# Thread: Calc - Extreme Values - Word Problem

1. ## Calc - Extreme Values - Word Problem

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

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

$\displaystyle 0 = 2.44 - 40 000x$

$\displaystyle 6.1^{-0.5} = x$

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

$\displaystyle P(0) = -5000$

$\displaystyle P(35 000) = 19 150$

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

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

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

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

$\displaystyle 0 = 2.44 - 40 000x$

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

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

$\displaystyle P(0) = -5000$

$\displaystyle P(35 000) = 19 150$

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

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

3. Check your derivative.

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

$\displaystyle \frac{61}{25}-\frac{x}{10,000}=0$

$\displaystyle x=24,400$

Now check the slope on either side.

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