Answer:
60 seconds
Step-by-step explanation:
In this problem, the altitude of Ying at time x is given by the equation
[tex]D(x)=\frac{1}{36}(x-60)^2-100[/tex]
where
x is the time in seconds
D is the altitude above sea level, in meters
Here we want to find the time x' at which Ying reach his lowest altitude. This means that we have to find the minimum of the function D(x). In order to do that, first we rewrite the function as
[tex]D(x)=\frac{1}{36}(x^2-120x+3600)-100\\D(x)=\frac{1}{36}x^2-\frac{10}{3}x+100-100\\D(x)=\frac{1}{36}x^2-\frac{10}{3}x[/tex]
This is an upward parabola, written in the form:
[tex]y=ax^2+bx+c[/tex]
We know that the minimum of an upward parabola corresponds to its vertex, which has x-coordinate of
[tex]x_b=-\frac{b}{2a}[/tex] (1)
In this case,
[tex]b=-\frac{10}{3}\\a=\frac{1}{36}[/tex]
Substituting into (1), we find
[tex]x_b=-\frac{-10/3}{2(1/36)}=\frac{10}{3}\cdot 18 =60[/tex]
So, the minimum of D(x) occurs at x = 60: therefore, Ying reaches his lowest altitude after 60 seconds.
