# P = 3.44x – 0.00005 x^2– 5000, 0 ≤ x ≤ 67,300.The profit (in dollars) made by a fast-food restaurant selling x hamburgers isP = 3.44x – 0.00005 x^2– 5000, 0 ≤ x ≤ 67,300.Find the...

P = 3.44x – 0.00005 x^2– 5000, 0 ≤ x ≤ 67,300.

The profit (in dollars) made by a fast-food restaurant selling x hamburgers is

P = 3.44x – 0.00005 x^2– 5000, 0 ≤ x ≤ 67,300.

Find the open intervals on which P is increasing or decreasing.

### 1 Answer | Add Yours

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).**

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