Question

In: Economics

A manufacturer of base units for home sensor protection (security, water, carbon monoxide, etc.) produces these...

  1. A manufacturer of base units for home sensor protection (security, water, carbon monoxide, etc.) produces these units in a small facility in northern Indiana. The firm’s engineer has set up “assembly cells” in the facility that she believes will be the most efficient means of producing the units. The following table displays the various costs per hour for producing the units.

Outputs per hour Total Cost Fixed Cost Variable Cost Avg Total Cost Avg Fixed Cost Avg Var Cost Marg Cost

0 $100 $100 $30

1 $130 $100 $30 $130    $100 30 $28

2 $158 $100 $58 $79 $50 $29 $25

3 $183 $100 $83 $61 $33.33 $27.67 $21

4 $204 $100 $104 $51 $25 $26.00 $18

5 $222 $100 $122 $44.40 $20 $24.40 $20

6 $242 $100 $142 $40.33 $16.67 $23.67 $23

7 $265 $100 $165 $37.86 $14.29 $23.57 $30

8 $295 $100 $195 $36.88 $12.50 $24.38 $38

9 $333 $100 $233 $37 $11.11 $25.89 $40

10 $373 $100 $273 $37.30 $10.00 $27.30

  1. Explain why Average Total Costs (ATC) declines through the production quantities. Only with the tenth unit per hour does the firm see a minor increase in ATC.
  2. Explain why Average Fixed Cost (AFC) declines across all levels of production. What do you call this phenomenon?
  3. Explain why Marginal Costs initially decline but then increase with the sixth unit of production. Which short run production principle is at work here in creating this pattern of Marginal Costs?

Solutions

Expert Solution

Ans. =

A)

As we start Utilizing the Resources, Initially there is a production on the Level which is Lesser than Total Capacity. which means the Per Unit Cost of Output would be Less, and Average Total Cost measures only the Per Unit Cost of Output.

and as we move towards employing more and more capacity, Per Unit Cost Of Rises as if we produce above the Optimum Level of Resources, Costs will Definitely Rise and that's what we see from the 9th to 10th level of Output. ATC starts Rising.

B)

Fixed Costs are those which must be Incurred to the Busniess in a Fixed Amount (No matter how much the Quantity Of Output is, Fixed Cost will remain the same even if we produce 1 Level of Output)

Hence, Average Fixed Cost is the Per Unit Fixed Cost of Output. So, If the Total Fixed Cost is not going to Increase or Decrease and we are producing more and more then Obviously Per unit Fixed Cost per Output would Definitely Decrease.

Average Fixed Cost never becomes Zero (Means it doesn't touch X-axis or Y-axis) Ut will get less and less but never be Zero. That's why its called as Rectangular Hyper-bola.

C)

Marginal Cost is the Additional Cost. which is Incurred with one more unit of Rise in Output.

Initially as the firm starts employing resources both fixed and variable, Marginal Cost would Decrease because Initially we are employing more and more resources and we are utilizing the Capacity available for Production, that would Definitely Decrease the Overall Costs and Marginal Cost too.

But as we move towards producing more and more Output We reach a Point of Optimum Utilization after that point there would be more costs incurred with every Increase in the Output due to Diseconomies of Scale.



_______________________________________

Please UPVOTE if you find this Answer Helpful and Post your Doubts (if any) in the comment section.

Thanks
_______________________________________


Related Solutions

The water gas shift reaction involves the reaction of carbon monoxide with water vapor to form...
The water gas shift reaction involves the reaction of carbon monoxide with water vapor to form carbon dioxide and hydrogen gas: CO(g) + H2O(g) → CO2(g) + H2(g) 28.01 g CO and 9.01 g water vapor are combined in a reaction vessel. At the end of the reaction, a total pressure of 945.3 torr is found. A.) Which gases are left at the end of the reaction? B.) What are mole fractions of the gases left in the reaction vessel...
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The...
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×10−3If a 1.55-L reaction vessel initially contains 143 mbar of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The...
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×10−3. . If a 1.55-L reaction vessel initially contains 153 torr of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium. mH2/mH2+mCO+mH2O =
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The...
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×10−3. If a 1.55-L reaction vessel initially contains 247 torr of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The...
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is K p =1.60× 10 −3 . If a 1.55- L reaction vessel initially contains 157 torr of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The...
Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is 1.6×10−3 Kp . Part A If a 1.55-LL reaction vessel initially contains 140 torr of water at 700.0 KK in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium. mH2mH2+mCO+mH2OmH2mH2+mCO+mH2O = %
If the concentration (solubility) of carbon monoxide in water is 0.27 M and the Henry's Law...
If the concentration (solubility) of carbon monoxide in water is 0.27 M and the Henry's Law constant for carbon monoxide is 9.5 x 10-4 M/atm, what is the partial pressure of the gas above the solution? A solution is prepared by adding 3.65 x 10-3 g of sodium carbonate to 575 g of water. Find ppb and molality
URGENT!! Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas....
URGENT!! Solid carbon can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp=1.60×10−3. Part A If a 1.55-L reaction vessel initially contains 121 mbar of water at 700.0 K in contact with excess solid carbon, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium. I've tried 4 times, I keep getting it wrong. Please show me how to do it!...
The steam reforming of methane is an industrial which produces carbon monoxide and hydrogen as feedstocks...
The steam reforming of methane is an industrial which produces carbon monoxide and hydrogen as feedstocks for other materials. This reaction is normally run anywhere between 750 and 1100 C. When the reaction is run in a large excess of water, it displays first order kinetics with respect to the concentration of methane and zeroeth order kinetics with respect to water. At 750 degrees, the half-life of this reaction under these conditoins is 67.5 minutes. If the activitation energy for...
Sometimes patients will recieve treatment for carbon monoxide, feel better, go home, and then all of...
Sometimes patients will recieve treatment for carbon monoxide, feel better, go home, and then all of the sudden have increased symptoms a few hours later. There is speculation that this is due to a delayed CO2 from myoglobin, which then binds to hemoglobin. Discuss why this might happen using an oxygen hemoglobin dissociation curve, as well as the potential partial pressures of CO2 and O2 once the patient returns home.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT