A series of locks manages the water height along a water source used to produce energy. As the locks are opened and closed, the water height between two consecutive locks fluctuates. The height of the water at point B located between two locks is observed. Water height measurements are made every 10 minutes beginning at 8:00 a. m.

It is determined that the height of the water at B can be modeled by the function f(x)=−11cos(πx/48−5π/12)+28 , where the height of water is measured in feet and x is measured in minutes.

What is the maximum and minimum water height at B, and when do these heights first occur? Drag a value or phrase into each box to correctly complete the statements.

See the attached figure.

Step-by-step explanation:

The given function is [tex]f(x) = -11 \ cos(\frac{\pi x}{48} -\frac{5 \pi}{12} )+28[/tex]

We should know that:e

y = cos x

So, the maximum is y = 1 at x = 0 and the minimum is y = -1 at x = π

So, for the given function

The maximum of f(x) will be at [tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = -1[/tex]

And f(x) = -11 * -1 + 28 = 11 + 28 = 39

[tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = -1[/tex]

∴ [tex](\frac{\pi x}{48} -\frac{5 \pi}{12}) = \pi[/tex]

[tex]\frac{\pi x}{48} =\pi + \frac{5 \pi}{12} = \frac{17}{12} \pi[/tex]

x = 48 * 17/12 = 68 minutes = 1 hour and 8 minutes

The results beginning at 8:00 a.m

So, the maximum will occurs at 9:08 a.m

The minimum of f(x) will be at [tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = 1[/tex]

And f(x) = -11 * 1 + 28 = -11 + 28 = 17

[tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = 1[/tex]

[tex]\frac{\pi x}{48} -\frac{5 \pi}{12}=0[/tex]

[tex]\frac{\pi x}{48} =\frac{5 \pi}{12}[/tex]

x = 48*5/12 = 20 minutes

The results beginning at 8:00 a.m

So, the minimum will occurs at 8:20 a.m

[tex]Please Help, I do not have much time! A series of locks manages the water height along a water sourc[/tex]