The monthly sales S (in hundreds of units) of baseball equipment for an Internet sporting goods site are approximated by
S = 60.1 − 43.9 cos πt/ 6
where t is the time (in months), with t = 1 corresponding to January. Determine the months when sales exceed 7700 units at any time during the month.
is c the correct answer?
60.1 − 43.9 cos πt/ 6 ≥ 77
− 43.9 cos πt/ 6 ≥ 16.9
cos πt/6 ≤ - .38496...
consider cos πt/6 = - .38496...
I know cos 1.1756.. = +.38496
so πt/6 = π - 1.1756 = appr 1.966
t = 3.755
or
πt/6 = π + 1.1756.. = 4.3172..
t = 8.245
draw you conclusion regarding the month
verification by Wolfram:
http://www.wolframalpha.com/input/?i=solve+y+%3D+60.1+%E2%88%92+43.9+cos+(%CF%80t%2F6)+,+y+%3D+77
look at the graph for t ≥ 0 at the first intersection
the months are
a. march through september
b. april through august
c. march through august
d. may through september
e. april through august
i think the answer is c
To determine the months when sales exceed 7700 units at any time during the month, we need to solve the equation:
S > 7700
Let's substitute the given equation for S:
60.1 − 43.9 cos (πt/6) > 7700
Now we can solve for t. Here's how:
Step 1: Subtract 60.1 from both sides to isolate the cosine term:
-43.9 cos (πt/6) > 7700 - 60.1
Simplify:
-43.9 cos (πt/6) > 7640.9
Step 2: Divide both sides by -43.9. Since we're dividing by a negative number, we need to reverse the inequality sign:
cos (πt/6) < (7640.9 / -43.9)
Simplify:
cos (πt/6) < -174.18
Step 3: Inverse the cosine function to get rid of the inequality:
(πt/6) > acos(-174.18)
Simplify:
t > (6/π) * acos(-174.18)
The result of acos(-174.18) is undefined because the cosine function operates between -1 and 1. This means there are no months when sales exceed 7700 units at any time during the month.