Answer:
the correct answer is $150
Step-by-step explanation:
TC=500 + 150q - 20q^2 + q^3
AVC=(150Q-20Q^2+Q^3)/Q
=150-20Q+Q^2
When AVC is at its minimum means that the marginal cost( CM) is igual to AVC, so we could consider this analysis:
CM= d(TC)/dq =150-40Q+3Q^2
CM=AVC
150-40Q+3Q^2=150-20Q+Q^2
Join similar terms:
150-150-40Q+20Q+3Q^2-Q^2=0
0-20Q+2Q^2=0
Q(-20+2Q)=0
Q_1=0 y Q_2=20/2=10
with q_1 with q_2
150-40*0+3*0=150-20*0+0 150-40*10+3*10^2=150-20*10+10^2
$150=$150 150-400+300 =150-200+100
$50= $ 50
We have two solution if we assume that q=0 like the minimum then the results is $150.
f we assume that q=10 like the minimum then the results is $50.