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