The sum of two positive integers is 9. What is the least possible sum of their reciprocals?
Printable View
The sum of two positive integers is 9. What is the least possible sum of their reciprocals?
the two integers would have to be 4 and 5. hence the sum being 9/20, which is the smallest.
Yes i would agree with Acidic. 4 + 5 = 9 --> 4/1 + 5/1 --> 1/4 + 1/5 --> 9/20.
i dont know anything about maths and dont know what a reciprocals is but why cant the 2 numbers be 2+7 = 9?
why do they have to be 4 + 5
1 + 8 = 1/1 + 1/8 = 9/8
2 + 7 = 1/2 + 1/7 = 9/14
3 + 6 = 1/3 + 1/6 = 9/18
4 + 5 = 1/4 + 1/5 = 9/20
Smallest is one with largest denominator, hence 4 + 5.
a reciprocal of a number is that number to the power -1Quote:
Originally posted by davebat
...dont know what a reciprocals is
so the reciprocal of 2 is 1/2, of 10 is 1/10 of n is 1/n or n-1
In general we should say that:
1/x + 1/y = (x+y)/xy
Now minimising 1/x + 1/y is equivalent to minimising (x+y)/xy
As (x+y) is constant, the problem reduces to maximizing xy.
We know (x+y)^2 = (x-y)^2 + 4xy
Thus if (x+y) is given, we can get xy to me maximum if x=y because in that case only (x-y)^2 will be zero, which is least for a square.
In short, if sum of two numbers is constant, their product is maximum when they are equal.
So making x and y towards equal, we will get 4 and 5.
This concept can be generalised however big the number is.
Similarly, vice versa, if product of two numbers is constant, their sum is minimum when they are equal.