Answer:
Option 1
Earn $50 every month.
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000
[tex]\implies y = 50x + 1000[/tex]
This is a linear function.
Option 2
Earn 3% interest each month.
(Assuming the interest earned each month is compounding interest.)
- Let x = number of months the money is left in the account
- Let y = the amount in the account
- Initial amount = $1,000
[tex]\implies y = 1000(1.03)^x[/tex]
This is an exponential function.
Table of values
[tex]\large \begin{array}{| c | l | l |}\cline{1-3} & \multicolumn{2}{|c|}{\sf Account\:Balance} \\ \cline{1-3} & \sf Option\:1 & \sf Option\:2 \\\sf Month & \sf \$50\:per\:mth & \sf 3\%\:per\:mth \\\cline{1-3} 0 & \$1000 & \$1000 \\\cline{1-3} 1 & \$1050 & \$1030 \\\cline{1-3} 2 & \$1100 & \$1060.90 \\\cline{1-3} 3 & \$1150 & \$1092.73 \\\cline{1-3} 4 & \$1200 & \$1125.51 \\\cline{1-3} 5 & \$1250 & \$1159.27 \\\cline{1-3} 6 & \$1300 & \$1194.05 \\\cline{1-3} 7 & \$1350 & \$1229.87 \\\cline{1-3}\end{array}[/tex]
From the table of values, it appears that Account Option 1 is the best choice, as the accumulative growth of this account is higher than the other account option.
However, there will be a point in time when Account Option 2 starts accruing more than Account Option 2 each month. To find this, graph the two functions and find the point of intersection.
From the attached graph, Account Option 1 accrues more until month 32. From month 33, Account Option 2 accrues more in the account.
Conclusion
If the money is going to be invested for less than 33 months then Account Option 1 is the better choice. However, if the money is going to be invested for 33 months or more, then Account Option 2 is the better choice.
