Optimization Review

Hello again, Kristina here…or Scribe-ina
according to Justus. Yes…so I see that
the slides are now up now that I’ve just
woken up from my nap.
Anyways, today’s class was solely based on
reviewing optimization questions since I
had asked for the review, along with my
other peers. Now let’s begin.

A Little Review…A Little Review…

What?! No Constants?!What?! No Constants?!
Like the title says, there
are no given constants
in this question. Which
is one of the things that
makes solving this hard.
First, we drew a
correctly drawn
diagram of the right
angled triangle. This is
to help us get a visual of
what’s going on in the
question.

Now, what we are
looking to optimize is
the little rectangle
inscribed within the
triangle, with corner
points “A” and “P”.
Since it is the area of
the rectangle we are
trying to optimize, that
means that our
optimization equation
will be “A = lw”. But
what are the values for
length and width you
ask? Well that’s where
things start getting
tricky.

Since we have no
known constants, we
will just choose to label
the base and height of
the triangle anything we
want, in this case we
used “h” and “w”. As
for point “P”, we can
just label that point as
(x,y) since we don’t
know the exact place
we should put “P” in
order to maximize the
rectangle’s area.

This should leave us
with variables for our
rectangle’s width and
length, “x” and “y”
respectively. Knowing
this, we can substitute
“x” and “y” into the
optimization equation
we found previously.
We aren’t finished yet.
After all, we will need
to take the derivative of
our optimization
equation, as said in Step
3 of the guide I posted
earlier.

The reason for that is
because when finding
the “maximum” value
for the area, we will
need to refer to the
first derivative test and
try to find the local
maximum. So we are
going to need an
equation for the area in
terms of one variable
first so we can
differentiate.

Well how are we going
to find any values for
“x” and “y” if we aren’t
given any values?
Simple..kind of. As you
can see in the diagram,
there are similar
triangles. With similar
triangles, we can form a
ratio between the
similar triangles, as
shown in the green and
red beside our
optimization equation.
As you can see, we can
now solve for one of
our variables.

Solving for “y” is the
cleanest and easiest, so
that is what we did
here. Now you can
input that “thing” into
the “y” of your
optimization equation,
leaving you with only
one variable so you can
now differentiate. To
make life easier though,
multiply the values out
before you differentiate,
unless you want to do
product rule, which is
uglier.

One thing to remember
before you start
differentiating.
Remember that “h” and
“w” are constants so
when you start
differentiating, don’t be
confused and treat
them the same way as
you would for the
variables. Usually we
would be given a value
for those constants but
not in this case, so it
can get pretty
confusing.

Once you’ve finished
differentiating, you first
have to solve for the
zeroes of the derivative.
After you’ve done that,
you can now use your
first derivative test and
determine whether that
root you’ve found is a
maximum or a
minimum value of the
original function.

As seen in the black
underneath the
diagram, the root we
found “w/2” is a
maximum since the
original function is
increasing to the left of
the root and decreasing
to the right.
Therefore, point P
should be placed at an
x-value halfway
between side AC.

No Constants…Again?!No Constants…Again?!
Yes, another question
with no given constants.
Although this one was a
bit easier to deal with
since it was just
working with squares.
As usual, start out with
drawing your diagram
and label accordingly.
Notice how the length
of the big square is
“L-2x” since we had cut
out those little squares,
with sides labeled “x”.

Now let’s find our
After reading through
the question, we can
see that we are trying
to maximize the volume
of the metal tray that
was formed after
cutting out the small
square corners.
Meaning, our equation
will be for the volume
of the tray, “A=lwh”.

Unlike the last question,
we were lucky this time
since we are only
working with one
variable so we can just
jump to differentiating
right off the hop,
although we should
multiply out the values
to make things easier.
Just another reminder,
remember that the
variable you used for the
length of the tray is a
constant, so differentiate
accordingly!

After differentiating,
you should end up with
that ugly quadratic, the
second green line. It
doesn’t factor out
nicely so just use the
quadratic formula in
order to find the roots.
Just sing the song Mr. K
taught us if you don’t
remember the formula!
Anyways, after using the
quadratic formula and
finding your roots you
can now use the first
derivative test.

Before that, if you’re
wondering what the
green on the bottom
right corner is, Mr. K
basically just simplified
the fractions we found
for the second root and
separated them in
order to make life
easier.
Do the same for the
first root there as well,
since you are not
finished after doing the
first derivative test with
just the second one
shown.

Once you’ve done the
first derivative test with
both zeroes, take the
maximums you found
and place them into the
This is to find which of
those maximums will
yield the largest possible
volume. After doing that,
whichever zero yielded
the largest volume, that
will be the value for “x”,
or in others words how
big the corner squares
should be to maximize
volume.

To Be Continued…To Be Continued…
Yep, we aren’t finished with optimization
yet. Tomorrow, we are going to continue
with our review on these tough little
buggers. Also, slides 4-6 are homework
I’m assuming. Although, we should’ve
gotten some work done with slide 4
during class. That is all. Have a nice day,
and make sure you study for the exam on
Wednesday .

Optimization Review

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