the depth of water, d metres, at a port entrance is given by the function d(t)= 4.5+1.5 sin(pi *t/12). A find the maxium , and the minimum

All that needs to be done to find the maximum and minimum depths is to note that sin has a maximum of 1 and a minimum of -1, so the maximum depth will be 4.5 + 1.5 = 6, and the minimum will be 4.5 - 1.5 = 3.

If the times at which they occur are needed then we can just set pi*t/12 = pi/2 or t=6 (and every 24 hours (?) thereafter) for the maximum and pi*t/12 = 3pi/2 or t = 18 (and every 24 hours thereafter) for the minimum.

The calculus approach, though perfectly valid, is overkill here!

***********Plug in t = 6 and t = 18 into the original function to find the depth maximum/minimum.************

Since this is function is periodic, there will be many maximums and minimums. We can find a couple and maybe make a general statement of where they will occur. Taking the derivative, we have:

1.5cos(pi*t/12)(pi/12) = 0

cos(pi*t/12) = 0

The first place that the cosine  = 0 is at pi/2. So, setting pi*t/12 = pi/2, we have that

t = 6

And the next place that the cosine = 0 is at 3pi/2...so

pi*t/12 = 3/2*pi

t = 18

Taking the second derivative, we have

-1.5(pi/12)sin(pi*t/12)(pi/12) =

[-1.5(pi/12)^2]sin(pi*t/12)

And when t = 6 , this is negative, so a maximum occurs at (6,6)

And when t = 18, the second derivative is positive, so this indicates a minimum at... (18, 3).

In general, a maximum depth of 6 m will occur every (6 ± 24t) (hrs??)  and a minimum depth of 3 m will occur every (18 ± 24t)(hrs??) where t is some integer.

Here's a graph of the function:

Usually it is me that turns a mole hill into a mountain. 

I am glad that other people do this sometimes too.  lol.