Find two consecutive natural numbers if the sum of the larger number and four times the reciprocal of the smaller number is a minimum.
can somone help me set it up ?? kinda confused what the question is saying
Try naming one of your variables and writing the others in terms of that one. I.e., let the smaller of the two consecutive numbers be n. Then the larger one is n+1 and their sum is \[n+(n+1)=2n+1.\]
what do u get for the solution ?
Nothing, I wasn't actually planning on solving it. Sorry for the confusion!
do u get 2 and 3?
No, I actually got something else. I'm also curious what course this is for since I ended up differentiating the starting expression to find the solution.
calc 1
i think u have the main equation wrong
Oh, nevermind, I did get 2 and 3 as a local minimum and -2 and -1 as another local minimum. I didn't bother checking which is the absolute min.
i dont see any resciprocal
The equation should be \[f(n)=n+1+\frac{ 4 }{ n },\] Right?
yea
Oh, natural numbers. Great, looks like you're done.
can u help me with one more
wait could u exlain me why the equation should be like that? ?
Sure. If n is the smaller number, then \[n+1\] is the next one and \[\frac{4}{n}\] is four times the reciprocal of the smaller one. The equation I gave is just the sum of those two expressions.
how does n+1 become a smaller number ?
What are you referring to? The smaller number is n in my setup.
A possible solution using Mathematica is attached.
Thank you for the medal.
why did u add +1 after the n ?
Because it's the number that's one more than n. That's what "consecutive" means.
Review the English in the problem statement. Let n be the larger number. The smaller number n-1, is in the fraction. n-1 and n are two consecutive numbers.
I guess that n could be defined as the smaller number and n+1 be the larger one. I chose to encode the problem statement as it was written.
i have another question
Find two consecutive natural numbers if the sum of the larger number and four times the reciprocal of the smaller number is a minimum.
in this question i got 10 and -6 for my solition but i am wrong and idk what i did wrong ?
ok so for S i substitued x=4-y , but in the solution they sunbitied y=4-x and that is the only difference
Wait, is this for the same question?
no
@slaw
actaully hold on it is
this is question i wanted to ask , The sum of two positive numbers is 4. If the sum of their cubes is a minimum, what are these numbers?
Your method is fine because the equation is symmetric in x and y. That is, you can replace every x with a y and every y with an x to arrive at the same expression.
but does it matter ???
beacuse i see no differenec
If something went wrong, it wasn't that step.
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