Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.4 - Introduction to Derrivatives - Exercise Set - Page 1177: 81


The investment with $4\%$ compounded is better by $ 179$ dollars.

Work Step by Step

Step 1. For the first account, we have $P=6000, t=5, r=0.04, n=12$ and the end balance is $A_1=P(1+\frac{r}{n})^{nt}=6000(1+\frac{0.04}{12})^{12(5)}\approx7326$ dollars. Step 2. For the second account, we have $P=6000, t=5, r=0.035$ and the end balance is $A_2=Pe^{rt}=6000(e)^{0.035(5)}\approx7147$ dollars. Step 3. Since $A_1\gt A_2$, the first investment is better by $A_1-A_2=179$ dollars.
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