1. ## Revenue

The revenue of McMenamy's fish shanty is approximately

R(t)=2(5-4cos (pi/6)t) (0<t<12)

during the t week (t=1 corresponds to the first week of June) where R is measured in thousands of dollars. On what week is the marginal revenue zero?

2. Originally Posted by lemontea
The revenue of McMenamy's fish shanty is approximately

R(t)=2(5-4cos (pi/6)t) (0<t<12)

during the t week (t=1 corresponds to the first week of June) where R is measured in thousands of dollars. On what week is the marginal revenue zero?
Marginal R is dR/dt = R'(t)

R(t) = 2{5 -4cos[(pi/6)t]}
R'(t) = 2[(-4)(-sin[pi/6)t] *(pi/6)]
R'(t) = (8pi/6)sin[(pi/6)t]

When R'(t) = 0.
0 = (8pi/6)sin[(pi/6)t]
0 = sin[(pi/6)t]
(pi/6)t = arcsin(0)
(pi/6)t = 0, or 2pi

When (pi/6)t = 0,
t = 0 -----------------**

When (pi/6)t = 2pi,
t = (2pi)/(pi/6) = 12 ----**

Therefore, the marginal revenue is zero at week zero and week 12. ----answer.

3. marginal revenue is the derivative of the revenue function right?

so, solve $\displaystyle t$ such that $\displaystyle R'(t)=0$.

EDIT: didn't ticbol's solution..

in addition, t=6 is also a solution since arcsin 0 = 0, pi, or 2pi..

EDIT: i mean, "didn't see ticbol's solution.."

4. Originally Posted by kalagota
marginal revenue is the derivative of the revenue function right?

so, solve $\displaystyle t$ such that $\displaystyle R'(t)=0$.

EDIT: didn't ticbol's solution..

in addition, t=6 is also a solution since arcsin 0 = 0, pi, or 2pi..
Yes, I forgot about arcsin(0) = pi also.

Thanks.