SOLUTION: For all positive integers x and y such that 1/x + 1/y = 1/12, find the greatest value that y can have. What is the way to do this? Thanks!

Algebra ->  Expressions -> SOLUTION: For all positive integers x and y such that 1/x + 1/y = 1/12, find the greatest value that y can have. What is the way to do this? Thanks!      Log On


   

Question 1048379: For all positive integers x and y such that 1/x + 1/y = 1/12, find the greatest value that y can have.
What is the way to do this? Thanks!
Found 2 solutions by josgarithmetic, robertb:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Do you need a specific way of dealing with this, or is making a graph acceptable and looking for integer points?

12xy is common denominator.

12xy%281%2Fx%2B1%2Fy%29=12xy%281%2F12%29

12y%2B12x=xy

xy-12y=12x

y%28x-12%29=12x

y=%2812x%29%2F%28x-12%29
but does this have a maximum?
dy%2Fdx=%28%28x-12%29%2A12-12x%2A1%29%2F%28x-12%29%5E2, derivative, Quotient Rule.

dy%2Fdx=%2812x-144-12x%29%2F%28x-12%29%5E2

dy%2Fdx=-144%2F%28x-12%29%5E2------THIS IS NEVER 0.
But you are looking for POSITIVE integers.

You could try positive integers x starting with 0, on y=%2812x%29%2F%28x-12%29. Would any acceptable y, integer, also be positive?

graph%28300%2C300%2C-8%2C8%2C-8%2C8%2C12x%2F%28x-12%29%29

NO.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
1%2Fx%2B1%2Fy+=+1%2F12
<===> y=%2812x%29%2F%28x-12%29 <===> y=12%281%2B12%2F%28x-12%29%29.
Since x and y are positive integers, x-12 must divide 12.
===> x-12 = 1,2,3,4,6,12
Obviously, the value of x-12 that would maximize y is 1, which would give x = 13. The corresponding value of y would be 156.