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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

4 Answers 4

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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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    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.