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

Answer:

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

Step-by-step 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

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

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