Question

P(t)= 100 sin((pi t)/2)+500
Calculate the maximum value for t.
I have derived the function to
P'= 50pi cos((pi t)/2)

Answer 1

Let P' equal 0 and solve for t.
0 = 50pi cos(pi t / 2)
cos(pi t / 2) = 0
pi t / 2 = arccos 0 = pi / 2 + 2kpi
t = 1 + 4k; where k is any integer

Answer 2

ok, i havnt learnt arccos but the textbook says "Take the first two values to see which is minimum" pi t/2 = pi/2 or 3pi/2
t= 1,3

I dont understand this part.

Answer 3

Well, the text is telling you to see which one produces a maximum value. Either t = 1 or t = 3 produces a maximum value. If you put t = 1 into the original equation P(1) = 100sin(pi (1)/2) + 500, you will get a maximum P = 600. If you put t = 3 into the original equation P(3) = 100sin(pi (3)/2) + 500, you will get a minimum P = 400. The t = 1 is clearly the maximum value.

Is this Calc 1 / AB? I'd imagine you would have learned how to calculate arc cosines in trigonometry/precalculus.

Answer 4

Uh, i live in Australia so i would imagine that the subjects would be different. But yeah, i want to know how the textbook got t=1,3?

Answer 5

Oh, I'm sorry. Anyway, I guess the values t = 1 and t = 3 are the first values that would come to mind that would produce a maximum/minimum. There's nothing really special about them though. If you plug any integer k into t = 1 + 4k to get a number t. That number t will produce a maximum value. Likewise, if you plug any integer k into t = 3 + 4k to get a number t, that number t will produce a minimum value.

Answer 6

Oh so, for any trig max/min problems the t= 1,3 would most likely be max/ win?

Answer 7

It depends on what is inside the sine. For example, if P = 100sin(t) + 500, then, the maxima would occur at t = pi / 2 + 2pi; the minima would occur at t = 3pi / 2 + 2pi.

It could also change if the sine was a cosine instead. It takes a fundamental understanding of trigonometry to see the relationship. But no, t = 1, 3 are not always the max and min t values...

Answer 8

oh right, guess i need to brush up on trig!

Answer 9

we cant find the max value of t . it will infinite as domain of this function is (-infinite, infinite) . we need not to find out maxima or minima of p(t)

Answer 10

Of course we can't find the max value of t. I assumed we were talking about relative extrema.