A community for students.
Here's the question you clicked on:
 0 viewing
ParthKohli
 2 years ago
How do I prove that if the sum of two numbers is \(x\), then \(\frac{x}{2}, \frac{x}{2}\) maximizes the product?
ParthKohli
 2 years ago
How do I prove that if the sum of two numbers is \(x\), then \(\frac{x}{2}, \frac{x}{2}\) maximizes the product?

This Question is Closed

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Let the two numbers be \((x  n)(x k) = x^2  nx  kx + nk= x^2  (n+k) x + nk\). I don't know if that helps.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0I just have to maximize \(x^2  (n + k)x + nk\) but have no idea how.

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1a+b=x a*b=max use calculus

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i would argue by symmetry call one \(a\) and the other \(b\) with \(a+b=x\) then since this equation is symmetric in \(a\) and \(b\), i.e. you cannot tell them apart, it must be largest in the middle

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0But it can also be the smallest

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1by which i mean the product must be largest in the middle but you can write as @Zarkon said maximize \(a(xa)\) using calculus

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1can't be the smallest if \(a=x\) or \(b=x\) you get \(ab=0\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i.e. it is smallest at the endpoint of the interval

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1or for that matter you can find the max of \(axa^2\) is at the vertex, which gives \(\frac{x}{2}\)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah, that's a good idea.

Thoughts
 2 years ago
Best ResponseYou've already chosen the best response.3Let those numbers be a and b a + b = x You have to prove that : \(ab < \cfrac{x^2}{4} \) a + b = x \(\cfrac{a+b}{2} = \cfrac{x}{2}\) Using A.M  G.M inequality : \(\cfrac{\alpha + \beta}{2} > \sqrt{\alpha \beta}\) Therefore : \(\cfrac{a+b}{2} > \sqrt{ab} \) \(\cfrac{x}{2} > \sqrt{ab}\) \(\cfrac{x^2 }{4} > ab\) or : \(ab < \cfrac{x^2}{4} \)

Thoughts
 2 years ago
Best ResponseYou've already chosen the best response.3Hence Proved (if that was to prove basically :) )

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0It should be \(\le\) and \(\ge\), but I get your point. Real good :)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1nah, i like symmetry more look, you call one \(a\) and the other \(b\) where \(a+b=x\) but i come along and call the first one \(b\) and the second one \(a\) and write \(b+a=x\) it is pretty clear that there is no difference between them so how could you favour \(a\) over \(b\) somehow, so that the answer would not be \(a=b\) ?

Thoughts
 2 years ago
Best ResponseYou've already chosen the best response.3Symmetry is good too. But I think using AM  GM inequality is also fine and good. :)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1then i am probably wrong

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1I'm referring to the AM  GM inequality

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1actually i think i am still right

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1happens about once a day, sometimes twice

Thoughts
 2 years ago
Best ResponseYou've already chosen the best response.3Done (in a clever 5th grade way) ;)

Thoughts
 2 years ago
Best ResponseYou've already chosen the best response.3Have a look at my solution :

Thoughts
 2 years ago
Best ResponseYou've already chosen the best response.3No problem then : \(\bf { a+b = x\\ \text{Now} \space , (ab)^2 \ge 0 \\ a^2 + b^2  2ab \ge 0 \\ a^2 + b^2 \ge 2ab \\ a^2 + b^2 + 2ab \ge 4ab \\ (a+b)^2 \ge 4ab \\ \cfrac{(a+b)^2}{4} \ge ab \\ \cfrac{x^2}{4} \ge ab \\ } \) \(\textbf{Proved }\) ....

Thoughts
 2 years ago
Best ResponseYou've already chosen the best response.3Looks better now. Isn't it @ParthKohli @Zarkon @satellite73 :)

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.0Ah! Great one again!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.