# The net monthly profit, in dollars, from the sale of a certain item is given by the formula P(x) = 10^6[1 + (x-1)e^0.001x], where x is the number ofitems sold. Determine the number of items that...

The net monthly profit, in dollars, from the sale of a certain item is given by the formula P(x) = 10^6[1 + (x-1)e^0.001x], where x is the number of

items sold. Determine the number of items that yield the maximum profit. At full capacity, the factory can produce 2000 items per month.

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`P(x)=10^6(1+(x-1)e^(0.001x))`

When in doubt take the derivative using the product property.

`P'(x)=10^6((x-1)(0.001e^(0.001x))+e^(0.001x))=10^6e^(0.001x)(0.001x+0.999)`

P'(x)=0 when (0.001x+0.999)=0

Solving for x =-999. This is the only critical points, but it is not possible to make -999 items.

We also have to check x=0 and x=2000

`P(0)=10^6(1+0e^(0.001x))=10^6`

`P(2000)=10^6(1+1999e^(0.001(2000))=10^6(1+1999e^2)~~1.477xx10^10`

` `**The solution is to produce 2000 items.**