SOLUTION: for 1980 to 1985, the anual sales S, in thousands of dollars, of a shoe store store can be modeled by S= -0.5t^2+3t+42, where t=0 represents 1980. question: A: During which years d

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: for 1980 to 1985, the anual sales S, in thousands of dollars, of a shoe store store can be modeled by S= -0.5t^2+3t+42, where t=0 represents 1980. question: A: During which years d      Log On


   



Question 65507: for 1980 to 1985, the anual sales S, in thousands of dollars, of a shoe store store can be modeled by S= -0.5t^2+3t+42, where t=0 represents 1980. question: A: During which years did the store have sales of more than $50,000? B: During which year did the store achieve its max sales?"
Solving it as a quadratic is not the real question here one can find
t = 3+/-(93)^1/2.
How is this quadratic set up to relate the Sales to the time period?
One can even find the vertex of the parabola as (3, 93/2)
There is a 5 year sales (earning) period involved
S does not relate well in solving for t
There appears to be a minimum for S when t=0 in 1980 of 42

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
S+=+-0.5t%5E2+%2B+3t+%2B+42
The point where the graph begins is the S-intercept, (0,42).
The maximum is at (-b+%2F+2a, S) t%5Bmax%5D+=+-3+%2F+%28-.5%29%2A2+=+3
S+=+-0.5%2A3%5E2+%2B+3%2A3+%2B+42
S+=+-4.5+%2B+9+%2B+42
S+=+46.5
so, the max is at (3,46.5)
I would interpret that as the sales for the 3rd year
I think what isn't mentioned in the problem is that
t is in years and
t = 1 is the end of the 1st year
t = 2 is the end of the 2nd year
....
t = 5 is the end of the 5th year
The annual sales for any year never exceeded $46.5K
The annual sales at the end of '85 was $34.5K
S+=+-0.5%2A5%5E2+%2B+3%2A5+%2B+42
S+=+-22.5+%2B+15+%2B+42
S+=+34.5