# 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.

• It might be helpful if you explain how you would find the maximum product. (Hint: there actually is no maximum product; the product of two numbers whose difference is $4$ can be made as large as you want. So somewhere in your solution to that problem you have made an error.) – mweiss Jan 18 '17 at 4:06
• I found out that I made a trivial error maximum solution, and correcting the error, found the solution. – suomynonA Jan 18 '17 at 4:08

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 $2(2-4)=2\cdot (-2)=\boxed{-4}.$

• I did the same mistake as last time -_- – suomynonA Jan 18 '17 at 4:10
• whoops :3 it happens though you should be fine – pie314271 Jan 18 '17 at 4:11

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 $b=-2$.

• thanks for spotting the mistake. – Siong Thye Goh Jan 18 '17 at 4:08
• 2-(-6)=8 unfortunately :\ – pie314271 Jan 18 '17 at 4:09

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.

• @mweiss It just happens that I know simple derivatives, but good thing to keep in mind in algebra-precalc questions – suomynonA Jan 18 '17 at 4:12

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 $-2, 2$.

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 $-2,2$.

$y-x=4(1)$

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) $y = 2$

$p(-2,2) = (-2)*2 = -4$