Ok I am not a mathematician and I got myself tripping on this question (which I think is not correctly formulated )
at first I thought "pfff thats easy its just an aslant line so I'll go to minus infinity on both sides and that would be my minimum product"
but if I remember correctly from school -infinity * -infinity = +infinity
and how could the minimum product of an equation like that be infinity since just by doing some over the top of my head calculation I could find quite a few x,y pairs from y = 3x – 6 of which the product is smaller than infinity!!
then I thought to to multiply everything with x and get xy = 3x^2 -6x and just find the minimum of 3x^2 -6 which is 1,-3 ….
but I can't make any inference from that since this may be the lowest point of that quadratic equation but these coordinates do not appear on the line 3x – 6…
So I am very confused…. is the question somehow not correct? what am I missing?
Best Answer
Given $x$, the product is $$ x(3x-6)=3(x^2-2x)=3(x^2-2x+1-1)=3((x-1)^2-1) $$ Since $(x-1)^2\ge0$, it should be clear that the minimum possible is $-3$, for $x=1$ and $y=-3$.
The trick is “completing the square”, that is, going from $x^2-2x$ to $x^2-2x+1-1=(x-1)^2-1$.