# Thread: Calc - Extreme Values - Word Problem

1. ## Calc - Extreme Values - Word Problem

The profit, P, in dollars for selling x hamburgers is modelled by $P(x)= 2.44x - \frac {x^2}{20 000} - 5 000$, where $0 \leq x \leq 35 000$. For what quantities of hamburgers is the profit increasing and decreasing?

$P'(x) = 2.44 - 40000x$

$0 = 2.44 - 40 000x$

$6.1^{-0.5} = x$

$P(6.1^{-0.5}) = -4999$

$P(0) = -5000$

$P(35 000) = 19 150$

Therefore, decreasing at $(0, -5000)$ and increasing at $(35 000, 19 150)$

increasing - $0 \leq x \leq 24 400$
decreasing - $24 400 \leq x \leq 35 000$

2. Hello,

Originally Posted by Macleef
The profit, P, in dollars for selling x hamburgers is modelled by $P(x)= 2.44x - \frac {x^2}{20 000} - 5 000$, where $0 \leq x \leq 35 000$. For what quantities of hamburgers is the profit increasing and decreasing?

$P'(x) = 2.44 - {\color{red}40000x}$ << SiMoon says : it should be $\frac{x}{10000}$ !!

$0 = 2.44 - 40 000x$

$6.1^{-0.5} = x$ << SiMoon says : if you were right, it would have been $6.1 \cdot {\color{red}10^5}$ !

$P(6.1^{-0.5}) = -4999$

$P(0) = -5000$

$P(35 000) = 19 150$

Therefore, decreasing at $(0, -5000)$ and increasing at $(35 000, 19 150)$

increasing - $0 \leq x \leq 24 400$
decreasing - $24 400 \leq x \leq 35 000$
Therefore, there are problems all along :/

$P'(x)=\frac{61}{25}-\frac{x}{10,000}$
$\frac{61}{25}-\frac{x}{10,000}=0$
$x=24,400$