Question

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

Answer 1

`r'(t) = 298t -850`

Explanation:

Given

`q(t) = 1000t - 150t^2`

`p(t) = 150t - t^2`

Required

`r'(t)`

First, we calculate the revenue

`r(t) = p(t) - q(t)`

So, we have:

`r(t) = 150t - t^2 - (1000t - 150t^2)`

Open bracket

`r(t) = 150t - t^2 - 1000t + 150t^2`

Collect like terms

`r(t) = 150t^2 - t^2 + 150t - 1000t`

`r(t) = 149t^2 -850t`

Differentiate to get the revenue change with time

`r'(t) = 2 * 149t -850`

`r'(t) = 298t -850`