Find the maximum and minimum values of the function g ( θ ) = 7 θ − 8 sin ( θ ) on the interval [ 0 , π /2 ]

Question

Find the maximum and minimum values of the function  on the interval 

I found the derivative to be 7-8cos(θ). I set it = to 0 and get 7/8 i plug f'(7/8) into 7-8cos(7/8) and get -.015348 which is wrong. then I plugged 0 and pi/2 into the prognonal function and get o and 7 which are both wrong. Please help!

Answer 1

The solution to your first derivative equation is not θ = 7/8. It is cos^-1(7/8) ≈ .50536. Evaluating g at that point gives g(.50536) = 7*.50536 - 8*sin(,50536) ≈ -.33546. Testing the endpoints of the interval, g(0) = 0, and g(π/2) = 7π/2 - 8 ≈ 3.0, which is the maximum.

Answer 2

When a function which is continuously differentiable is defined on a closed interval, we can find the minima and maxima by finding where the first derivative is 0 as well as checking the values at the endpoint.

f(θ) = 7θ - 8 sin(θ)

f'(θ) = 7 - 8 cos(θ)

8 cos(θ) = 7

cos(θ) = 7/8

As cos(θ) ranges from 0 to 1 in the first quadrant, there will be a solution.

As we are in the first quadrant, sin(θ) will be non-negative.

Then, when cos(θ) = 7/8, sin(θ) = √(1 - (7/8)²) = √(15/64) = 1/8 √15

The second derivative is 8 sin(θ). Thus, the second derivative is positive in the interior of the interval, so the function will be a minimum where cos(θ) = 7/8

θ = cos^-1(7/8)

Then, the minimum is 7 cos^-1(7/8) - 8 * 1/8 √15 or 7 cos^-1(7/8) - √15 = -0.335459774218317

The maximum will be at one of the 2 endpoints.

f(0) = 7 * 0 + 8 * sin(0) = 7 * 0 + 8 * 0 = 0

f(π/2) = 7 * π/2 - 8 * sin(π/2) = 7 π/2 - 8 * 1 = 7 π/2 - 8 is approximately 2.99557428756428

The maximum is 7 π/2 - 8, which is approximately 2.99557428756428

The minimum is 7 cos^-1(7/8) - √15, which is approximately -0.335459774218317

Answer 3

First make sure your calculator is in radians.

You were right with the step

f '(θ) = 7 - 8cos(θ)

Next to solve for θ rearage it to be

and cos(θ) = 7/8 at f '(θ) = 0

This means acos(7/8) = θ

θ = 0.5053

But the next step is to plug in θ = 7/8 into the original function

g (0.5053) = 7 *0.5053 − 8 sin (0.5053 ) = -0.3355

The other max and min could occur on the edges of the interval. (0 and π/2)

Plugging these in

g (0) = 7 *0− 8 sin (0) = 0

g (π/2) = 7 *π/2− 8 sin (π/2) = 2.9956

Remember sin(π/2) = 1

Max 2.9956

Min -0.3355