Maximize two variable equation

[malaclemys]

Hello,

This will probably sound like a HW question, but it's not. There is a certain phone system that I can use to call someone. There is an outgoing charge per call and a per minute charge.

The per minute charge is $ 7/18
The outgoing call charge is $299/90

Say I have $50 on the phone system. I'd like to maximize the number of calls I can make to this person, but also have a good balance on the duration of those calls. What is the value that maximizes both the number of calls and the duration of those calls?

I set up an equation like x=\frac{50}{299/90+7/18y}, where x is the number of calls and y is the number of minutes of the call.

The question is, what is the value in which x and y are maximum in respect to one another...

 

[Simon Bridge]

You need another equation that describes what you are trying to maximize in math terms.
If you just want a balance, then you are probably thinking of some sort of weighted average ... i.e. maximizing (ax+by): a+b=1.

 

[LCKurtz]

Hello,

This will probably sound like a HW question, but it's not. There is a certain phone system that I can use to call someone. There is an outgoing charge per call and a per minute charge.

The per minute charge is $ 7/18
The outgoing call charge is $299/90

Say I have $50 on the phone system. I'd like to maximize the number of calls I can make to this person, but also have a good balance on the duration of those calls. What is the value that maximizes both the number of calls and the duration of those calls?

I set up an equation like x=\frac{50}{299/90+7/18y}, where x is the number of calls and y is the number of minutes of the call.

The question is, what is the value in which x and y are maximum in respect to one another...

You can't maximize x and y "with respect to each other", whatever that means.

Let's call your connection charge c and your per minute charge m and your $50 limit L. If you make x calls averaging y minutes each your total cost is x(c+my), and if you are to use up all your money you have$$
x(c+my) = L$$which agrees with your equation. The extreme possibilities are to use up all your money with connections and no talking, so $$x=\frac L c$$ is your theoretical maximum number of calls. With your numbers, $\frac L c\approx 15$. I'm assuming $15$ calls of $0$ duration wouldn't be satisfactory. The other extreme is to use up all your money in one call. That call could last $\frac{L-c} m$ which is about $120$ minutes. That may or may not be satisfactory to you, it's up to you. The point is, as Simon as noted, you have to decide what is important and what, exactly, you want to maximize. Maybe a weighted sum of $x$ and $y$ or something like $xy$.

 

[malaclemys]

Thank you very much for both of your responses. Let's say I wanted to maximize the term xy. How would I go about doing that? How would I find the values of x and y for which the term xy is maximized? Would I have to take partial derivatives in respect to x and y?

Sorry for all the questions, but it's been many years since I've been out of school!

 

[LCKurtz]

Thank you very much for both of your responses. Let's say I wanted to maximize the term xy. How would I go about doing that? How would I find the values of x and y for which the term xy is maximized? Would I have to take partial derivatives in respect to x and y?

Sorry for all the questions, but it's been many years since I've been out of school!

Well, if it isn't a HW question and is just something you want the answer to, I wouldn't use calculus at all. You already know the only possible number of phone calls is between 1 and 15. Just make a spreadsheet with the first column the $x$ values from 1 to 15. Make the next column the formula solved for y:$$
y = \frac{L-cx}{mx}$$Make the third and following columns whatever you want to try like xy or ax + by with weights or whatever. Play with it until you find something you like.

 

[Simon Bridge]

The quantity you want to maximize is $z: z=xy$ and you know that $x$ and $y$ are related.
Since x is a whole number, you can solve for x and sub into the relation for z, then find where dz/dx=0 ... round to the nearest whole number etc.

But I'm with LCKurtz: there are only 15 values of x you need to know y for - just crunch the numbers 15 times.