Respuesta :
The path of the ball is given by [tex] y=\frac{-1}{20}x^{2}+3x+5 [/tex] where y is the height in feet and x is the horizontal distance in feet.
We will use the second derivative test which states,
If a function has a critical point for which f′(x) = 0 and the second derivative is positive at this point, then f has a local minimum. If, the function has a critical point for which f′(x) = 0 and the second derivative is negative at this point, then f has local maximum.
Firstly, we will calculate the first derivative of the given function,
[tex] y'=\frac{-2x}{20}+3 [/tex]
To find the horizontal distance, let y'=0
[tex] \frac{-2x}{20}+3=0 [/tex]
[tex] \frac{-2x}{20}=-3 [/tex]
[tex] {-2x} =-60 [/tex]
[tex] x=30 [/tex]
Now let us consider the second derivative of the given function,
[tex] y''=\frac{-2}{20} [/tex]
Since the second derivative that is y'' is negative.
Therefore, x=30 is the maximum horizontal distance.
Substituting x =30 in the given function to get the maximum height,
[tex] y=\frac{-1}{20}x^{2}+3x+5 [/tex]
[tex] y=\frac{-1}{20}(30)^{2}+3(30)+5 [/tex]
y=50 is the maximum height in feet.
