Given: TC = 2994 + 388Q -30Q2 + 3Q3 a. Find equations for TFC, TVC, AFC, AVC, and MC.

Given: TC = 2994 + 388Q -30Q2 + 3Q3 a. Find equations for TFC, TVC, AFC, AVC, and MC. b. the levels of output at which both AVC and MC are minimum. c. Find the AVC and MC for the level of output at which the AVC curve is minimum. They should be the same $ amount Note: the turning point of any curve is where it’s rate of change; i. e., it’s derivative equals zero.

Solutions

Expert Solution

TC = 2994 + 388Q -30Q² + 3Q³

TFC = 1994

TVC = 388Q -30Q² + 3Q³

AFC = TFC/Q = 1994/Q

AVC = TVC/Q = 388 -30Q + 3Q²

MC = d(TC)/dQ = 388 - 60Q + 9Q²

Finding Q at which AVC is minimum

differentiating AVC with respect to Q and equate to zero and solve for Q

d(AVC)/dQ = - 30 + 6Q = 0

Q = 5

At Q =5; AVC = 388 -30*5 + 3*25

AVC = 313

MC at Q=5

= 388 - 60*5 + 9*25

= 313

Hence, at Q = 5, MC and AVC are same.

Rahul Sunny answered 1 year ago

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