SOLUTION: The following equation models the rate at which a worker produces so many units after x hours when the worker starts at 8AM. (Assume 8AM means x=0) f(x)=-2x^2+8x+50 At what

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Question 1182599: The following equation models the rate at which a worker produces so many units after x hours when the worker starts at 8AM. (Assume 8AM means x=0)
f(x)=-2x^2+8x+50

At what time is the worker working most efficiently and how many units is the worker producing at this time?

Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
The following equation models the rate at which a worker produces so many units after x hours when the worker starts at 8AM (Assume 8AM means x=0)

At what time is the worker working most efficiently and how many units is the worker producing at this time?
-> this is downward parabola and max is at vertex
so, write equation in vertex form
...........factor out
.......complete square
.......



=>, => vertex is at (,)
Assuming means , will be

the worker working most efficiently at and producing units at this time




Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!

The following equation models the rate at which a worker produces so many units after x hours when the worker starts at 8AM. (Assume 8AM means x=0)
f(x)=-2x^2+8x+50

At what time is the worker working most efficiently and how many units is the worker producing at this time?
IGNORE the RUDDISH that woman is "feeding" you. You don't need to Complete the Square unless of course, you're required
to or want to! That woman has a LOT to learn, but she never will because she's hard-headed and doesn't believe she should
change, and learn to do mathematics the right way.
With x being time, the worker works most efficiently when he/she puts in the number of hours that produces the maximum number
of units. Those maximum units occur when time, or
From equation , we then get:
Now, with maximum units produced when x = 2 hours, we get maximum units of
AS [x, f(x)], or (x, y) is (2, 58), time when he/she works most efficiently is: 8:00 a.m. + 2 hours = 10:00 a.m.
At 10:00 a.m., he/she will have produced maximum units of 58.
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