Given the equation:
[tex]S(t)=20(7-\frac{7}{2+t})[/tex]Where t is the time in months.
Let's solve for the following:
• (a). Average rate of change of S(t) during the first year.
During the first year, the time interval, t is from 0 to 12 months.
Now, to find the average rate of change for the first year, apply the formula:
[tex]S(t)_{avg}=\frac{S(12)-S(0)}{12-0}[/tex]Now, let's solve for S(12) and S(0):
[tex]\begin{gathered} S(12)=20(7-\frac{7}{2+12})\Longrightarrow20(7-\frac{7}{14})\Longrightarrow20(7-0.5)=130 \\ \\ S(0)=20(7-\frac{7}{2+0})=\Rightarrow20(7-\frac{7}{2})\Longrightarrow20(7-3.5)=70 \end{gathered}[/tex]Hence, to find the average rate of change, we have:
[tex]\begin{gathered} S(t)_{\text{avg}}=\frac{S(12)-S(0)}{12-0} \\ \\ S(t)_{\text{avg}}=\frac{130-70}{12-0} \\ \\ S(t)_{\text{avg}}=\frac{60}{12} \\ \\ S(t)_{\text{avg}}=5 \end{gathered}[/tex]Therefore, the average rate of change during the first year is 5
• (b). During what month of the first year does S (1) equal the average rate of change?
Let's first find the derivative of S(t):
[tex]\begin{gathered} S^{\prime}(t)=20(0+\frac{7}{(2+t)^2}) \\ \\ S^{\prime}(t)=20(\frac{7}{(2+t)^2}) \\ \\ S^{\prime}(t)=\frac{140}{(2+t)^2} \end{gathered}[/tex]Now, we have:
[tex]\begin{gathered} S^{\prime}(t)=S_{avg} \\ \\ 5=\frac{140}{(2+t)^2} \\ \\ 5(2+t)^2=140 \\ \\ (2+t)^2=\frac{140}{5} \\ \\ (2+t)^2=28 \end{gathered}[/tex]Take the square root of both sides:
[tex]\begin{gathered} \sqrt[]{(2+t)^2}=\sqrt[]{28} \\ \\ 2+t=5.3 \\ \\ t=5.3-2 \\ \\ t=3.3\approx4 \end{gathered}[/tex]Therefore, the month will be the 4th month which is April.
ANSWER:
(a). 5
(b). April