To calculate the maximum distance that the ball takes, we can use the first and second derivatives of y to find out this. We have:
[tex]\begin{gathered} y=-\frac{1}{14}x^2+4x+3 \\ \Rightarrow y^{\prime}=-\frac{2}{14}x+4=-\frac{1}{7}x+4 \\ \Rightarrow y^{\doubleprime}=-\frac{1}{7} \end{gathered}[/tex]
Using the second derivative criterion, we have that y'' < 0, therefore, we have a maximum in the root of the first derivative. For that, we get the following:
[tex]\begin{gathered} y^{\prime}=0 \\ \Rightarrow-\frac{1}{7}x+4=0 \\ \Rightarrow\frac{1}{7}x=4 \\ \Rightarrow x=7\cdot4=28 \\ x=28 \end{gathered}[/tex]
Therefore, at x=28 is where the maximum distance is. Now we only substitute x=28 in y to find out:
[tex]\begin{gathered} y=-\frac{1}{14}(28)^2+4(28)+3 \\ \Rightarrow y=-\frac{1}{14}(784)+112+3 \\ \Rightarrow y=-56+112+3=59 \\ y=59 \end{gathered}[/tex]
Finally, we have that the ball will go 59 feet from the child
