What is the minimum possible product of two numbers that differ by $8$ ?

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Answer 1 · 2016-11-06T08:15:39

The minimum product is $-16$ and the two numbers are $-4$ and $4$

Let the two numbers be $x$ and $x+8$ .

Then their product is:

$f(x) = x(x+8)$

$color(white)(f(x)) = x^2+8x$

$color(white)(f(x)) = x^2+8x+16-16$

$color(white)(f(x)) = (x+4)^2-16$

For any Real value of $x$ we will have $(x+4)^2 >= 0$

Hence $f(x)$ attains its minimum value $-16$ when $(x+4)^2 = 0$

That is when $x = -4$

So the minimum product is $-16$ and the two numbers are $-4$ and $4$

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