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From: rpg16 on 15 Jan 2010 06:23
here's the problem

a and b are positive real numbers
b > a

find, a,b such that
(b - a) takes a minimum value and
a/b takes a maximum value

i have no clue how to proceed
From: scattered on 15 Jan 2010 06:34
On Jan 15, 6:23 am, rpg16 <roupam.gh...(a)gmail.com> wrote:
> here's the problem
>
> a and b are positive real numbers
> b > a
>
> find, a,b such that
> (b - a) takes a minimum value and
> a/b takes a maximum value
>
> i have no clue how to proceed

Are you sure that there even is a solution? Is {(a,b)| b > a > 0} a
compact set? If not, why do you think that those expressions even have
maximum and minumum values?

Did you mean b >= a > 0?
From: Henry on 15 Jan 2010 06:37
On 15 Jan, 11:34, scattered <still.scatte...(a)gmail.com> wrote:
> On Jan 15, 6:23 am, rpg16 <roupam.gh...(a)gmail.com> wrote:
>
> > here's the problem
>
> > a and b are positive real numbers
> > b > a
>
> > find, a,b such that
> > (b - a) takes a minimum value and
> > a/b takes a maximum value
>
> > i have no clue how to proceed
>
> Are you sure that there even is a solution? Is {(a,b)| b > a > 0} a
> compact set? If not, why do you think that those expressions even have
> maximum and minumum values?
>
> Did you mean b >= a > 0?

Even if you did allow equality, there would be a large number of
possible values for a and b achieving the extremes for the
expressions. There must be more to the question.
From: Tim Little on 15 Jan 2010 07:53
On 2010-01-15, rpg16 <roupam.ghosh(a)gmail.com> wrote:
> a and b are positive real numbers
> b > a
>
> find, a,b such that
> (b - a) takes a minimum value and
> a/b takes a maximum value
>
> i have no clue how to proceed

Clearly b-a > 0 and a/b < 1. Suppose a = x-1/x, b = x for some x > 1.
Then (b - a) = 1/x, and a/b = 1 - 1/x^2. By choosing a suitably large
x, (b-a) can be arbitrarily close to 0 and a/b arbitrarily close to 1.

So there is no minimum for (b-a) and no maximum for a/b.


- Tim
From: rpg16 on 15 Jan 2010 08:11
On Jan 15, 5:53 pm, Tim Little <t...(a)little-possums.net> wrote:
> On 2010-01-15, rpg16 <roupam.gh...(a)gmail.com> wrote:
>
> > a and b are positive real numbers
> > b > a
>
> > find, a,b such that
> > (b - a) takes a minimum value and
> > a/b takes a maximum value
>
> > i have no clue how to proceed
>
> Clearly b-a > 0 and a/b < 1.  Suppose a = x-1/x, b = x for some x > 1.
> Then (b - a) = 1/x, and a/b = 1 - 1/x^2.  By choosing a suitably large
> x, (b-a) can be arbitrarily close to 0 and a/b arbitrarily close to 1.
>
> So there is no minimum for (b-a) and no maximum for a/b.
>
> - Tim

Thanks Tim
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