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How do I find the minimum of 5a + b?

  • #26
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@verty largest b given a. b2 equal to 20a or less than that.
 
  • #27
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got the function B = b2/4 + b by taking the minimum value of 5a
underlined part is unclear. I assumed ## b^2=20a## holds and parametrised the expression ##5a+b## with respect to ##b##. The graph of the resulting function ##B## is just a square parabola so there's no secret what its extremum point is.
 
  • #28
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underlined part is unclear. I assumed ## b^2=20a## holds and parametrised the expression ##5a+b## with respect to ##b##. The graph of the resulting function ##B## is just a square parabola so there's no secret what its extremum point is.
5a ≥ b2/4 right? So the minimum value of 5a is b2/4. Can't neglect that possibility.
Hence, the B ≥ b2/4 + b. Again, B|min = b2/4 + b.
 
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  • #29
verty
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I think we need to clear up whether ##a## and ##b## are meant to vary. There was no "given ##a##, ##b##, if ##ax^2 - bx + 5##", it just said "if ##ax^2 - bx + 5##", those variables are unbound, IMHO. But it could be a convention that ##a##,##b## don't vary but ##x##,##y## do.

Ignore this, I was confusing myself.
 
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  • #30
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To elaborate more, B ≥ b2/4 + b is a family of parabolas for all values of b and the 'greater than part' represents the locus of the points outside the interval (α,β) if they're the roots of B
 
  • #31
Ray Vickson
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The answer they've given is an integer.
That is just an accident. The question itself did not specify that ##a## or ##b## need be integers. Likewise, it did not specify that ##5a+b## need be an integer, but possibly with non-integer ##a,b##. Of course, I am assuming you wrote out a complete statement of the question that was given to you.
 
  • #32
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That is just an accident. The question itself did not specify that ##a## or ##b## need be integers. Likewise, it did not specify that ##5a+b## need be an integer, but possibly with non-integer ##a,b##. Of course, I am assuming you wrote out a complete statement of the question that was given to you.
Yes, I did. But B when differentiated gives -1 as the extremum has some connection for sure.
 
  • #33
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You have the condition ##|b| < 2\sqrt{5a} ## to consider. Given the initial post, ##a \geq 0?## may be arbitrary.
For which value of ##b## is the vertex of the resulting parabola lowest possible?
b=-2
 
  • #34
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Sorry, my reply was unnecessarely complicated. Given ##b^2< 20a ## it holds that ##5a+b > \frac{b^2}{4} + b ##, thus it suffices to minimise ##B##.
 
  • #35
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That is just an accident. The question itself did not specify that ##a## or ##b## need be integers. Likewise, it did not specify that ##5a+b## need be an integer, but possibly with non-integer ##a,b##. Of course, I am assuming you wrote out a complete statement of the question that was given to you.
I'm not sure I'm getting your point.
 
  • #36
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What he means is in general ##5a+b\in\mathbb Z\implies a,b\in\mathbb Z ## is false.
 
  • #37
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To elaborate more, B ≥ b2/4 + b is a family of parabolas for all values of b and the 'greater than part' represents the locus of the points outside the interval (α,β) if they're the roots of B
I get it now. A little imagination did the trick
 
  • #38
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I get it now. A little imagination did the trick
how do you order arbitrary complex numbers? The expression ##(x,y) ## as an interval makes sense in this context if they are comparable.
 
  • #39
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What he means is in general ##5a+b\in\mathbb Z\implies a,b\in\mathbb Z ## is false.
My computer isn't recognising these as symbols, so they're just visible as plain text. Do you mean to say that 5a + b is an integer does not imply a,b are intergers?
 
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  • #40
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I just noticed your #28. Yes, ##5a > b^2 /4 ## is precisely why you may assume ##b^2=20a##. Thus the problem is solved.
 
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  • #41
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how do you order arbitrary complex numbers? The expression ##(x,y) ## as an interval makes sense in this context if they are comparable.
complex numbers or not, for simpliclity, I imagined this
picture-of-vertex-of-a-parabola.jpg
 

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  • #42
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The first parabola will be B|min = b2/4 + b
 
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  • #43
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Ok. Thanks to @verty @ehild @Mark44 @Ray Vickson & @nuuskur for your inputs. The rest tuned out of the discussion near it's end. So if there's any post you don't agree with...or have some more suggestions to give, please tell me.
Cheers
 
  • #44
epenguin
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The answer they've given is an integer.
Is it 35? Just to know whether I've got the right answer before I make a fool of myself telling everyone it's easy! :oldbiggrin:
 
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  • #46
Ray Vickson
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:oldsurprised:
Please respond using words. I cannot figure out what you are attempting to convey here, and likewise, I cannot figure out which post you are replying to.
 
  • #47
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Please respond using words. I cannot figure out what you are attempting to convey here, and likewise, I cannot figure out which post you are replying to.
Nothing, just reacting to epenguin's post (#44). Humor--plain and simple
 

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