Two positive numbers have a sum of 150. Minimize the sum the of one mumber
and the reciprocal of the other.

This question doesn't seem to make sense to me.

Let's say the numbers are a and b then a + b = 150 so a = 150 - b

The sum of a and the reciprocal of b is then 150 - b + 1/b

Now if f(b) = 150 - b + b^-1 then f'(b) = -1 - b^-2 or -1 - 1/(b^2) but that
can never be zero so there is no minimum.

Is there an interpretaion of this question which makes sense?

On Friday, 26 May 2023 at 04:00:41 UTC+2, Math Student wrote:
> Two positive numbers have a sum of 150. Minimize the sum the of one mumber
> and the reciprocal of the other.
>
> This question doesn't seem to make sense to me.

> Now if f(b) = 150 - b + b^-1 then f'(b) = -1 - b^-2

Last time I have checked, the derivative of 1/b was not -1/b^2...

Julio

On Friday, 26 May 2023 at 04:22:47 UTC+2, Julio Di Egidio wrote:
> On Friday, 26 May 2023 at 04:00:41 UTC+2, Math Student wrote:
> > Two positive numbers have a sum of 150. Minimize the sum the of one mumber
> > and the reciprocal of the other.
> >
> > This question doesn't seem to make sense to me.
> >
> > Now if f(b) = 150 - b + b^-1 then f'(b) = -1 - b^-2
>
> Last time I have checked, the derivative of 1/b was not -1/b^2...

No, sorry, that was correct. But do study/draw the
graph of the function, there is a minimum since the
domain is bounded...

Julio

On 26/05/2023 02:58, Math Student wrote:
> Two positive numbers have a sum of 150. Minimize the sum the of one mumber
> and the reciprocal of the other.
>
> This question doesn't seem to make sense to me.
>
> Let's say the numbers are a and b then a + b = 150 so a = 150 - b
>
> The sum of a and the reciprocal of b is then 150 - b + 1/b
>
> Now if f(b) = 150 - b + b^-1 then f'(b) = -1 - b^-2 or -1 - 1/(b^2) but that
> can never be zero so there is no minimum.
>
> Is there an interpretaion of this question which makes sense?
>
>

Your derivative calculation looks right, so f(b) is decreasing as b increases.

If b had a maximum allowable value, f(b) would be be a minimum at that value for b.

From your description, 0 < b < 150, so if a,b are real numbers there is no maximum for b, so no
solution. Possibly a,b are intended to be whole numbers? Then there would be a reasonable answer.

Regards,
Mike.

On Friday, 26 May 2023 at 04:35:15 UTC+2, Julio Di Egidio wrote:
> On Friday, 26 May 2023 at 04:22:47 UTC+2, Julio Di Egidio wrote:
> > On Friday, 26 May 2023 at 04:00:41 UTC+2, Math Student wrote:
> > > Two positive numbers have a sum of 150. Minimize the sum the of one mumber
> > > and the reciprocal of the other.
> > >
> > > This question doesn't seem to make sense to me.
> > >
> > > Now if f(b) = 150 - b + b^-1 then f'(b) = -1 - b^-2
> >
> > Last time I have checked, the derivative of 1/b was not -1/b^2...
>
> No, sorry, that was correct. But do study/draw the
> graph of the function, there is a minimum since the
> domain is bounded...

Mike Terry is right, the domain is bounded but not a
closed interval (as long as "positive" means strictly
positive), so in the real numbers there is no minimum.

Thanks for the exercise...

Julio

"Mike Terry" <news.dead.person.stones@darjeeling.plus.com> wrote in message
news:5didnR9PAfLeg-35nZ2dnZfqnPqdnZ2d@brightview.co.uk...
> On 26/05/2023 02:58, Math Student wrote:
>> Two positive numbers have a sum of 150. Minimize the sum the of one
>> mumber
>> and the reciprocal of the other.
>>
>> This question doesn't seem to make sense to me.
>>
>> Let's say the numbers are a and b then a + b = 150 so a = 150 - b
>>
>> The sum of a and the reciprocal of b is then 150 - b + 1/b
>>
>> Now if f(b) = 150 - b + b^-1 then f'(b) = -1 - b^-2 or -1 - 1/(b^2) but
>> that
>> can never be zero so there is no minimum.
>>
>> Is there an interpretaion of this question which makes sense?
>>
>>
>
> Your derivative calculation looks right, so f(b) is decreasing as b
> increases.
>
> If b had a maximum allowable value, f(b) would be be a minimum at that
> value for b.
>
> From your description, 0 < b < 150, so if a,b are real numbers there is no
> maximum for b, so no solution. Possibly a,b are intended to be whole
> numbers? Then there would be a reasonable answer.
>
> Regards,
> Mike.

Thanks Mike, and Julio.
I'm trying to find out what the teacher thinks the answer should be.
It won't surprise me if an overworked teacher got the question wrong.

>