Can some one help with this one? I am new to derivatives.

Question

A company finds that the profit 𝑃(𝑥) where 𝑥 represents thousands of units, is given by
𝑃(𝑥)= $−x^3+9x^2−15x−9$

If the company can only make a maximum of 6000 units, what is the absolute maximum profit?

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Answer 1 · 2018-06-15T07:57:37

At 5,000 units, the absolute maximum profit is $P = 16$ .

$P(x) = −x^3+9x^2−15x−9$

$x in [0,6]$

1st derivative to optimise :

$P'(x) = −3x^2+ 18x −15 = -3 (x - 5) (x - 1) = 0$

$P' = 0 implies x = 1,5$

2nd derivative test for max/min :

$P''(x) = −6x+ 18$

So profit is maximised at $x = 5, P = 16$

That is, at 5,000 units, the profit is $16$ .