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Question 380855: Worker efficiency. In a study of worker efficiency at Wong
Laboratories it was found that the number of components
assembled per hour by the average worker t hours after
starting work could be modeled by the formula
N(t) = 3t^3 + 23t^2 + 8t
a) Rewrite the formula by factoring the right-hand side
completely.
b) Use the factored version of the formula to find N(3).
c) What is the time at which the workers are most efficient.
d) What is the maximum number of components assembled per
hour during an 8-hour shift.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula you gave is N(t) = 3t^3 + 23t^2 + 8t
I'll make x = t so this can be graphed and referenced easily.
the equation becomes n(x) = 3x^3 + 23x^2 + 8x
the graph of this formula is shown below:
the formula as given doesn't make any sense.
there is no maximum when x is greater than 0.
x cannot be less than 0 so any values on the graph where x is less than 0 are to be ignored.
your graph would make more sense if the equation was:
N(t) = -3t^3 + 23t^2 + 8t which I would translate to:
n(x) = -3x^3 + 23x^2 + 8x.
that graph is shown below:
now you have a maximum when x is greater than 0 which is more realistic.
I'll solve for n(x) = -3x^3 + 23x^2 + 8x.
x can be factored out to get:
n(x) = x * (-3x^2 + 23x + 8)
(-3x^2 + 23x + 8) factors out to be (3x+1) * (-x+8)
the completely factored equation becomes:
n(x) = x * (3x+1) * (-x+8)
when x = 0, n(x) = 0 as it should because no time worked = no components produced.
when x = 3, n(x) = (3) * (10) * (5) = 150 using the factored equation.
when x = 3, n(x) = -3(3^3) + 23*(3^2) + 8*3 = -81 + 207 + 24 = 150
you get the same answer with the factored equation and the non factored equation which confirms that the factorization is good.
I believe we need to use calculus to find the maximum / minimum point of this cubic equation.
I couldn't find any formula not using calculus on the web, so we'll use calculus.
The maximum point occurs when the derivative of the equation equals 0.
the equation is -3x^3 + 23x^2 + 8x
The derivative of this equation is -9x^2 + 46x + 8
This derivative is 0 when x = 5.279477927
That could not be factored easily, so I used the quadratic formula of:
x = (-b +/- sqrt(b^2-4ac))/(2a)
When x = 5.279477927, n(x) = 241.8493507
I'll add a line at n(x) = 241.8493507 to the graph of the equation and it should confirm that is the maximum point.
the graph looks like this:
The graph confirms the maximum point.
Since x represents t which represents the number of hours after starting work, then the maximum amount of output per hour occurs 5.279477927 hours after work starts.
The maximum amount of output per hour during the 8 hour shift becomes 241.8493507 units per hour which occurs at that time.
Best I can do in the short period of time allotted to the solving of this problem.
I'm not a calculus guru, but I know a little bit about it, and the little bit I know was helpful in this case.
If you were not to use calculus, then I don't really know how to find the maximum point of the cubic equation other than by iteration.
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