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

Oct 25, 2016

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

$y - x = 40$

$y = 40 + x$

Let $P$ be the product.

$P = x y$

$P = x \left(40 + x\right)$

$P = {x}^{2} + 40 x$

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

$P = 1 \left({x}^{2} + 40 x + 400 - 400\right)$

$P = 1 \left({x}^{2} + 40 x + 400\right) - 400$

$P = 1 {\left(x + 20\right)}^{2} - 400$

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

Hopefully this helps!