dairy costs example p.c
DESCRIPTION
Example of shifts in cost curves when a variable cost changes.TRANSCRIPT
Price
Quantity
P=MR=AR =D*
ATC*
MC*
Qf*
Farmer
Here is a graph of a Dairy Farmer in “Long Run Equilibrium”. He/She is making “Normal Profits/Breaking Even” (in Economic Terms).
“A”
Price
Quantity
P=MR=AR =D*
ATC*
MC*
Qf*
ATC 1
If a Variable Cost, like the input Corn for cattle feed decreases in price then that lowers the Cost of Production. This is going to DECREASE the Average Total Cost of Producing ALL units. ATC* curve will shift DOWN to ATC 1.
“A”
Farmer
Price
Quantity
P=MR=AR =D*
ATC*
MC*
Qf*
ATC 1
MC 1
“A”
Qf 1
Farmer
Corn as an input is a Variable Cost to the Dairy Farmer. Variable Cost affect the Marginal Cost of Production.
The Marginal Cost curve is also the firms SUPPLY CURVE!
If the cost of production DECREASES then the Marginal Cost (MC*) curve will shift to the RIGHT (MC 1), Increasing Quantity Supplied at every Price.
Price
Quantity
P=MR=AR =D*
MC*
Qf*
ATC 1
MC 1
“B”“A”
Qf 1
Farmer
Lots going on here! First thing you want to do is locate the PROFIT MAXIMIZING QUANTITY.
MR = MCThat is Point “B” at “Qf1”
Price
Quantity
P=MR=AR =D*
MC*
Qf*
ATC 1
MC 1
“B”“A”
Qf 1
“C”ATC 1
Farmer
From Point “B” and “Qf1” locate where the Profit Maximizing Quantity intersects the ATC 1:
Point “C”
Price
Quantity
P=MR=AR =D*
Qf*
ATC 1
The Area of ECONOMIC PROFITS is shaded in BLUE. I deleted the old ATC* and MC* curves to clean it up.
Bottom line: With a decrease in production costs the firm increases production by the difference between Qf* and Qf1 and earns ECONOMIC PROFITS over and above Normal Profits EVEN AT THE SAME MARKET PRICE as before.
MC 1
“B”
Qf 1
“C”ATC 1
Area of Economic Profit
Farmer
P=MR=AR =D*
Price
QuantityQf*
MC 1
“B”
Qf 1
“C”ATC 1
Area of Economic Profit
Farmer
D*
S*Market for Milk
Price
Quantity
P*
Q*
The Price of Milk is established on commodity exchanges/ co-operatives OUTSIDE the control of the Dairy Farmer. This is shown with a Market Graph on the RIGHT. The Dairy Farmer is a “Price Taker” because any single Dairy Farmer cannot influence the price of milk. P* established in the Market is the price the farmer has to take for each gallon of milk produced. This is reflected in the HORIZONTAL Demand Curve represented by “P=MR=AR=D*” on the graph on the LEFT.
“A”
ATC 1
P=MR=AR =D*
Price
QuantityQf*
MC 1
“B”
Qf 1
“C”ATC 1
Area of Economic Profit
Farmer
D*
S*Market for Milk
Price
Quantity
P*
Q*
The article says that the Demand for Milk as an input for Cheese making AND for Export has increased and the MARKET PRICE for Milk has INCREASED. This is reflected in the graph on the RIGHT.
D1
Q 1
P1 “A”
“B”
ATC 1
P=MR=AR =D*
Price
QuantityQf*
MC 1
“B”
Qf 1
“C”ATC 1
Area of Economic Profit
Farmer
D*
S*Market for Milk
Price
Quantity
P*
Q*
Now that the Market Price of Milk is HIGHER, the Dairy Farmer now will receive “P1” as the price for each gallon of Milk produced.
D1
Q 1
P1 “A”
“B”
P=MR=AR =D*
P=MR=AR =D1
ATC 1
P=MR=AR =D*
Price
QuantityQf*
MC 1
“B”
Qf 1
“C”ATC 1
Area of Economic Profit
Farmer
D*
S*Market for Milk
Price
Quantity
P*
Q*
Locate the new PROFIT MAXIMIZING QUANTITY where “MR =MC”. That will be at Point “D” and Quantity “Qf 2”
D1
Q 1
P1 “A”
“B”
P=MR=AR =D*
P=MR=AR =D1
Qf 2
“D”
ATC 1
P=MR=AR =D*
Price
QuantityQf*
MC 1
“B”
Qf 1
ATC 1
Area of Economic Profit
Farmer
D*
S*Market for Milk
Price
Quantity
P*
Q*
From Point “D” and Quantity “Qf 2” locate where they intersect the ATC 1. That will be at Point “E”
D1
Q 1
P1 “A”
“B”
P=MR=AR =D*
P=MR=AR =D1
Qf 2
“D”
“E”ATC 1
Price
Quantity
ATC 1
MC 1
ATC 1
Area of Economic Profit (after a decrease in inputs cost AND an
increase in Market Price)
Farmer
D*
S*Market for Milk
Price
Quantity
P*
Q*
Let’s clean this up and find our area of ECONOMIC PROFIT. Compare this slide with the last one and compare Price, Firm Quantity Supplied, and Economic Profit. All have increased and the Dairy Farmer is enjoying nice profits IN THE SHORT RUN. HOWEVER, because of the nature of the Market we know in THE LONG RUN….No wait, YOU TELL ME WHAT HAPPENS!
D1
Q 1
P1 “A”
“B”P=MR=AR =D1
“D”
“E”
Qf 2