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

1 Answer
Nov 6, 2016

Answer:

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

Explanation:

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#