we have the equation
[tex]y=31sin5x+32[/tex]
Find out the first derivative
[tex]\begin{gathered} y^{\prime}=(5)(31)c0s5x \\ y^{\prime}=155cos5x \end{gathered}[/tex]
equate to zero the derivative, to find out the critical points
[tex]\begin{gathered} 155cos5x=0 \\ cos5x=0 \end{gathered}[/tex]
The value of cosine is zero when the angle is 90, 270 degrees (pi/2 and 3pi/2)
so
[tex]\begin{gathered} 5x=\frac{\pi}{2} \\ \\ x=\frac{\pi}{10} \end{gathered}[/tex][tex]\begin{gathered} 5x=\frac{3\pi}{2} \\ \\ x=\frac{3\pi}{10} \end{gathered}[/tex]
For x=pi/10
Find out the y-coordinate
[tex]\begin{gathered} y=31s\imaginaryI n\frac{5\pi}{10}+32 \\ \\ y=31sin\frac{\pi}{2}+32 \\ \\ y=31+32=63\text{ ft} \end{gathered}[/tex]
Verify for x=3pi/10
[tex]\begin{gathered} y=31s\imaginaryI n5(\frac{3\pi}{10})+32 \\ \\ y=31s\imaginaryI n(\frac{3\pi}{2})+32 \\ y=-31+32=1\text{ ft} \end{gathered}[/tex]
The maximum height is 63 ft
