How To Calculate Maximum Revenue With Pictures

For more information on graphing supply and demand data, see Find and Analyze Demand Function Curve.

p=500−150q{\displaystyle p=500-{\frac {1}{50}}q} p = price q = demand, in number of units This function sets the “zero price” at $500. For each unit sold, the price decreases by 1/50th of a dollar (two cents).

R(q)=p∗q{\displaystyle R(q)=p*q} R(q)=[500−150q]∗q{\displaystyle R(q)=[500-{\frac {1}{50}}q]*q} R(q)=500q−150q2{\displaystyle R(q)=500q-{\frac {1}{50}}q^{2}}

Suppose the revenue function, in terms of number of units sold, is R(q)=500q−150q2{\displaystyle R(q)=500q-{\frac {1}{50}}q^{2}}. The first derivative, therefore, is: R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q} For a review of derivatives, see the wikiHow article on how to Take Derivatives.

R′(q)=500−250q{\displaystyle R^{\prime }(q)=500-{\frac {2}{50}}q} 0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q}

0=500−250q{\displaystyle 0=500-{\frac {2}{50}}q} 250q=500{\displaystyle {\frac {2}{50}}q=500} 150q=250{\displaystyle {\frac {1}{50}}q=250} q=50∗250{\displaystyle q=50*250} q=12,500{\displaystyle q=12,500}

p=500−150q{\displaystyle p=500-{\frac {1}{50}}q} p=500−15012,500{\displaystyle p=500-{\frac {1}{50}}12,500} p=500−250{\displaystyle p=500-250} p=250{\displaystyle p=250}

R=p∗q{\displaystyle R=p*q} R=(250)(12,500){\displaystyle R=(250)(12,500)} R=3,125,000{\displaystyle R=3,125,000}

p=100−0. 01q{\displaystyle p=100-0. 01q}

R(q)=[100−0. 01q]∗q{\displaystyle R(q)=[100-0. 01q]*q} R(q)=100q−0. 01q2{\displaystyle R(q)=100q-0. 01q^{2}}

R(q)=100q−0. 01q2{\displaystyle R(q)=100q-0. 01q^{2}} R′(q)=100−(2)0. 01q{\displaystyle R^{\prime }(q)=100-(2)0. 01q} R′(q)=100−0. 02q{\displaystyle R^{\prime }(q)=100-0. 02q}

R′(q)=100−0. 02q{\displaystyle R^{\prime }(q)=100-0. 02q} 0=100−0. 02q{\displaystyle 0=100-0. 02q} 0. 02q=100{\displaystyle 0. 02q=100} q=100/0. 02{\displaystyle q=100/0. 02} q=5,000{\displaystyle q=5,000}

p=100−0. 01q{\displaystyle p=100-0. 01q} p=100−0. 01(5,000){\displaystyle p=100-0. 01(5,000)} p=100−50{\displaystyle p=100-50} p=50{\displaystyle p=50}

R(q)=p∗q{\displaystyle R(q)=pq} R(q)=50∗5,000{\displaystyle R(q)=505,000} R(q)=250,000{\displaystyle R(q)=250,000}