solution
cost function
c(q)="\\frac{1}{3}q^3-5q^2+3q+10"
MCP=6
for maximum profit will be at the point where
P=MCP=MC
and
MC="\\frac{dC(q)}{dq}"
"MC=\\frac{d}{dq}(\\frac{1}{3}q^3-5q^2+3q+10)"
MC="q^2-10q+3"
and MPC=6
so
"q^2-10q+3=6"
"q^2-10q-3=0"
by solving it
"q=-0.29\\space\\space ,\\space10.29"
by ignoring negative value , the profit maximising level of output will be q=10.29.
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