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P = 3.44x – 0.00005 x^2– 5000, 0 ≤ x ≤ 67,300.The profit (in dollars) made by a...
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You should remember that the first derivative of function helps you to find the intervals where the function increases or decreases.
You should find the first derivative such that:
`P'(x) = (3.44x - 0.00005 x^2 -5000)'`
`P'(x) = 3.44 - 0.0001x`
You need to determine the zero of `P'(x)` such that:
`P'(x) = 0 =gt 3.44 - 0.0001x = 0 =gt x = 3.44/0.0001`
`x = 34400`
Hence, the values of P'(x) are positive if `x in (0,34400) ` and negative if `x in (34400 , 67300), ` hence, the function increases over (0,34400) and it decreases over (34400 , 67300).
Posted by sciencesolve on June 26, 2012 at 5:21 AM (Answer #1)
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