word equation

[wildcornstalk]

annual profit in thousands of dollars is given by the function, P(x)=5000-(1000/(x-1)), where x is the number of items sold in the thousands, x>1.

question: describe the meaning of 5000 in the equation and describe the number of 1 in the equation.

i think the 5000 is the total profit???? and i have no clue what the 1 is for...

Thank you for your time.

 

[galactus]

Remember, Profit=Revenue-Cost

You have \(\displaystyle 5000-\frac{1000}{x}-1\) written, but I assume you mean:

\(\displaystyle P(x)=\underbrace{5000}_{\text{revenue}}-\underbrace{\frac{1000}{x-1}}_{\text{cost}}\).

What if x=1?. We have division by 0. What does this tell us?. Any asymptotes involved perhaps?.

 

[mmm4444bot]

P(x) = 5000 - (1000/x-1))

You typed mismatched parentheses on the righthand side.

Please proofread and correct your typing; otherwise, we can only guess.

Galactus guessed that the expression defining function P is 5000 - 1000/(x - 1).

If this is what you meant to type, then that's bad news for the company selling less than 1,201 items.


For example, if your company sold 1,005 items during the year, then you lost $195,000.


Are you in a beginning algebra class? Functions with asymptotic behavior are not discussed in such a class, so I'm wondering what's going on here.

 

[soroban]

Hello, wildcornstalk

\(\displaystyle \text{Annual profit in thousands of dollars is given by the function: }\(x)\:=\:5000 -\frac{1000}{x-1}\)
\(\displaystyle \text{where }x\text{ is the number of items sold in the thousands, }x\,>\,1.\)

\(\displaystyle \text{Describe the meaning of 5000 in the equation, and describe the meaning of 1 in the equation.}\)

Crank out a few values:

\(\displaystyle \begin{array}{cccccccc} P(2) &=& 5000-\frac{1000}{1} &=& 4000.00 \\ \\[-3mm] P(3) &=& 5000-\frac{1000}{2} &=& 4500.00 \\ \\[-3mm] P(5) &=& 5000 - \frac{1000}{4} &=& 4750.00 \\ \\[-3mm] P(11) &=& 5000 - \frac{1000}{10} &=& 4900.00 \\ \\[-3mm] \vdots && \vdots && \vdots \\ \\[-3mm] P(101) &=& 5000 - \frac{1000}{100} &=& 4990.00 \\ \\[-3mm] P(2001) &=& 5000 - \frac{1000}{2000} &=& 4999.50 \\ \\[-3mm] P(20,001) &=& 5000 - \frac{1000}{20,000} &=& 4999.95 \end{array}\)


We see that the profit approaches $5000 as a limit.
. . It is the maximum possible profit.


Here's my interpretation of the "1".

The first one thousand items sold are not considered in the profit function.
. . (Perhaps they cover the cost of production?)
Then any excess produces a profit for the company.

 

[mmm4444bot]

Oops, I missed the given units on P(x). This symbol represents the number of thousands of dollars.

So, in my example with P(1.005), the annual loss is much worse than previously reported: $195,000,000. :shock:

Again, you posted on the beginning algebra board. Is this the course that you're taking?

Soroban's reasoning is as good as we can do, I think, without knowing more about the actual revenue and cost functions.

 

[wildcornstalk]

I apologize, yes it was meant to be P(x)=5000-(1000/(x-1)). I am in college algebra and the topic for the section is functions and their graphs, in other problems there are asymptotes involved but I see nothing about it in this question. I thank you all for your time!!!

 

[tkhunny]

Well, you do have your asymptotes at x = 1 and P = 5000.

 

[shadiq]

Actually the formula is : P(x)=5000-1000/(x-1).