Respuesta :
Following are the calculation to the given question:
[tex]\to TC = 4,000 + 5Q + 10 \ Q2\\\\\to MC = 5 + 20\ Q\\\\[/tex]
For point A)
[tex](a)\ TFC = 4,000\\\\(b)\ AFC = \frac{TFC}{ Q} = \frac{4,000}{ Q}\\\\(c)\ TVC = 5Q + 10\ Q2\\\\(d)\ AVC = \frac{TVC }{Q} = 50 + 10\ Q\\\\(e)\ ATC = \frac{TC }{ Q} = (\frac{4,000}{ Q}) + 50 + 10Q \ \text{Also, ATC = AVC + AFC}\\\\[/tex]
For point B)
TFC remains unchanged at 4,000, regardless of the price of Q.
i)
[tex]\to Q = 0[/tex]
AFC, AVC, and ATC cannot be calculated (division by zero is not possible).
ii)
[tex]Q = 1\\\\AFC =\frac{4,000}{ 1} = 4,000\\\\TVC = (5 \times 1) + (10 \times 1) =5 + 10 = 15\\\\AVC = \frac{TVC}{ Q} = \frac{15}{1} = 15\\\\ATC = 4,000 + 15 = 4,015\\\\MC = 5 + (20 \times 10 = 5 + 20 = 25[/tex]
iii)
[tex]Q = 2\\\\AFC = \frac{4,000}{ 2} = 2,000\\\\TVC = (5 \times 2) + (10 \times 2 \times 2) = 10 + 40 = 50\\\\AVC = \frac{50}{2} = 25\\\\ATC = 2,000 + 25 = 2,025\\\\MC = 5 + (20 \times 2) = 5 + 40 = 45\\\\[/tex]
iv)
[tex]Q = 3\\\\AFC = \frac{4,000}{ 3} = 1,333.33\\\\TVC = (5 \times 3) + (10 \times 3 \times 3) = 15 + 90 = 105\\\\AVC = \frac{105}{3} = 35\\\\ATC = 1,333.33 + 35 = 1,368.33\\\\MC = 5 + (20 \times 3) = 5 + 60 = 65\\\\[/tex]
For point C)
i)
[tex]ATC[/tex] is minimized when [tex]\frac{dATC}{dQ} = 0[/tex]
[tex](- \frac{4,000}{Q2} ) + 10 = 0\\\\\frac{4,000}{Q2} = 10\\\\Q2 = 400\\\\Q = 20\\[/tex]
ii)
Part (B) shows that as MC increases from Q = 0 to Q = 3, ATC decreases, validating the link.
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