🚀 go-pugleaf

"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
> >
> >
>
>
>