How to adjust a ratio to create desired proportion Let's say I have a bunch of apples and oranges. Specifically 500 apples and 760 oranges. 
I need to create a big basket of fruit. For every 7 apples I include, I need to include 3 oranges. 
I want to make the biggest basket possible using as much of my fruit as I can while honoring the desired apple to orange ratio.
I'm not mathematically inclined and I've been just taking bunches of 7 and 3 at a time until I run out. There has to be some kind of formula I can use to tell me upfront the max number of each that I should pull.
Thanks in advance.
 A: Suppose you put $n$ groups of 7-apples-and-3-oranges into the big basket. Then you will have $7n$ apples and you will have $3n$ oranges in the basket.
If all of the apples are used up, you would have $7n=500$, which means you would have $n=500/7 = 71\frac37$ groups. That means you can have at most $71$ complete groups, as far as the apples are concerned.
On the other hand, if all of the oranges are used up, you would have $3n=760$, which means you would have $n=760/3=253\frac13$ groups. That means you can have at most $253$ complete groups, as far as the oranges are concerned.
So you will not be able to make any more groups once you have made $71$ groups (the apples will be exhausted then).
A: You have more oranges than apples at your disposal, but you need more than twice as many apples in the basket as you do oranges.
So, you'll run out of apples and have a lot of oranges left over.
If you have $7$ apples, you need $3$ oranges.
If you have $14$ apples, you need $6$ oranges.
If you have $7n$ apples, where $n$ is a positive integer, then you'll need $3n$ oranges.
What value of $n$ will let you use most of the apples?  Well, we can calculate
$$7n = 500 \to n = 71\text{ }3/7.$$
So $n = 71$, which means the number of apples is $7n = 7 \cdot 71 = 497$ and the number of oranges is $3n = 3 \cdot 71 = 213.$
A: We assume that fractional fruits are not sensible things to include in our basket. We need $7$ apples per $3$ oranges, but we can see that we have more oranges than apples. So, we know that apples are the limiting reagent. Because we only want natural numbers of fruits, and $7,3$ are relatively prime, we know that the largest multiple of $7$ which fits in $500$ is the number of apples we can have. From there we can determine the number of oranges.
As it turns out, we can have $71\times 7=497$ apples in the basket, and so $71\times3=213$ oranges. That is, we can have at most $710$ pieces of fruit: $497$ apples and $213$ oranges. This maximizes the ratio-honoring basket size.
A: HINT: 
$a\colon o = 7 \colon 3$ and the fraction $7\colon 3$ is in lowest terms so $a = 7 q$, $o = 3 q$ with the conditions $7 q \le 500$ and $3 q \le 760$ so $q \le \frac{497}{7} = 71$ and $q \le \frac{759}{3} = 253$, so the maximal amounts are given by the maximal $q$, $71$ and are 
$$a = 7 \cdot 71 = 497 \\
o = 3 \cdot 71 = 213$$
${\bf Added:}$ In general, if the proportion of apples and oranges is given 
$a\colon o = m\colon n$ a fraction in lowest terms and the available quantities of fruits are $A$ and $O$, then the largest amounts are $a = m q$, $o = n q$ where
$$q = \min (\left[ \frac{A}{m}\right], \left[ \frac{O}{n}\right])$$
the largest integer $\le$ both $\frac{A}{m}$ and $\frac{O}{n}$.
