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I don't know what is wrong with me but I can't even do the first problem.

Find two numbers whose difference is 100 and whose product is a minimum.

To me this means that I want $x-y = 100$

So I want to work in terms of one variable I believe and that gives me

$y=x-100$

This gives me $x-x+100 = 0$

I don't know what to do.

I am incredibly bad at this.

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    You actually want to minimize their product: $xy$.2012-04-01

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You want two numbers $x$ and $y$ such that $x-y=100$, and $xy$ is as small as possible. The first thing to do is to get rid of one of the variables to make it a one-variable problem: as you said, $y=x-100$, so the quantity $xy$ that you’re trying to minimize can be written $x(x-100)$.

Now consider the function $f(x)=x(x-100)=x^2-100x$; you want to find the value of $x$ that minimizes $f(x)$. How do you find the minimum value of a function?

(I’ll add to this later, if necessary, but you really should attempt to solve it yourself from this much of a head start getting it set up properly.)

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    Ok so I think what I need to do then is find the zeroes of the derivative and then find the minimum by finding where the zeroes of the derivative change from decreasing to increasing. According to what I did that is -50 so that means my numbers are -50 and 502012-04-01
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    @Jordan: I think that you have a typo there: the minimum is at $x=50$, not $-50$, but yes, then $x=50$ and $y=50-100=-50$, and the minimum possible product is $-2500$.2012-04-01
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    I meant -50, I probably messed up the math, I am really good at that.2012-04-01
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You are correct that you are looking for two numbers, $x$ and $y$ that satisfy $x-y=100$, or $y=x-100$.

But you seem to have forgotten about the second part of the problem, which says:

... whose product is a minimum.

That is, among all numbers $x$ and $y$ that satisfy $x-y=100$, you are looking for two with $xy$ smallest possible.

For instance, one possibility would be $x=100$, $y=0$; then $xy=0$. Or you could take $x=90$, $y=-10$, in which case $xy=-90$; clearly, $x=90$, $y=-10$ is a "better" choice than $x=100$, $y=0$, because you are looking for a choice that makes the product as small as possible.

So you are trying to minimize (optimization problem alert!) the product $xy$. Since $y=100-x$, that means that you are looking for the minimum of the function $$p(x) = xy = x(100-x).$$

So you want to find the absolute minimum of $p(x)$. Can you take it form here?

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In these optimization problems there are usually two equations. In your case, $x=100+y$, and $A=xy$. The area, $A$, is the quantity you are trying to optimize. Plug $x$ in to get $A=(100+y)y$. You can solver this be completing the square, realizing the minimum is halfway between the roots, or by taking a derivative and setting it equal to 0.

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Here's another approach, which may help with the intuition. Sketch the line $y = x-100$ on some paper. Now sketch some contours of the function $x y$ for example, start with $x y = 1$, which would be the function $ y = 1/x$ (ignoring $0$, of course). Repeat for $x y = -1$, and a few other values, just to get an idea of what the contours look like.

Now look and guess what point on the line will hit the maximum contour of $xy$. Hopefully it will be clear just by symmetry that the solution must satisfy $y = -x$. Then figure out what points on the line satisfy this symmetry.

This can be made rigorous using Lagrange multipliers, but that is secondary to the intuition.