p=500−150q,where p=price and q=demandR(q)=p×q=500q−150q2,textwhereR=revenueAnd, maximum revenue is given by derivative of Revenue functionR′(q)=500−250q0=500−250qq=12500p=250Therefore, Revenue R is maximum when price p=250.p=500-\frac{1}{50}q,\\ \text{where p=price and q=demand}\\ R(q)=p×q=500q-\frac{1}{50}q^2, text{where R = revenue}\\ \text{And, maximum revenue is given by derivative of Revenue function}\\ R'(q)=500-\frac{2}{50}q\\ 0=500-\frac{2}{50}q\\ q=12500\\ p=250 \text{Therefore, Revenue R is maximum when price p=250.}p=500−501q,where p=price and q=demandR(q)=p×q=500q−501q2,textwhereR=revenueAnd, maximum revenue is given by derivative of Revenue functionR′(q)=500−502q0=500−502qq=12500p=250Therefore, Revenue R is maximum when price p=250.
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