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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.

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  • $\begingroup$ 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.) $\endgroup$ – mweiss Jan 18 '17 at 4:06
  • $\begingroup$ I found out that I made a trivial error maximum solution, and correcting the error, found the solution. $\endgroup$ – suomynonA Jan 18 '17 at 4:08
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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 $2(2-4)=2\cdot (-2)=\boxed{-4}.$

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  • $\begingroup$ I did the same mistake as last time -_- $\endgroup$ – suomynonA Jan 18 '17 at 4:10
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    $\begingroup$ whoops :3 it happens though you should be fine $\endgroup$ – pie314271 Jan 18 '17 at 4:11
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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 $b=-2$.

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

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  • $\begingroup$ @mweiss It just happens that I know simple derivatives, but good thing to keep in mind in algebra-precalc questions $\endgroup$ – suomynonA Jan 18 '17 at 4:12
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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 $-2,2$.

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$y-x=4(1)$

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 $

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