🚀 go-pugleaf

"Jonathan" <jp_wells@optusnet.com.au> wrote in message
news:3d1eedb4$0$25797$afc38c87@news.optusnet.com.au...
> As part of a school assignment I have to complete these two problems, but
I
> have no idea as to how to do them, I was wondering anyone knows how to do
> them?
>
> 1)  The product of two positive number is 12. Find the least value of
their
> sum. Is there a largest value?  (10 marks)

I seem to recall that the product of two numbers means 'multiply two numbers
together'.

Therefore 12 x 1 fits the bill as this equals 12.

So would 6 x 2 and 4 x3.

So the sum of 12 + 1 is 13, the sum of 6 and 2 is eight, so we have a
smaller sum and we are getting closer...the sum of 4 and 3 is 7, the
smallest so far!

Please note that 3.75 x 3.2 = 12 and this has a sum of 6.95 which is smaller
still.

Also, 3 1/3 (three and one-third) x 3.6 = 12 and sums to 6.93 (with the
three recurring), even smaller yet.

What about 3.409 (with the 0.409 part recurring) x 3.52?  This product
equals twelve and the sum is 6.92909 (the last two digits recurring) which
is even smaller still!

At this point I am getting bored of trying out numbers.  There is almost
certainly a technique for finding the least sum, but I do not know it, but I
have given you a start.

I don't believe that there is a 'largest number' since you could always make
one side double the size and compensate by making the other side half the
size

e.g. 12 x 1 = 12 (which sums to 13)

but then 24 x 0.5 = 12 and sums to 24.5
and then 48 x 0.25 = 12 and sums to 48.25 and so on and so on, until one
side is infinitely large and the other side is infinitely small.

If there is an answer it looks like infinitesimal calculus is involved, but
I don't know much about it!


--
MESSAGE ENDS.
John Porcella

Thank you for sharing this messagee on lots of newsgroups...
Theoretically, you don't have to speak english on french newsgroups : there
is no reason to post there...

The answer to your question :

you can take B very Big and :
B * (12/B) = 12
B + (12/B) > B which can be taken as big as you want...
So, there is no maximum value...

You intend to minimize a + b when a * b = 12
You have so to study f : x -> x + 12/x
and find its minimum value for x > 0...

f ' (x) = 1 - 12/x�
f ' (x) = 0 if x = sqrt(12) and this is a minimum...

So the minimum is : 4 sqrt(3)

Please : DON'T POST on FRENCH groups when you can only SPEAK ENGLISH !

(sorry for my poor english...)

--

          Mot

PS : you have to remove "ANTISPAM" in my e-mail address

"Mot" <thomas.sirvent+news@ANTISPAMpolytechnique.org> wrote in message
news:afnacq$10f8$1@f1node01.rhrz.uni-bonn.de...
> Thank you for sharing this messagee on lots of newsgroups...
> Theoretically, you don't have to speak english on french newsgroups :
there
> is no reason to post there...
>
> The answer to your question :
>
> you can take B very Big and :
> B * (12/B) = 12
> B + (12/B) > B which can be taken as big as you want...
> So, there is no maximum value...
>
> You intend to minimize a + b when a * b = 12
> You have so to study f : x -> x + 12/x
> and find its minimum value for x > 0...
>
> f ' (x) = 1 - 12/x�
> f ' (x) = 0 if x = sqrt(12) and this is a minimum...
>
> So the minimum is : 4 sqrt(3)

Erm, if sqrt(12) = sqrt (4x3), then surely the answer is 2 sqrt (3)?


--
MESSAGE ENDS.
John Porcella

"Mot" <thomas.sirvent+news@ANTISPAMpolytechnique.org> wrote in message
news:afnacq$10f8$1@f1node01.rhrz.uni-bonn.de...
> Thank you for sharing this messagee on lots of newsgroups...
> Theoretically, you don't have to speak english on french newsgroups :
there
> is no reason to post there...
>
> The answer to your question :
>
> you can take B very Big and :
> B * (12/B) = 12
> B + (12/B) > B which can be taken as big as you want...
> So, there is no maximum value...
>
> You intend to minimize a + b when a * b = 12
> You have so to study f : x -> x + 12/x
> and find its minimum value for x > 0...
>
> f ' (x) = 1 - 12/x�
> f ' (x) = 0 if x = sqrt(12) and this is a minimum...

To ascertain whether it is a minimum, a maximum or simply a point of
inflection, you need to take the second derivative (or so I have just been
taught):

f" (x): 12/x^3...and since THIS is positive, then 2 sqrt (3) is a minimum.


--
MESSAGE ENDS.
John Porcella

> Erm, if sqrt(12) = sqrt (4x3), then surely the answer is 2 sqrt (3)?

You are looking for the minimum of the sum so :
x = sqrt(12) = 2 sqrt(3)
answer = x + 12/x = 2 sqrt(3) + 6 / sqrt(3) = 2 sqrt(3) + 2 sqrt(3)
answer = 4 sqrt(3)

So the minimum you are looking for is 4 sqrt(3)

-- 

          Mot

PS : you have to remove "ANTISPAM" in my e-mail address

> To ascertain whether it is a minimum, a maximum or simply a point
> of inflection, you need to take the second derivative

Yes... or you can simply say that f ' is negative before sqrt(12) and
positive after sqrt(12).

--

          Mot

PS : you have to remove "ANTISPAM" in my e-mail address

"Mot" <thomas.sirvent+news@ANTISPAMpolytechnique.org> wrote in message
news:afo2m8$ii4$1@f1node01.rhrz.uni-bonn.de...
> > Erm, if sqrt(12) = sqrt (4x3), then surely the answer is 2 sqrt (3)?
>
> You are looking for the minimum of the sum so :
> x = sqrt(12) = 2 sqrt(3)
> answer = x + 12/x = 2 sqrt(3) + 6 / sqrt(3) = 2 sqrt(3) + 2 sqrt(3)
> answer = 4 sqrt(3)
>
> So the minimum you are looking for is 4 sqrt(3)

I understand now!  Many thanks!


--
MESSAGE ENDS.
John Porcella

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"Jonathan" <jp_wells@optusnet.com.au> wrote in message =
news:3d1eedb4$0$25797$afc38c87@news.optusnet.com.au...
> As part of a school assignment I have to complete these two problems, =
but I
> have no idea as to how to do them, I was wondering anyone knows how to =
do
> them?
>=20
> 1)  The product of two positive number is 12. Find the least value of =
their
> sum. Is there a largest value?  (10 marks)
x*y x+y 2*6 =3D 12 2+6=3D8 try this with all - try using a spreadsheet. =
1*12 2*6 3*4 etc make a column for x and for y then make a column for =
the sum of the two.=20

> 2)  The sum of the base radius and perpendicular height of a cylinder =
is
> 12cm. Find the maximum volume of the cyclinder.  (10 marks)
Make a list of all possible combinations of numbers - lets call them r =
(for radius) and h (for height)=20
we know r+h=3D12
we also know  that the area of a circle is  =CE=A0r2 so times the number =
you have as r by pie then square it. U can do this on a spreadsheet.
We also know that the area of the base of a cylinder times the height of =
the cylinder makes the area of a cylinder. Imagine loads of slices of =
saliami squished together !
The maxuim volume is the numbers which multiply together to make the =
highest number.=20

I imagine for both of these you would need to show your workings, =
produce graphs if you can, formulaes and if it is a courseworky type =
thing - like we have here in britain then extending it to numbers other =
than twelve for the two questions and seeing if you can (probably) =
algerbricaly explain for all numbers.

Good luck - I notice allot of people have posted answers here - or more =
likely help so maybe you could tell us how you got along.   =20
>=20
> Thankyou
>=20
>=20

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<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>"Jonathan" &lt;</FONT><A=20
href=3D"mailto:jp_wells@optusnet.com.au"><FONT face=3DArial=20
size=3D2>jp_wells@optusnet.com.au</FONT></A><FONT face=3DArial =
size=3D2>&gt; wrote in=20
message </FONT><A=20
href=3D"news:3d1eedb4$0$25797$afc38c87@news.optusnet.com.au"><FONT =
face=3DArial=20
size=3D2>news:3d1eedb4$0$25797$afc38c87@news.optusnet.com.au</FONT></A><F=
ONT=20
face=3DArial size=3D2>...</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>&gt; As part of a school assignment I =
have to=20
complete these two problems, but I<BR>&gt; have no idea as to how to do =
them, I=20
was wondering anyone knows how to do<BR>&gt; them?<BR>&gt; <BR>&gt; =
1)&nbsp; The=20
product of two positive number is 12. Find the least value of =
their<BR>&gt; sum.=20
Is there a largest value?&nbsp; (10 marks)<BR>x*y x+y 2*6 =3D 12 2+6=3D8 =
try this=20
with all - try using a spreadsheet. 1*12 2*6 3*4 etc make a column for x =
and for=20
y then make a column for the sum of the two. </FONT></DIV>
<DIV><BR><FONT face=3DArial size=3D2>&gt; 2)&nbsp; The sum of the base =
radius and=20
perpendicular height of a cylinder is<BR>&gt; 12cm. Find the maximum =
volume of=20
the cyclinder.&nbsp; (10 marks)<BR>Make a list of all possible =
combinations of=20
numbers - lets call them&nbsp;r (for radius)&nbsp;and&nbsp;h (for=20
height)&nbsp;</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>we know r+h=3D12</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>we also&nbsp;know&nbsp; that the area =
of a circle=20
is&nbsp;&nbsp;<FONT face=3DAndy><SPAN lang=3DDE=20
style=3D"FONT-SIZE: 18pt; FONT-FAMILY: 'Comic Sans MS'; =
mso-bidi-font-size: 12.0pt; mso-fareast-font-family: 'Times New Roman'; =
mso-bidi-font-family: 'Times New Roman'; mso-ansi-language: DE; =
mso-fareast-language: EN-US; mso-bidi-language: AR-SA"><FONT=20
size=3D3>=CE=A0</FONT></SPAN><SPAN lang=3DDE=20
style=3D"FONT-SIZE: 18pt; FONT-FAMILY: 'Comic Sans MS'; =
mso-bidi-font-size: 12.0pt; mso-fareast-font-family: 'Times New Roman'; =
mso-bidi-font-family: 'Times New Roman'; mso-ansi-language: DE; =
mso-fareast-language: EN-US; mso-bidi-language: AR-SA"><SUP><FONT=20
size=3D3>r2</FONT> </SUP></SPAN></FONT></FONT><FONT face=3DArial=20
size=3D2>so&nbsp;times the number you have as r by pie then square =
it.&nbsp;U can=20
do this on a spreadsheet.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>We also know that the area of the base =
of a=20
cylinder times the height of the cylinder makes the area of a cylinder. =
Imagine=20
loads of slices of saliami squished together !</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>The maxuim volume is the numbers which =
multiply=20
together to make the highest number.&nbsp;</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>I&nbsp;imagine for both&nbsp;of these =
you would=20
need to show your workings, produce graphs if you can, formulaes and if =
it is a=20
courseworky type thing - like we have&nbsp;here in britain =
then&nbsp;extending=20
it to numbers other than twelve for the two&nbsp;questions and seeing if =
you=20
can&nbsp;(probably) algerbricaly&nbsp;explain for all =
numbers.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Good luck - I notice allot of people =
have posted=20
answers here - or more likely help so maybe you could tell us how you =
got=20
along.&nbsp;&nbsp;&nbsp;&nbsp;<BR>&gt; <BR>&gt; Thankyou<BR>&gt; =
<BR>&gt;=20
</FONT></DIV></BODY></HTML>

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I'd do these on a graph, unless you're doing this at postgrad level on an
engineering course or something

1)
if one of the numbers is 1, the other is 12, sum = 13
                                    2                     6               8
                                    3                     4               7
                                    4                     3               7
                                     5                   2.4             7.4
                                     6                    2                8
looks like a minimum value somewhere between  x=3 & x=4


2)
similarly  using   pi times r^2 times h
and  r + h = 12
so if r=1, h=11
       r=2, h=10
and so on, work out the values and graph them

enjoy

Jon


"Jonathan" <jp_wells@optusnet.com.au> wrote in message
news:3d1eedb4$0$25797$afc38c87@news.optusnet.com.au...
> As part of a school assignment I have to complete these two problems, but
I
> have no idea as to how to do them, I was wondering anyone knows how to do
> them?
>
> 1)  The product of two positive number is 12. Find the least value of
their
> sum. Is there a largest value?  (10 marks)
>
> 2)  The sum of the base radius and perpendicular height of a cylinder is
> 12cm. Find the maximum volume of the cyclinder.  (10 marks)
>



>
> Thankyou
>
>

"teddy boy" <jonool@ntlworld.com> wrote in message
news:xw0Y8.10628$rU2.1105690@news11-gui.server.ntli.net...
> I'd do these on a graph, unless you're doing this at postgrad level on an
> engineering course or something
>
> 1)
> if one of the numbers is 1, the other is 12, sum = 13
>                                     2                     6
8
>                                     3                     4
7
>                                     4                     3
7
>                                      5                   2.4
7.4
>                                      6                    2
8
> looks like a minimum value somewhere between  x=3 & x=4

The problem asks to find the least value of the sum.  It does not ask to
guess which two integers this value lies between.  The chart gets one
nowhere...an analytic approach is appropriate.

The two numbers can be represented as x and 12/x, where x>0.  We desire to
find the least value of their sum, i.e. we need to minimize:

f(x) = x + 12/x, x>0

This can be done by the usual method of finding the minimum of f:

f'(x) = 1 - 12/x^2 = 0

This yields x = 2sqrt(3) as a critical value.  We confirm this to be a
minimum w/ 1st derivative test.  I beleive I read somewhere in this trhread
that the 2nd derivative test is required...not so.  It can be done but not
required by any means.



sign of f'(x)        ---            +++
               <-------------|--------------->
                          2sqrt(3)

So the two numbers are 2sqrt(3) and 12/2sqrt(3)=6/sqrt(3).
Their sum is 12/sqrt(3).

For the second part of the question----since x=2sqrt(3) is the only critical
value of f, and we have already determined that it produces a minimum, f has
no maximum.

> 2)
> similarly  using   pi times r^2 times h
> and  r + h = 12
> so if r=1, h=11
>        r=2, h=10
> and so on, work out the values and graph them

Again...an analytic approach is appropriate:

r+h=12  ==>  h=12-r

We maximize:

V(r) = pi(r^2)(12-r) = 12pi(r^2)-pi(r^3)

V'(r) =  24pi(r)-3pi(r^2) = (3pi)(r)(8-r) = 0

3pi <> 0
r <> 0 , else not a cylinder

8 - r = 0

r = 8

V(8) = pi(8^2)(12-8) = 256pi

--
Darrell

>
> enjoy
>
> Jon
>
>
> "Jonathan" <jp_wells@optusnet.com.au> wrote in message
> news:3d1eedb4$0$25797$afc38c87@news.optusnet.com.au...
> > As part of a school assignment I have to complete these two problems,
but
> I
> > have no idea as to how to do them, I was wondering anyone knows how to
do
> > them?
> >
> > 1)  The product of two positive number is 12. Find the least value of
> their
> > sum. Is there a largest value?  (10 marks)
> >
> > 2)  The sum of the base radius and perpendicular height of a cylinder is
> > 12cm. Find the maximum volume of the cyclinder.  (10 marks)
> >
>
>
>
> >
> > Thankyou
> >
> >
>
>
>

In article <NqsY8.13902$Kx3.1560@newsread1.prod.itd.earthlink.net>,
Darrell <dr6583@hotmail.com> wrote:
>The two numbers can be represented as x and 12/x, where x>0.  We desire to
>find the least value of their sum, i.e. we need to minimize:
>f(x) = x + 12/x, x>0
>This can be done by the usual method of finding the minimum of f:
>f'(x) = 1 - 12/x^2 = 0

	Does no-one do the AM-GM inequality these days?  For x > 0,
{ x + 12/x } / 2 >= sqrt (x,12/x) == sqrt (12), with equality iff
x == 12/x, ie x == sqrt (12).  So x + 12/x has a minimum of 2.sqrt(12)
[or 4.sqrt(3) if you prefer] at that value of x.  No need for the
sledgehammer of calculus!

>Again... [...]
>We maximize:
>V(r) = pi(r^2)(12-r) = 12pi(r^2)-pi(r^3)

	OK, so for 0 < r < 12, we note that r/2 + r/2 + 12-r is
constant, and so cbrt (r/2 . r/2 , (12-r)) <= (r/2 + r/2 + 12-r)/3
== 4, so r/2 . r/2 , (12-r) <= 4^3 == 64, so that V(r) <= 256 pi,
with equality iff r/2 == 12-r, ie r == 8.

	The standard inequalities are your friends for these sorts
of problems!  [I seem to write this article twice per year -- is
there any point having an education group where people don't learn?!
Harrumph.]

	We used to have a standard interview question -- farmer has
400m of fencing with which to enclose a rectangular field, blah, blah.
Applicants would let the sides be a and b, then get into a mess 'cos
they couldn't do calculus with two unknowns, but eventually they got
there.  There were follow-ups where one side was against a cliff, so
only three sides were needed, etc.  But I usually asked them if they
had known what the answer would be before they "proved" it -- "Oh yes,
it's a square with side 100".  OK, so if one side is 100+h, what's
the other side?  "100 - h."  So the area is?  "100^2 - h^2."  So,
what can you say?  "???"  Well, what do you know about h^2?  "It's
positive?"  So -h^2 is?  "Negative?"  And 10000-h^2?  "Positive?"
NO!  "Ah, less than 10000."  With equality when?  "h == 0 [triumph!]."
Happy days.

-- 
Andy Walker, School of MathSci., Univ. of Nott'm, UK.
anw@maths.nott.ac.uk