2022-07-28

The price p and the quantity x sold of a certain product obey the demand equation: P=-1/8X+100, {X/0 </ X </ 800}
What is the revenue, R(x) = x . p(x) ,to the nearest dollar when 700 units are sold?

wintern90

Expert

$P=-\frac{1}{8}+100$
revenue $=xp\left(x\right)=-\frac{{x}^{2}}{8}+100x$
revenue when units sold = 700 is $-\frac{{700}^{2}}{8}+100×700=8750$

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