among all pairs of numbers whos difference is 12, find a pair whose product is as small as possible. what is the minimum product?

Question

anonymous · Answer

12

anonymous · Answer

my bad 24

kinggeorge · Answer

Well, he have a number $x$ and $x+12$ so that $(x+12)-x=12$ Now we need to minimize $$x\cdot (x+12)=x^2+12x$$

anonymous · Answer

so its 24

anonymous · Answer

so the first part is 12 & the second part is 24?

anonymous · Answer

sure, I dunno. Im only in ALgebra 2

kinggeorge · Answer

There are two ways to do it. Option one: Do calculus. 

Option two: Since it's an upward opening parabola, we know it has a minimum at the vertex.  We can find this vertex using the formula $$-{b \over 2a}$$Since $a=1, b=12$, the vertex  occurs at $$x=-{12 \over 2}=-6$$If we plug this value back into the equation, we find that $$y=(-6)^2+12(-6)=36-72=-24$$So the minimum product should be -36

kinggeorge · Answer

Namely, you get a product of $-36$ and a difference of 12 when you use the numbers $6$ and $-6$.

kinggeorge · Answer

I made a typo in my explanation. When I said $$y=(−6)^2+12(−6)=36−72=−24$$It should have been$$y=(−6)^2+12(−6)=36−72=−36$$

anonymous · Answer

thank you soooooo much!!

kinggeorge · Answer

You're welcome.