How do you find two real numbers whose difference is 40 and whose product is a minimum?

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Answer 1 · 2016-10-25T18:30:35

Let the numbers be $x$ and $y$ , with $y > x$ .

$y - x = 40$

$y = 40 + x$

Let $P$ be the product.

$P = xy$

$P = x(40 + x)$

$P = x^2 + 40x$

We can determine the minimum product by completing the square and finding the coordinates of the vertex.

$P = 1(x^2 + 40x + 400 - 400)$

$P = 1(x^2 + 40x + 400) - 400$

$P = 1(x + 20)^2 - 400$

So, the minimum product is $-400$ and the two numbers that would give this product are $20$ and $-20$ .

Hopefully this helps!