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 optimization equation. 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 optimization equation. 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

Education

given constants

known constants

optimization equation

maximum value

rectangles area

little rectangle

root youve

little squares