Question

The water level, measured in feet above mean sea level, of Lake Lanier in Georgia, USA, during 2012 can be modeled by the function

\(L\left( t \right) = {\bf{0}}{\bf{.01441}}{t^{\bf{3}}} - {\bf{0}}{\bf{.4177}}{t^{\bf{2}}} + {\bf{2}}{\bf{.703}}t + {\bf{1060}}{\bf{.1}}\)

Where \(t\) is measured in months since Janurary 1, 2012. Estimate when the water level was highest during 2012.

Short answer

The water level was highest during 2012 about 4.1 months after Janurary 1.

Step-by-step solution

Differentiate the equation of L

Differentiate the equation \(L\left( t \right) = 0.01441{t^3} - 0.4177{t^2} + 2.703t + 1060.1\) with respect to \(t\).

\(\begin{aligned}{c}\frac{{{\rm{d}}L}}{{{\rm{d}}t}} &= \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {0.01441{t^3} - 0.4177{t^2} + 2.703t + 1060.1} \right)\\ &= 0.01441\left( {3{t^2}} \right) - 0.4177\left( {2t} \right) + 2.703\\ &= 0.04323{t^2} - 0.8354t + 2.703\end{aligned}\)

Find the root of the equation \(L'\left( t \right) = {\bf{0}}\)

Solve the equation \(L'\left( t \right) = 0\).

\(\begin{aligned}{c}0.04323{t^2} - 0.8354t + 2.703 &= 0\\t &= \frac{{0.8354 \pm \sqrt {{{\left( {0.8354} \right)}^2} - 4\left( {0.04323} \right)\left( {2.702} \right)} }}{{2\left( {0.04323} \right)}}\\ &\approx 4.1\;{\rm{or}}\,{\rm{ }}15.2\end{aligned}\)

Find the time when water level is highest

As \(0 \le t \le 12\), the value of \(L\left( t \right)\) can be calculated as,

For \(t = 0\),

\(L\left( 0 \right) = 1060.1\)

For \(t = 4.1\),

\(L\left( {4.1} \right) \approx 1065.2\)

And,

\(L\left( {12} \right) \approx 1057.3\)

Thus, the water level was highest during 2012 about 4.1 months after Janurary 1.