# A small company of science writers found that its rate of profit (in thousands of dollars) after...

## Question:

A small company of science writers found that its rate of profit (in thousands of dollars) after {eq}t {/eq} years of operation is given by the function {eq}P'(t)=(9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}}. {/eq}

a) Find the total profit in the first three years.

b) Find the profit in the fourth year of operation.

c) What is happening to the annual profit over the long run?

## Definite Integrals:

Given the rate of profit {eq}P'(t) {/eq}, the profit function {eq}P(t) {/eq} can be derived by using definite integrals.

The total profit after {eq}t {/eq} years of operation is given by

{eq}\displaystyle \int_{0}^{t} P'(t) \; \mathrm{d}t {/eq}

where

{eq}P'(t) {/eq} is the rate of profit as a function of the time {eq}t {/eq}

Its solution is achieved through the use of the fundamental theorem of calculus.

{eq}\displaystyle \int_{0}^{t} P'(t) \; \mathrm{d}t = P(t) - P(0) {/eq}

## Answer and Explanation: 1

a) The total profit in the first three years is equivalent to the value of the definite integral {eq}\displaystyle \int_{0}^{3} (9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}} \; \mathrm{d}t {/eq}

Using FTC:

{eq}\begin{align*} \displaystyle \int_{0}^{3} (9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}} \; \mathrm{d}t & = \displaystyle \frac{9}{2}\int_{0}^{3} (2t+2)(t^2+2t+2)^{\textstyle\frac{1}{3}} \; \mathrm{d}t\\ & = \frac{9}{2}\left[ \frac{(t^2+2t+2)^{\textstyle\frac{4}{3}}}{\frac{4}{3}} \right]_{0}^{3}\\ & = \frac{9}{2}\left[ \frac{3(t^2+2t+2)^{\textstyle\frac{4}{3}}}{4} \right]_{0}^{3}\\ & = \frac{9}{2}\left( \frac{3(3^2+2(3)+2)^{\textstyle\frac{4}{3}}}{4} - \left( \frac{3(0^2+2(0)+2)^{\textstyle\frac{4}{3}}}{4} \right) \right)\\ & = \frac{27(17)^{\textstyle\frac{4}{3}}}{8} - \frac{27(2)^{\textstyle\frac{4}{3}}}{8} \\ & \approx 139.023\\ \end{align*} {/eq}

{eq}P(3) = (139.023)(1000)= 139,023 {/eq} since {eq}P(t) {/eq} is in thousands of dollars.

a) Hence ,the total profit in the first three years of operation is {eq}139,023 {/eq} dollars.

b) The total profit in the fourth year of operation can be determined by assessing

the value of the difference between the total profit in the first four years to the total profit in the first three years:

{eq}P(4) - P(3) = \displaystyle \int_{0}^{4} (9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}} \; \mathrm{d}t - \displaystyle \int_{0}^{3} (9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}} \; \mathrm{d}t {/eq}

Using the result from part a, and solving for the value of the remaining definite integral:

{eq}\begin{align*} P(4) - P(3) & =\displaystyle \int_{0}^{4} (9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}} \; \mathrm{d}t - \displaystyle \int_{0}^{3} (9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}} \; \mathrm{d}t\\ & = \frac{9}{2}\left[ \frac{3(t^2+2t+2)^{\textstyle\frac{4}{3}}}{4} \right]_{0}^{4} - 139023\\ & = \frac{9}{2}\left( \frac{3(4^2+2(4)+2)^{\textstyle\frac{4}{3}}}{4} - \left( \frac{3(0^2+2(0)+2)^{\textstyle\frac{4}{3}}}{4} \right)\right) - 139023\\ & = \frac{27(26)^{\textstyle\frac{4}{3}}}{8} - \frac{27(2)^{\textstyle\frac{4}{3}}}{8} - 139023\\ & = (251.455)(1000) - 139023\\ & = 251455 - 139023\\ & = 112432\\ \end{align*} {/eq}

b) Thus, the profit in the fourth year of operation is {eq}112,432 {/eq} dollars.

c) The rate of profit {eq}P'(t)=(9t+9)(t^2+2t+2)^{\textstyle\frac{1}{3}} {/eq} is positive when {eq}t > -1 {/eq}.

So {eq}P(t) {/eq} is an increasing function.

It has been increasing since the start of the operation ({eq}t \geq 0 {/eq}).

c) Thus, the annual profit is increasing over the long run.