What is the max value of the function 12/(2sin(x+60)+4) and when does this occur? What working should I show?

We need to minimize the denominator to make the function max :

-1 ≤ sin (x + 60) ≤ 1 => if sin(x + 60) = -1, =>

2sin(x + 60) + 4 = 2

=> 12/[2sin(x + 60) + 4] = 12/2 = 6

The maximum value of the function is 6.

When we have this maximum value? At what x values?

sin(x + 60) = -1

x + 60 = 360n - 90 , n any integer

x = 360n - 150 , n any integer

if n = 1, x = 210 degrees

n = 2, x = 510 degrees

n = -1, x = -570 degrees

n = 0, x = -210 degrees

Since sine is a periodic function, there are infinitely many x values so that the function has a maximum value.

Hope this helps.

Put u = x + 60 and f(u) = 6/[2 + sin(u)] = 6(2 + s)^(-1). where s = sin(u). Now f'(u) = -6c(2 + s)^(-2), c = cos(u).

Now f'(u) = 0 iff c = 0, ie., u = - pi/2 or + pi/2, ie., x + pi/3 = - pi/2 or + pi/2, ie., x = - pi/2 - pi/3 or + pi/2 - pi/3, ie.,

x = - (5/6)pi or (1/6)pi. Now f''(u) = 6s(2 + s)^(-2) + 12c^2(2 + s)^(-3) = 6(2 + s)^(-3)[s^2 + 2s + 2c^2]

= 6(1 + 2s + c^2)(2 + s)^(-3).Now f''(-pi/2) = 6(1 + 2[-1] + [0]^2) = - 6 ,< 0-->rel max at u = - pi/2. f''(pi/2) = 6(1 + 2

+ 0)(3)^(-3) = 1/9 , >0--> rel min at u = pi/2. Therefore f(u) has a rel max at (u , f(u)) = (- pi/2 , 6) and f(x) has a rel

max at (x , f(x)) = (-[5/6]pi , 6).Clearly, examination of the function reveals that the max of the function will occur

when the denominator is a minimum, ie., when sin(x + 60) = -1, thus reducing the denominator to 2 and giving

a max of 12/2 = 6.

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