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We're going to look at the following problem:
Maximize x
subject to
sine x equal to a half,
x greater than or equal to 0.
If you remember your trigonometry,
sine of
pi over 6 is one-half.
So if this angle here is
pi
over six radians, then
the sine of that
will be one-half.
Sine of
pi over 6
is a half.
But that's not the only angle whose sine is equal to a half.
sine function is periodic.
So if we're just
drawing from
x greater than or equal to 0, onwards,
sine kind of looks like this
a sine wave.
And a half is roughly here.
The sine crosses the
x-axis
at pi
2 pi
at integer multiples of pi.
So what we can see is that
sine of
pi over 6
plus the period
an integer multiple of the period
is going to be a half as well.
that means that there are infinitely many values for x for which the sine is
a half.
And they are all nonnegative.
Now if we are looking for the maximum possible x such that sine x is
a half,
x greater than or equal to zero,
we will never find a maximum because
if you look at the
angles that give sine equal to a half,
they can be as large as we want.
So what we can say is
no
maximum value
exists
for our problem
since x
equal
pi over 6 plus
2 k pi
satisfy
sine x equal to a half
and
x greater than or equal to 0.
So no maximum exists
and we call this problem unbounded.