# Thread: deravitive and profit analysis

1. ## deravitive and profit analysis

The question is: A fast food restaurant determines the cost and revenue models for its hamburgers to be:

c = .06X = 7500 ; 0 <= X <= 50000

r= (1/20000)(65000x - x^2) ; 0 <= X <= 50000

a. Write the profit function for this situation

b. determine the intervals the function decreases / increases

c. determine how many hamburgers the restaurant needs to sell to obtain max profit.

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I have found the profit function to be p= r-c

so I got : P = (1/20000)(65000x - x^2) - (.6x + 7500)

and P'= (65000x - x^2) + (1/20000)(65000 - x) - .6

but this is not the answer in the book. I'm not sure what I am doing wrong.

Thanks!

2. Here's a hint: if $\displaystyle p=r-c=\frac{1}{20000}(65000x-x^2)-0.6x-7500$, then

\displaystyle \begin{aligned} p'&=\frac{d}{dx}\left(\frac{1}{20000}(65000x-x^2)-0.6x-7500\right) \\ &=\frac{1}{20000}(65000-2x)-0.6. \\ \end{aligned}

As $\displaystyle p=0$ at the endpoints of the domain $\displaystyle x=0$ and $\displaystyle x=50000$, we know that the maximum profit will be attained at a point where

$\displaystyle p'=0$.

3. Originally Posted by craziebbygirl
The question is: A fast food restaurant determines the cost and revenue models for its hamburgers to be:

c = .06X = 7500 ; 0 <= X <= 50000

r= (1/20000)(65000x - x^2) ; 0 <= X <= 50000

a. Write the profit function for this situation

b. determine the intervals the function decreases / increases

c. determine how many hamburgers the restaurant needs to sell to obtain max profit.

__
I have found the profit function to be p= r-c

so I got : P = (1/20000)(65000x - x^2) - (.6x + 7500)

and P'= (65000x - x^2) + (1/20000)(65000 - x) - .6

but this is not the answer in the book. I'm not sure what I am doing wrong.

Thanks!
first of all, is it ...

$\displaystyle c = .06x + 7500$ or

$\displaystyle c = .6x + 7500$ ?

you set your profit function up correctly, but your derivative is incorrect.

If $\displaystyle P = (1/20000)(65000x - x^2) - (.6x + 7500)$ , then

$\displaystyle P' = (1/20000)(65000 - 2x) - .6$

4. yes it is a .6

So you do not need to do a product rule for th (1/20000)(65000 - 2X)? I thought that you needed to do the (f' of g) + (g' of f) for that.

5. Originally Posted by craziebbygirl
yes it is a .6

So you do not need to do a product rule for th (1/20000)(65000 - 2X)? I thought that you needed to do the (f' of g) + (g' of f) for that.
$\displaystyle \frac{1}{20000}$ is just a konstant.

$\displaystyle \frac{d}{dx}[k \cdot f(x)] = k \cdot f'(x)$

6. When you use the product rule, the result is

\displaystyle \begin{aligned} \frac{d}{dx}\left(\frac{1}{20000}(65000x - x^2)\right) &= \frac{1}{20000}(65000 - 2x)+0\cdot(65000x - x^2) \\ &= \frac{1}{20000}(65000 - 2x).\\ \end{aligned}

The general rule for derivatives of constant multiples of functions is

$\displaystyle \frac{d}{dx}(af(x))=af'(x).$