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x - y = 100 and xy is a minimum.
1st we solve `x-y=100` for y to get `y = x - 100` . substituting we get `xy = x(x-100) = x^2 - 100x`
`f(x) = x^2 - 100x` , `f'(x) = 2x - 100` , this is zero when x = 50, so y = -50. We can use the second derivative test to see `f''(x) = 2` which is positive, so (50, -50) is a minimum.
Answer: 50, -50 are the two numbers.
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