Question #288316

A firm has the following information on production and costs from past data:

Output (Y) 0 6 12 18

Total Cost (TC) 9 2775 5361 8199

If the total cost function is known to be

TC =aY3 +bY2 +kY + f , and the demand for the product of the firm is Y = 320 − (1 2)⋅ P answer the following:

• Determine the coefficients of the cubic cost function.

• Derive all cost and revenue curves and the profit function.• Show that the MC cuts the AVC when AVC is at its minimum point. Plot the relevant graph indicating all points.

• Calculate the break even and profit maximizing levels of output and price.

• What is the relationship between price, marginal revenue and own price elasticity of demand at the profit maximization point.


1
Expert's answer
2022-01-18T10:09:07-0500

Step 1

Since you have posted a question with multiple sub-parts, we will solve first three subparts for you. To get the remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.

1.TC=aY3+bY2+kY+fTC=aY^3+bY^2+kY+f

When Y=0, TC=f=9

TC=aY3+bY2+kY+9TC=aY^3+bY^2+kY+9

When Y=6, TC=2775

2775=(6)3a+(6)2b+6k+92775=(6)^3a+(6)^2b+6k+9

2766=216a+36b+6k2766=216a+36b+6k

461=36a+6b+k461=36a+6b+k ....(1)

When Y=12,TC=5361

5361=(12)3a+(12)2b+12k+95361=(12)^3a+(12)^2b+12k+9

5352=(12)3a+(12)2b+12k5352=(12)^3a+(12)^2b+12k

Step 2

446=(12)2a+12b+k446=(12)^2a+12b+k ....(2)

When Y=18, TC=8199

8199=(18)3a+(18)2b+18k+98199=(18)^3a+(18)^2b+18k+9

8190=(18)3a+(18)2b+18k8190=(18)^3a+(18)^2b+18k

455=(18)2a+(18)b+k455=(18)^2a+(18)b+k .....(3)

3 equations and 3 variables, Solving them , we get:

a=1/3

b=-8.5

c=500

TC=13Y38.5Y2+500Y+9TC=\frac{1}{3}Y^3-8.5Y^2+500Y+9

TVC=13Y38.5Y2+500YTVC=\frac{1}{3}Y^3-8.5Y^2+500Y

TFC=9TFC=9


2.AVC=TVC/Y=1/3Y28.5Y+500AVC=TVC/Y=1/3Y^2-8.5Y+500

ATC=TC/Y=1/3Y28.5Y+500+9/YATC=TC/Y=1/3Y^2-8.5Y+500+9/Y

AFC=TFC/Y=9/YAFC=TFC/Y=9/Y

MC=d(TC)/dY=Y217Y+500MC=d(TC)/dY=Y^2-17Y+500

 

Step 3

Y=320P2Y=320-\frac{P}{2}

P(Y)=6402YP(Y)=640-2Y

Profit=P(Y).Y-TC

=(6402Y)Y(13Y38.5Y2+500Y+9)=(640-2Y)Y-(\frac{1}{3}Y^3-8.5Y^2+500Y+9)

=13Y36.5Y2+140Y9=\frac{-1}{3}Y^3-6.5Y^2+140Y-9

MC=AVC

Y217Y+500=13Y28.5Y+500Y^2-17Y+500=\frac{1}{3}Y^2-8.5Y+500

=23Y2=172Y=\frac{2}{3}Y^2=\frac{17}{2}Y

Y=514=12.75Y=\frac{51}{4}=12.75

AVC=13Y28.5Y+500AVC=\frac{1}{3}Y^2-8.5Y+500

dAVCdY=23Y8.5=0(MinimumAVC)\frac{dAVC}{dY}=\frac{2}{3}Y-8.5=0(Minimum AVC)

Y=8.5×32=12.75Y=8.5\times\frac{3}{2}=12.75

MC cuts AVC at the lowest point


TC=13Y38.5Y2+500Y+9TC=\frac{1}{3}Y^3-8.5Y^2+500Y+9

Demand: Y=32012PY=320-\frac{1}{2}P

2Y640=P2Y-640=P

P=6402YP=640-2Y

TR=PY=(6402Y)Y=640Y2Y2TR=PY=(640-2Y)Y=640Y-2Y^2


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