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

• baldbrain
In summary: I don't know what it is. But they've given an answer. I think I know what it is. But proving it is the problem.In summary, The minimum value of 5a + b can be found by expressing a in terms of b and using the relation between a and b in the equation. It is important to note that the equation may have complex roots and the crucial observation is that b^2 = 20a. The problem does not specify that a and b have to be integers, so the answer may not be an integer.
What he means is in general ##5a+b\in\mathbb Z\implies a,b\in\mathbb Z ## is false.

baldbrain said:
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

baldbrain said:
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.

nuuskur said:
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?

nuuskur
I just noticed your #28. Yes, ##5a > b^2 /4 ## is precisely why you may assume ##b^2=20a##. Thus the problem is solved.

baldbrain
nuuskur said:
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

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• picture-of-vertex-of-a-parabola.jpg
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The first parabola will be B|min = b2/4 + b

nuuskur
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

baldbrain said:
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!

nuuskur

baldbrain said:
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.

Ray Vickson said:
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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