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A product is introduced into the market. Suppose a product's sales quantity per month q ( t ) is a function of time t in months is given by q ( t ) = 1000 t − 150 t 2 And suppose the price in dollars of that product, p ( t ) , is also a function of time t in months and is given by p ( t ) = 150 − t 2 A. Find, R ' ( t ) , the rate of change of revenue as a function of time t

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MrRoyal

Answer:

[tex]r'(t) = 298t -850[/tex]

Step-by-step explanation:

Given

[tex]q(t) = 1000t - 150t^2[/tex]

[tex]p(t) = 150t - t^2[/tex]

Required

[tex]r'(t)[/tex]

First, we calculate the revenue

[tex]r(t) = p(t) - q(t)[/tex]

So, we have:

[tex]r(t) = 150t - t^2 - (1000t - 150t^2)[/tex]

Open bracket

[tex]r(t) = 150t - t^2 - 1000t + 150t^2[/tex]

Collect like terms

[tex]r(t) = 150t^2 - t^2 + 150t - 1000t[/tex]

[tex]r(t) = 149t^2 -850t[/tex]

Differentiate to get the revenue change with time

[tex]r'(t) = 2 * 149t -850[/tex]

[tex]r'(t) = 298t -850[/tex]

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