Max Profit & Number of Items for Maximum Profit: Find Out Now!

[recoil33]

The total profit $P, generated from the production and marketing of n items of a certain product is given by:

P = -10800*n-4*n³+600*n²-166

How many items should be made for maximum profit? What is the maximum profit?

Firstly, I think i would find the derivative of the function.

P' = (-10800)(12n²) + 1200n
P' = (-900)(n²)+100

P'' = (24n + 1200)

Therefore because the second derivative is positive, it means that it would a relative minimum?

Although, I'm stuck on how i should go upon figuring out the initial question?
I figured out earlier, that (x = 10, x = 90). Can't remember how, because my book with my working out is not with me.

Any help would be appreciated, thank you.

 

[SteamKing]

I would double check the calculation of P' and P".

 

[recoil33]

I would double check the calculation of P' and P".

P' = -12(x² - 100x - 900)
(I converted n to x because i find it easier to work with)

Expand the brackets:
P' = -12x² + 100x + 900

Therefore:

P'' = -24x + 100

 

[SteamKing]

You're still having problems doing your algebra correctly. Double check your latest expression for P' when expanding and removing the brackets.

 

[recoil33]

You're still having problems doing your algebra correctly. Double check your latest expression for P' when expanding and removing the brackets.

That was a mistake, i must have typed it without thinking..

P'(x) = -12(n² - 100n - 166)
P'(x) = -12n² + 1200n - 1080

P''(x) = -24n² - 1200

When P'(0) = -12(n² - 100n - 166)

x = 10, x = 90

P''(10) = -24(10)² - 1200
= -1440

P''(90) = -24(90)² - 1200
= -3360

 

[SteamKing]

In order to find out the number of items which give the maximum profit, you will have to substitute the two solutions from P' = 0 back into the original expression P(x) for profit. One of these values of x (or n) will give a profit which is greater than the other value of x will.