SOLUTION: The profit P for a product whose sales fluctuate with the seasons is estimated to be P= 14 + 5sin(pit/52), where t is given in weeks and P is in thousands of dollars. Determine the

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: The profit P for a product whose sales fluctuate with the seasons is estimated to be P= 14 + 5sin(pit/52), where t is given in weeks and P is in thousands of dollars. Determine the      Log On

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Question 1012885: The profit P for a product whose sales fluctuate with the seasons is estimated to be P= 14 + 5sin(pit/52), where t is given in weeks and P is in thousands of dollars. Determine the number of weeks it would take for the profit to initially reach $18,000.
I attempted this problem, but when I plug the sin inverse of a number into my calculator, it says there is a math error. Thank you!

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+P+=+14+%2B+5%2Asin%28+pi%2A%28+t%2F52+%29+%29+
The greatest positive value for ANY
sin function is +1+, so +P%5Bmax%5D+ =
+P%5Bmax%5D+=+14+%2B+5%2A1+
+P%5Bmax%5D+=+19+ in thousands, so
that is $19,000.
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The problem makes sense, so I can continue
+P+=+18+
+18+=+14+%2B+5%2Asin%28+pi%2A%28+t%2F52+%29+%29+
++5%2Asin%28+pi%2A%28+t%2F52+%29+%29+=+4+
+sin%28+pi%2A%28+t%2F52+%29+%29+=+4%2F5+
Let +pi%2A%28+t%2F52+%29+=+theta+
+sin+%28+theta+%29+=+4%2F5+
+arc+sin+%28+4%2F5+%29+=+.9273+
( note that I had to get the angle measured in radians
because of the +pi+ given
+theta+=+.9273+
+pi%2A%28+t%2F52+%29+=+.9273+
+t%2F52+=+.9273+%2F+3.1416+
+t%2F52+=+.2952+
+t+=+15.3488+
In the 15th week, the profit will reach $18,000
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check:
+P+=+14+%2B+5%2Asin%28+pi%2A%28+t%2F52+%29+%29+
+P+=+14+%2B+5%2Asin%28+pi%2A%2815.3488%2F52+%29+%29+
+P+=+14+%2B+5%2Asin%28+.9273+%29+
+P+=+14+%2B+5%2A.8+
+P+=+14+%2B+4+
+P+=+18+
OK