<p>It's an optimization type problem:</p>
<p>"Find two positive numbers that satisfy the given requirements:</p>
<p>The second number is the reciprocal of the first and the sum is a minimum"</p>
<p>I started out like:</p>
<p>x = 1/y</p>
<p>S = x + 1/y</p>
<p>S = 2x</p>
<p>S'=2</p>
<p>but this doesn't make sense...there's no critical points to S'=2</p>
<p>Any help would be great!!</p>
<p>Your process really doesn’t make sense :)<br>
x = 1/y
S = x + 1/y
The two numbers aren’t necessarily equal to each other, why would you substitute? </p>
<p>Your process should be:</p>
<p>x+(1/x)</p>
<p>Then find the derivative and set it equal to zero.</p>
<p>1 - 1/(x^2) = 0
(x^2 - 1)/x^2 = 0
((x+1)(x-1))/x^2 = 0</p>
<p>Therefore, the function has critical values when the first derivative is zero, (at 1 and -1) or when the derivative doesn’t exist, which is when x=0.</p>
<p>Since the problem asked for two positive numbers, you can eliminate -1 and 0, leaving you with x=1. You can also check that this number is a minimum by plugging in values left and right of -1 into the derivative (as long as the left side is greater than zero, because that’s another one of your critical values), and noting the sign change from negative to positive.</p>
<p>Therefore, x = 1, and the other number is 1/x, so that number is also 1. </p>
<p>Hope this helps :)</p>
<p>Wow, I’m ■■■■■■■■. hahah</p>
<p>Thanks for the help!</p>
can you please explain your steps from 1-(1/x^2) = 0 to (x^2 - 1)/x^2 = 0
@knuttknocker you generally shouldn’t revive old threads…but I guess starting a new thread with the question you asked isn’t worth a thread either.
This is just basic algebra (rewriting 1 as x^2/x^2).
Lol wow what a douche response yeah I’m in calc 3 I just didn’t see it. Go on being your disrespectful self and one day you will find out nobody cares to be around you
@knuttknocker geez, you don’t have to be so harsh and generalizing. I was not trying to degrade you.
All I was pointing out was you can replace 1 with x^2/x^2.