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find two numbers a and b (with a≤b) whose difference is 38 and whose product is minimized

Mathematics
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if one number is \(x\) and the other is \(y\) then their difference is \(x-y=38\) making \(y=x-38\)
you want to minimize the product \[x(x-38)=x^2-38x\]
minimum will be at the vertex \(-\frac{b}{2a}=-\frac{-38}{2}=19\)

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Other answers:

Or it can be found by finding the critical point. That is, where the derivative of the above found product is equal to zero. So the derivative of \[x ^{2} -38x\] is \[2x - 38\], set it equal to zero and solve for x, x = 19.
got it! 19 and -19 (: thanks!

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