Solution
Cost function
Tc=13q3−5q2+20q+50Tc=\frac{1}{3}q^3-5q^2+20q+50Tc=31q3−5q2+20q+50
At maximize level of price
MC=MCP
MC=d(Tc)dqMC=\frac{d(Tc)}{dq}MC=dqd(Tc)
MC=ddq(13q3−5q2+20q+50)MC=\frac{d}{dq}(\frac{1}{3}q^3-5q^2+20q+50)MC=dqd(31q3−5q2+20q+50)
MC=q2−10q+20MC=q^2-10q+20MC=q2−10q+20
And MCP=4
So
q2−10q+20=4q2−10q+16=0q^2-10q+20=4\\q^2-10q+16=0q2−10q+20=4q2−10q+16=0
q=8 and 2
Maximize level of price will be 2 because d2(Tc)dq<0\frac{d^2(Tc)}{dq}<0dqd2(Tc)<0 for 2.
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We are good luck
Thank you