ) ( 150 ) – ( $ 20 ) ( 200 ) = $ 2. 000 Part B Price addition to $ 50:
Q = Dresses per hebdomad
L= Number of labour hours per hebdomad
Q = L –L2/800
MCL= $ 20
P= $ 50

A house maximizes net income when it equates MRPL = ( MR ) * ( MPL ) = MCL MPL= dQ/dL =1 – L/400
Therefore ( 50 ) * ( 1-L/400 ) = 20. The solution is L = 240.
In bend. Q = 240 – ( 2402/800 ) . The solution is Q = 168.
The houses net income is ( $ 40 ) ( 168 ) – ( $ 20 ) ( 240 ) = $ 1. 920
Optimal end product of the house would increase from 150 to 168. and labour would increase from 200 to 240. ensuing in a lessening in net income to $ 1. 920. Part B rising prices in labour and end product monetary value:
Assuming a 10 % addition IN LABOR COST AND OUTPUT PRICE…
Q = Dresses per hebdomad
L= Number of labour hours per hebdomad
Q = L –L2/800
MCL= $ 20. 20 ( 20* . 10 )
P= $ 40. 40 ( $ 40* . 10 )

A house maximizes net income when it equates MRPL = ( MR ) * ( MPL ) = MCL MPL= dQ/dL =1 – L/400
Therefore ( 40. 40 ) * ( 1-L/400 ) = 20. 20. The solution is L = 200. In bend. Q = 200 – ( 2002/800 ) . The solution is Q = 150.
The houses net income is ( $ 40. 40 ) ( 150 ) – ( $ 20. 20 ) ( 200 ) = $ 2. 020 Optimal end product of the house would stay the same at 150. and labour would stay the same at 200. nevertheless. there would be an addition in net income to $ 2. 020 to match to the per centum addition in end product monetary value and labour cost ( in this illustration 10 % ) . Part C 25 % addition in MPL:

The fringy cost of labour would increase by the same per centum sum as monetary value ( 25 % ) . therefore the Marginal Cost of labour would increase from 20 to 25. Therefore 50 – L/8 =25 and L=200
End product and hours of labour remain unchanged due to the fact that monetary value and cost of labour addition by same per centum sums ALSO SEE PART B ABOVE INFLATION EXAMPLE I MADE DENOTING 10 PERCENT INCREASE IN LABOR AND OUTPUT. Chapter 5 Question 12 Page 220

Part A:
Q = 100 ( 1. 01 ) . 5 ( 1 ) . 4 = 100. 50. Compare this to the original of Q=100 and we can find that Output additions by. 5 % . The power coefficient measures the snap of the end product with regard to the input. A 1 % addition in labour green goodss a ( . 5 ) ( 1 ) = . 5 % addition in end product.

Part B:
Dr. Ghosh- per my electronic mail I was a spot baffled with this inquiry based on your talk notes ( as your notes province that BOTH inputs must alter for a returns to scale to be determined ) . so I have two different sentiments. Opinion 1- The nature of returns to scale in production depends on the amount of the advocates. ?+? . Decreasing returns exist if ?+?? 1. The amount of the power coefficients is. 5 + . 4 & lt ; 1. the production map exhibits diminishing returns to scale where end product additions in a smaller proportion than input. This is reflected in Part A of this job where a 1 % addition in labour ( input ) consequences in a. 5 % addition in end product. Opinion 2- BOTH
inputs must be changed in the same proportion ( harmonizing to your talk notes ) . Therefore. in this inquiry I am confused. Merely one of the inputs are being changed. Does this construct non use. and is my original reply incorrect? I don’t see any graduated table where merely one of the inputs are changed…As such. if both inputs MUST be changed so returns to scale can non be determined for this inquiry as merely L was originally changed. Chapter 6 Question 6 Part B Page 265 ( portion A non required )

Demand is P = 48 – Q/200
Costss are C = 60. 000 + . 0025Q2.

Therefore the TR= 48Q-Q2/200. and the derivative MR map would be MR = 48 – Q/100. The house maximizes net income by puting MR = MC. Therefore. MR = 48 – Q/100 and MC = . 005Q. Puting MR = MC ( 48 – Q/100 ) = . 005Q consequences in: Q* = 3. 200. In bend. P* = $ 32 ( where 48-3200/200 ) . Chapter 6 Question 8 Page 265

CE= 250. 000 +1. 000Q + 5Q2
$ 2. 000= Cost of Frames and assembly
P= 10. 000-30Q

Part A:
Fringy Cost of bring forthing an extra engine…
CE = 250. 000 +1. 000Q +5Q2
MCE = d/dQ ( 250. 000 +1. 000Q + 5Q2 )
=10Q + 1. 000
MCCycle=MCEngine +MCframes and assembly ; hence
MCCylce = 1. 000+ 2. 000 +10Q
The reverse demand map provided in the text was P= 10. 000-30Q TR = ( P ) * ( Q )
= ( 10. 000-30Q ) *Q
=10. 000Q – 30Q2

Obtain the derived function of this map to happen MR:
MR=d/dQ
= ( 10. 000Q – 30Q2 )
MR=10. 000 – 60Q

MR = MC
10. 000 – 60Q = 1. 000 + 2. 000 +10Q
7. 000 = 70Q
Q=100 ( net income maximizing end product )

P= 10. 000 – 30Q
=10. 000 -30 ( 100 )
Net income Maximizing Price=7. 000

Therefore the Marginal Cost of bring forthing an engine
=1. 000 + 10Q ( q=100 from work outing above )
=2. 000 MCEngine

Fringy Cost of Producing a Cycle
From equation developed above…
MCCycle = 1. 000 +2. 000 +10Q
=1. 000 +2. 000 + 10 ( 100 )
= $ 4. 000 MCCycle

Part B:
Since the house can bring forth engines at a Fringy Cost of $ 2. 000. the chance to purchase from another house at a greatly reduced Marginal Cost of $ 1. 400 would be reasonable. MCEngine= $ 1. 400
MR = MC
10. 000 – 60Q = 2. 000 +1. 400
10. 000- 60Q = 3400
Q=110 ( net income maximizing end product )

P = 10. 000 – 30 ( 110 )
=6. 700 net income maximizing monetary value
Therefore the house should purchase the engine since the engine produced by the house is more than the engine provided by the other house. Chapter 6 Question 10 Page 266
Part A:
Gross is P*Q.
Obtain Marginal Cost map through
160 + 16Q + 0. 1Q2
FOC ( derivative of above equation )
16 + 0. 2Q= Megahertz

From the P= 96 – . 4Q we can find that entire gross = 96Q – . 4Q2 and the derivative or FOC is therefore 96 – . 8Q= Mister

Set MC = MR
16 + 0. 2Q = 96 – 0. 8Q
Q=80

We solve for P by stop uping this into our original equation
P= 96- . 4 ( 80 )
P=64

Net income = 5. 120 ( 80*64 ) – 2. 080 ( 160 + 16*80 + . 1 ( 80 ) 2 ) = $ 3. 040

Part B:
C =160 + 16Q + . 1Q2
AC= ( 160+16Q+ . 1Q^2 ) /Q
MC=d/dQ ( 160 + 16Q + . 1Q2 )
MC=16 + . 2Q
AC=MC
160/Q + 16 + . 1Q = 16 + . 2Q
160/Q = . 1Q
. 1Q2 =160
Q= 40
Average cost of production is minimized at 40 units. she is right as AC = MC ( see below ) . AC = 960/40 =24
MC = 16 + ( . 2 ) ( $ 40 ) = $ 24
However. optimum end product is Q=80 where MR = MC. hence her 2nd claim of 40 units as the firm’s net income maximising degree of end product is wrong. P =
96 – . 4 ( 40 )
P= $ 80
TR = 80*40 =3. 200
C = 160 + 16Q + . 1Q2
=960
Net income = Revenue – Cost = 3. 200 – 960 = 2. 240 hence end product at 80 is greater than the net income at 40. Part C:
We learned from portion a the individual works cost is $ 2. 080 or ( 160 + 16*80 + . 1 ( 80 ) 2 ) . If two workss were unfastened each bring forthing the minimal degree of end product detailed in portion B ( Q=40 ) so entire cost would be ( Q ) * ( AC ) = 24*80 = $ 1. 920. You can compare this to the cost in portion A of $ 2. 080 and find it is cheaper to bring forth utilizing the two workss.