Question

Demand Equation The price (in dollars) and the quantity x sold of a certain product obey the demand equation p= -1/10x+150, 0≤x≤1500
Revenue is x*p.
(a) Express the revenue R as a function of x.
(b) What quantity x maximizes revenue? What is the maximum revenue?

Answer 1

\(p = -\dfrac{1}{10}x + 150 \)

\(R(x) = px = (-\dfrac{1}{10}x + 150 )x \)

\( R(x) =-\dfrac{1}{10}x^2 + 150x\)

This is a parabola that opens downward.
To find the value of x that maximizes R, find the equation of the axis of symmetry.
Then plug in that value of x into the equation for R.

Answer 2

thank you

Answer 3

wlcm