Question

A ferry makes a trip from a port in one city to a port in another city in 30 minutes. The distance traveled s, in miles, t minutes after departing the first port is given by s(t)= -0.00077t^3 +0.03351t^2 +0.069t. At what time was the ferry moving the fastest?

Answer 1

$\bf s(t)=-0.00077t^3+0.03351t^2+0.069t\\\\ -----------------------------\\\\ \cfrac{ds}{dt}=-\cfrac{231}{1000000}t^2+\cfrac{3351}{50000}t+\cfrac{69}{1000} \\\\\\ 0=-\cfrac{231}{1000000}t^2+\cfrac{3351}{50000}t+\cfrac{69}{1000}$

notice, is a quadratic with a negative leading term's coefficient, so, is opening downwards

so, the highest point of it, will be at the vertex, so that'd be when the derivative is the highest

$\bf \textit{vertex of a parabola}\\ \quad \\ y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

so the ferry was going the fastest at $\bf -\cfrac{{{ b}}}{2{{ a}}} \ minutes$