SOLUTION: Given positive integers x and y such that 1/x,+1/y =1/12 , what is the smallest possible value for x + y?

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Question 1075905: Given positive integers x and y such that 1/x,+1/y =1/12 , what is the smallest possible value for x + y?
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
The equation 1%2Fx%2B1%2Fy+=+1%2F12+ can be re-arranged:
+%28x%2By%29%2Fxy+=+1%2F12+
+12%28x%2By%29+=+xy+ (1)

If we think of xy as the area of a rectangle, then the right hand side of (1) is the area, and the left hand side is 6 times the perimeter. The maximum area for minimal perimeter is when the shape is a square (x=y), so let's try x=y:
+12%282x%29+=+x%5E2+
++24x+=+x%5E2+
+x%5E2+-+24x+=+0+
+x%28x-24%29+=+0+
Discard x=0 because x and y are positive integers.
++x=24+ —> +y=24+
So +highlight%28x%2By+=+48%29+ is the minimum value of x+y.