maximizing a trig function
elodie
Can somebody please help me maximimize
f(x)=8/sin(x) +5/cos(x), where 0 < x < pi/2 ?
When I differentiate, I get
sin(x)/cos(x)^{2}-8cos(x)/sin(x)^{2} (eq 1)
When I set equation 1 to 0 and solve for x, my software program does
not return a root.
thanks in advance for your help.
achille
Are you sure you want maximize instead of minimize?
your f(x) diverges at 0 and pi/2 because sin(x) -> 0
as x -> 0 and cos(x) -> 0 as x -> pi/2.
Robert H. Lewis
>
> Can somebody please help me maximimize
> f(x)=8/sin(x) +5/cos(x), where 0 < x < pi/2 ?
>
> When I differentiate, I get
> sin(x)/cos(x)^{2}-8cos(x)/sin(x)^{2} (eq 1)
What happened to the 5?
> When I set equation 1 to 0 and solve for x, my
> software program does not return a root.
Heavens! There ought to be a law. Whatever shall we do?
Step 1: put away the computer and work it out for yourself.
Step 2: what is the definition of tan(x)?
Robert H. Lewis
Fordham University
Virgil
<58143025-552d-45d6...@c3g2000yqd.googlegroups.com>,
elodie <elodie....@gmail.com> wrote:
What makes you think that there must be a maximum?
While your function is continuous on the given OPEN interval, it is not
continuously extendable to the closure of that interval.
As x decreases towards 0, 8/sin(x) increases without bound while
5/cos(x) remains bunded, and as x increases towards pi/2, 5/cos(x)
increases without bound while 8/sin(x) remains bounded.
jdturner
> In article
> <58143025-552d-45d6-95b0-854243b45...@c3g2000yqd.googlegroups.com>,
Try plotting the function. It is then obvious that the min is near
0.8634
Robert H. Lewis
> >
> > elodie <elodie.gill...@gmail.com> wrote:
> > > Hi everyone
> >
> > > Can somebody please help me maximimize
> > > f(x)=8/sin(x) +5/cos(x), where 0 < x < pi/2 ?
> >
>
> Try plotting the function. It is then obvious that
> the min is near
> 0.8634
Your decimal approximation is correct. But let's hope the OP can derive the analytic expression for this number. That is, of course, far more interesting.
Virgil
<94d6375c-eecf-48ea...@e37g2000yqn.googlegroups.com>,
jdturner <jamesdan...@hotmail.com> wrote:
If the minimum were being asked for, it might be of some interest, but
it is not being asked for.
Lynne Vickson
> Hi everyone
>
> Can somebody please help me maximimize
> f(x)=8/sin(x) +5/cos(x), where 0 < x < pi/2 ?
>
> When I differentiate, I get
> sin(x)/cos(x)^{2}-8cos(x)/sin(x)^{2} (eq 1)
Forget forever the idea that you ALWAYS find optima of functions on
intervals by setting the derivative to zero. First: if the interval is
open (as is yours) there may not be a max or a min for a smooth
function f(x). Second, if you have a closed interval, and if the
function f(x) is bounded on that interval, the max or min may occur at
an endpoint, where the derivative is not necessarily zero. For
example, max f(x) = x on a <= x <= b has solution at x = b, while min f
(x) = x on a <= x <= b has solution at x = a. At both solutions the
derivative = 1. Also, if you have an open interval a < x < b, neither
problem has a solution: there is no smallest or largest x in the
interval.
In your case the function has no maximum; it is unbounded as x --> 0
from above or x --> pi/2 from below. However, it does have a minimum
on the interval, and that is obtained by setting the derivative to
zero.
Note: you should go back and review _exactly_ what the relevant
theorem says about finding optima by setting derivatives to zero. This
is not just an academic quibble: in real-world operations research/
engineering applications we are always having to find optima in
constrained problems, and we need to know how to deal with such
problems precisely.
R.G. Vickson