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The chief financial officer of a company reports that profits for the past fiscal year were $15.3...

Question:

The chief financial officer of a company reports that profits for the past fiscal year were $15.3 million. The officer predicts that profits for the next 4 years will grow at a continuous annual rate somewhere between 3.5 % and 4%. Estimate the cumulative difference in total profit over the 4 years based on the predicted range of growth rates. (Round your answer to two decimal places.)

Continuous Exponential:

In continuous compounded interest, we will often see the value of the Euler's number {eq}e {/eq} which is approximately 2.71828. This is derived from the fact that if evaluate the limit {eq}\lim _{x\to \:\infty }\left(1+\frac{1}{x}\right)^x = e {/eq}, which derives from the idea of compounded not every minute nor second but everytime, that is, continuosly.

Answer and Explanation:

We want to get the area under the curve of the function of the profit. Given that the profit is growing continuously, then by the equation of continuous compounding

$$\begin{align} P&=P_0e^{rt} \\ P_1 &= 15.3e^{r_1t} \\ P_1&=15.3e^{\left(\frac{3.5}{100}\right)t} \\ P_2 &= 15.3e^{r_2t}dt \\ P_2&=15.3e^{\left(\frac{4}{100}\right)t} \\ \end{align} $$

Now, we have {eq}P_0 = 15.3 {/eq} million, then the cumulative difference would be the integral of the difference of the two annual rates

$$\begin{align} P&=\int _0^4P_2-P_1dt \\ P&=\int _0^415.3e^{r_1t}-15.3e^{r_2t}dt \\ P&=\int _0^415.3e^{\left(\frac{4}{100}\right)t}-15.3e^{\left(\frac{3.5}{100}\right)t}dt \\ \end{align} $$

Performing the integration

$$\begin{align} P&=\int _0^415.3e^{\left(\frac{4}{100}\right)t}-15.3e^{\left(\frac{3.5}{100}\right)t}dt \\ P&=\int _0^415.3e^{\left(\frac{4}{100}\right)t}\times \frac{\frac{4}{100}}{\frac{4}{100}}dt-15.3e^{\left(\frac{3.5}{100}\right)t}\times \frac{\frac{3.5}{100}}{\frac{3.5}{100}}dt \\ P&=\int _0^4\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)t}\times \frac{4}{100}dt-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)t}\times \frac{3.5}{100}dt \\ P&=\left[\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)t}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)t}\right] \Biggr|_0^4 \\ P&=\left[\left(\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)\left(4\right)}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)\left(4\right)}\right)-\left(\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)\left(0\right)}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)\left(0\right)}\right)\right] \\ P&=\left[\left(\frac{15.3}{\frac{4}{100}}e^{\left(\frac{4}{100}\right)\left(4\right)}-\frac{15.3}{\frac{3.5}{100}}e^{\left(\frac{3.5}{100}\right)\left(4\right)}\right)-\left(\frac{15.3}{\frac{4}{100}}\left(1\right)-\frac{15.3}{\frac{3.5}{100}}\left(1\right)\right)\right] \\ P &\approx \left[\left(-53.96606\right)-\left(-54.64285\right)\right] \\ P &\approx 0.67679 \\ \end{align} $$

$$\boxed { P \approx \$ 0.68 \hspace{1mm} million } $$


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