What is the minimum product of two numbers whose difference is $4$? What are the numbers? 
What is the minimum product of two numbers whose difference is $4$? What are the numbers?

I'm looking for an answer that uses equations to solve the problem. 
 A: Are you sure that's the maximum product? I can just do like $(-10000, -10004)$ and my product is already larger than yours. :\
But to find the minimum you would note that one number is $x$ and the other $x-4$ so you want to minimize $x(x-4)=x^2-4x.$ So $x=2$ is easily the minimum value for $x$ and the minimum product is $2(2-4)=2\cdot (-2)=\boxed{-4}.$
A: We want to minimize $ab$ subject to $a-b=4$.
Hence we want to minimize $a(a-4)$ which is a quadratic equation that is symmetrical about $a=2$. Hence $a=2$ and $b=-2$.
A: We have to minimise the product $xy $ while keeping the difference $x-y=4$ as constant. Manipulating, we have, $$f (y)=4y+y^2=(y+2)^2-4 $$ Clearly, $\operatorname {min} f (y) $ occurs when $y=-2$. Thus, $x=2$. Hence the minimum product is $-4$. Hope it helps.  
A: Quick Answer
If a minimum of the product exists, then by symmetry one number has to be positive and the other negative, and both numerically the same, hence $-2, 2$. 
Detailed Answer
Let the midpoint be $x$. We want to minimize the product $y=(x-2)(x+2)$ or $y=x^2-4$ which clearly has a minimum of $-4$ at $x=0$, hence the numbers are $-2,2$. 
A: $y-x=4(1)$
Let $p(x,y) = x*y (2)$
from (1) $y=4+x(3)$ 
combine (2) and (3)
$p(x) = x(4+x) = x^2+4x (4)$
Minimum of the function $p(x)$ will be achieved if its differential equals $0$
$p^{'}(x)=0 (5)$
$p^{'}(x) = (x^2+4x)^{'} = 2x + 4 = 0(6)$
from (6) $x = -2$ and from (3) $y = 2$  
$p(-2,2) = (-2)*2 = -4 $
