solution
cost function
c(q)=31q3−5q2+3q+10
MCP=6
for maximum profit will be at the point where
P=MCP=MC
and
MC=dqdC(q)
MC=dqd(31q3−5q2+3q+10)
MC=q2−10q+3
and MPC=6
so
q2−10q+3=6
q2−10q−3=0
by solving it
q=−0.29 , 10.29
by ignoring negative value , the profit maximising level of output will be q=10.29.
Comments