Asked by Candi
Demand Equation:
The price p, in dollars, and the quantity x sold of a certain product obey the demand equation
x = -5p + 100, 0 (less than or equal to) p (less than or equal to) 20
a.) Express the revenue R as a funtion of x.
b.) What is the revenue if 15 units are sold?
c.) What quantity x maximizes revenue? What is the maximum revenue?
d.) What price should the company charge to maximize revenue?
Revenue= P*x
but p= 1/5 x +20 check that, then substitute to find Revenue as a function of x.
Maximizing Revenue. In calculus, there is a very elegant way. In algebra, we want find when Revenue is max.
Revenue= P*x
but p= 1/5 x +20 check that, then substitute to find Revenue as a function of x.
Maximizing Revenue. In calculus, there is a very elegant way. In algebra, we want find when Revenue is max.
The price p, in dollars, and the quantity x sold of a certain product obey the demand equation
x = -5p + 100, 0 (less than or equal to) p (less than or equal to) 20
a.) Express the revenue R as a funtion of x.
b.) What is the revenue if 15 units are sold?
c.) What quantity x maximizes revenue? What is the maximum revenue?
d.) What price should the company charge to maximize revenue?
Revenue= P*x
but p= 1/5 x +20 check that, then substitute to find Revenue as a function of x.
Maximizing Revenue. In calculus, there is a very elegant way. In algebra, we want find when Revenue is max.
Revenue= P*x
but p= 1/5 x +20 check that, then substitute to find Revenue as a function of x.
Maximizing Revenue. In calculus, there is a very elegant way. In algebra, we want find when Revenue is max.
Answers
Answered by
Anonymous
The revenue function for a bicycle shop is given by R(x) = x · p(x) dollars where x is the number of units sold and p(x) = 100 − 0.5x is the unit price. Find the maximum revenue.
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