More Optimization Problems

This class was focused on some more optimization problems.


By following the 6 step guide, it makes solving the problems a little easier.

Find the maximum volume of a right circular cylinder that can be inscribed in a cone of altitude 12 cm, and base radius of 4 cm, if the centres of the cylinder coincide.

Step 1)
The question tells us what to optimize, which is the volume.

Step 2)
We are optimizing volume so the formula for the volume of a cylinder is:

v=pir²h

Step 3a)
Through our picture of an inscribed cylinder in a cone, we see that we have similar triangles going on. So the height of the big triangle is proportioned to the height of the smaller triangle. As will the base of the big triangle be proportioned to the base of the smaller triangle, which is the radius of the cylinder.

Step 3b)
From this we can make the equation:

4/r = 12/12-h

From here, all we have to do is isolate one of the variables but we found out that there'll be less work if we isolate h. By isolating h we get :

h=12-3r

Step 3c)
After plugging in h into the optimization formula we'll get:

v(r)=12pi r²-3pi r³

Step 4)
We found the derivative of the optimization equation which is:

^{v'(t)=3pir(8-3r)}

Step 5)
The critical numbers is when r=8/3. We also checked if this was a max by using the first derivative test.

Step 6)

^{We find the max volume by substituing 8/3 into r.}Next Question.

So we always start with drawing a diagram first. From the diagram we can see that we have a triangle. X represents the point when he starts to walk. Using pythagoras we can find the hypotenuse, which is the distance of his boat ride. Then we can make a table of the values we know for distance, rate, and time. Time can be found by doing d/r. We have yet to find the optimization equation yet. What we need to minimize is the time. With the info in our table we can make a function t. The function t(x) is just the time he took to walk and row addeded together. Now all that is needed to do is to find the derivative and find critical points.

Yeah sorry this was kind of rushed but I've yet to eat dinner and I'm really hungry. As for scribe I'll pick Shelly. I know you said you were busy but I really don't know who hasn't scribed yet so yeah.

Today's Slides: December 8

Here they are ...

AP Calculus December 8, 2008

View SlideShare presentation or Upload your own. (tags: apcalc2008 math)

Optimization Problems II

A long rectangular sheet of metal, 12 inches wide, is to be made into a rain gutter by turning up two sides so that they are perpendicular to the sheet. How many inches should be turned up to give the gutter its greatest capacity?
1. We drew the diagram and assigned the variables, base (b) and height (h). We realized that the b=12-2h, since the rain gutter takes the shape of an open rectangle with only two sides.

2. We also realized that this is an area problem, not a volume problem.

A=b*h
A=(12-2h)h..........The definition of the base is given in terms of height. From the equation that we set up, we can see that the area is a function of height.
A(h)=12+2h^2
A'(h)=12-4h.........Differentiated the previous equation so that we may find the critical numbers by finding the x-intercepts of the derivative of A.
0 = 12-4h
h = 3......................Determine if the critical numbers are a maximum or a minimum using the 1st derivative test.

3. We then determined the domain of which the function can exist using the context in the problem. The lowest possible input of the graph would be 0, which would give us an area of 0 (imagine two walls stuck together). The highest possible input of the graph would be 6, which would also give us an area of 0 (imagine two walls stuck together).

Domain: [0, 6]

4. We realized that to the left of h=3 on the graph of A', the function is positive, and to the right, negative. We know this by plugging a number <3>3 into the equation of A'. Therefore, h=3 is a local max. This is the first derivative test.

Answer: 3 inches


A North-South highway intersects an East-West highway at a point P. A car crosses P at 10:00 a.m., traveling east at a constant speed of 20 mph. At that same instant another car is 2 miles north of P, traveling south at 50 mph. Find the time at which they are closest to each other and approximate the minimum distance between the automobiles.

1. Drew the diagram and laid out the info.

• Point P is the intersection.
• Car A is moving 20mph west.
• Car B is moving 50mph south, north of P.
• We can write designate the distance as a function of time. The derivative of the distance function is the velocity function. So (dB)/(dt) = 50 and (dA)/(dt) = 20.
• Designated the hypotenuse of the triangle, d.

2. Using the info provided, we organized the info onto a chart, where d=distance, r=velocity (a.k.a. rate), and t=time of cars A and B.

Using the chart, we were able to generate a formula for d, the distance or the hypotenuse or the hypotenusal distance. Yes, that's my adjective for hypotenuse; it's been copyrighted by none other than me, zeph.

3. Because d is a radical function and we're looking for the critical numbers, uh oh, I forgot. (Someone, help me out here.) Then we expand the equation we generated for d, so we can apply the derivative rules as minimum as possible. (Who wants to use chain rule 20 times for deriving an equation? I don't.) We did this so we can find the critical number(s), determine if the critical number(s) is a max or min using the 1st derivative test, then solve for t.

Answer: t=1/29 hr, which is approx, 2 min, so at 10:02am the cars are closest.

4. Since we know t=1/29 hr, we put that into d to determine the minimum distance at 10:02am.

Answer: d(1/29)=0.7428 miles.


END NOTES

• Homework: 5.4 #7-12
• Slide #7 has a question for you to solve.
  
• We will be continuing optimization problems on Monday.
• Passing the baton to Joyce.

Today's Slides: December 5

Here they are ...

AP Calculus December 5, 2008

View SlideShare presentation or Upload your own. (tags: apcalc2008 math)

Big and Tall...Oops, I mean Small :P

Today's class started off with Mr. K giving us a small chat over the tags on our blog. Since we had so many due to most of us tagging our work wrong, he went over what the right tags to use for our posts were. Here are the main three tags we are to use every time we post something onto the blog:

• Name
• Type of Post (Scribe Post, BOB, On My Mind, or Links for Learning - Explained in some detail later)
• Title of Current Unit (They are shown in the book, or, you can see the slides for them)

Now to get into more detail about the Types of Post, especially the last two which are indeed new to all of us.

• Scribe Post: Self-explanatory, this is what we use if we're posting the scribe for the day
• BOB: Use this to tag Pre-Test thoughts and such, we should know how to use this by now
• On My Mind: When posting something of interest that isn't a scribe post or BOB, use this tag. A good example of this tag would be one of Benchmen's standalone posts that wasn't a BOB or scribe post (See post after the last test date where he was asking for opinions on the test)
• Links for Learning: Mr. K will mostly be using this tag to post links that will help us with current or past units (Will have the unit tag along with it). I believe students may use this as well.
  

We then moved on to talking about the video assignment, "What is a Derivative?", that was previously talked about on the day Mr. K came back. The details are as followed:

• Create a commercial describing what a derivative is
• Must describe using little to no algebra
• Cannot use derivative rules
• Must be 30 seconds MAX, there is a 10% deduction for every second over the max time limit
• Must be posted as a video response to Mr. K's video which will be posted on youtube (Will be posted on blog with the tag "Calculus Commercials"
• This will not only be open to us, but to the whole youtube community. Who knows? We might even go viral!
• No specific due date but must be done by the holidays!

After that, we then moved onto a few questions that was review for yesterday's class on limits.

Like yesterday's questions, first start off by expanding the numerator and denominator. Once done with that, you multiply the numerator and denominator by 1/x² / 1/x² since that is the highest degree in this question. This, as a reminder, reduces the highest degree to 0, leaving us only with 6/2, along with the remaining parts of the numerator and denominator divided by a degree of x. Those go away as the limit reaches infinity and we are now left with the horizontal limit equaling 3.

An easier way, as shown boxed in red would be to just take the coefficients of the leading terms and dividing them since all we care about are the coefficients. It saves us time and thus makes a seemingly long question a lot shorter.

Now onto a quick review of the 1^st and 2^nd Derivative Tests.

• 1^st Derivative Test: Finds critical numbers, as well as looking for where the parent graph is increasing or decreasing
• 2^nd Derivative Test: Determines the concavity of parent graph along with finding inflection points

In reality, these two tests both do the same job. That same job being the fact that they both can be used to max or the mins of the parent graph. Although, they do require different procedures to go about it.

From there, we then started the new stuff. Optimization Problems. These are basically problems where we are looking for where a function is big and small (max and mins). An example would be owning a business and trying to make as much money as possible while keeping production costs at a minimum.

There are six steps to solving Optimization Problems. They are as follows (Click to enlarge image for easy viewing):




We then went on to using these steps to solve a problem.

An open box with a rectangular base is to be constructed from a rectangular piece of cardboard 16 inches wide and 21 inches long by cutting a square from each corner and then bending up the resulting sides. Find the size of the corner square the will produce a box having the largest possible volume.

First use Step 1 to find what we are trying to maximize or minimize, in this case it is the volume.

We then follow Step 2 and write an equation for what we found in Step 1, which is the volume. So...



I've included a diagram so you can get a better grasp of the situation, plus they look pretty :). We then move onto Step 3, which is take the optimization equation (which was made in Step 2) and turn it into a two variable equation so we can easily get the derivative. So just substitute in the values from the diagram into the optimization equation and you'll get this:

We need the domain of c first so we can then use the quadratic formula on the derivative function and then take the value within the domain. That will be the size of the corner of the square which will provide a maximized volume, thus answering our question. If we continue on and do the quadratic formula, we would then find out that the only value that fits into our domain is 3. Thus, the size of the square that will provide us the maximized volume is a 3x3 one. Tada, we're done!


Well that was fun, the next scribe shall be, following Francis' lead on taking the next person on the list, Joseph. I refuse to call you Master Joseph :D. Oh yeah, homework is Chapter 5.4 Questions 1-6. Alrighty then *puts in PROPER tags*...*leaves*

Today's Slides: December 4

Here they are ...

AP Calculus December 4, 2008

View SlideShare presentation or Upload your own. (tags: math apcalc2008)

Inifinite Limits

Today in class we spent the majority of our time learning, and expanding our understanding of infinite limits and asymptotes, of both the horizontal and vertical type. First off, I'd suggest you watch the video found on the slides of today's slide show, because it's very informative and useful when it comes to better understanding this unit, which will help you understand this scribe post (hopefully). We started out about stating how infinite limits come in 2 "flavours" when a limit of "x" goes to some value, and that equals infinity, or approaches infinity, since infinity isn't a number. The other "flavour" is when the value of x approaches infinity and equals to some value. Those are the 2 main rules.

We also mentioned sequences and series. Where sequences are a series of numbers, that follow some rule, and are related to each other by this rule, and a series is the sum of the numbers in a given sequence. I'm not entirely sure why we mentioned this, but I think we brought it up to further understand the value of infinity, or the not-value of infinity, since infinity isn't a value at all, if that makes sense.

Okay so found in slide 2 of the slide show from today's class, we can see the equation that looks something like this : f(x) = (x-a)(x-b)/(x-c)(x-d), this is pretty much the general equation we used for the whole class. With this equation, we looked at how to find asymptotes, of the horizontal and vertical type. A little note about asymptotes: the graph doesn't necessarily leave these sections "untouched" it depends on what section of the graphed function you're looking at, because there can be a section of the graph, that might actually touch or cross where the graph isn't shown.

In a rational function when the denominator equals 0, we have a vertical asymptote. With this said, the denominator determines the vertical asymptotes, and the numerator determines the roots of the graph. Vertical asymptotes are related to limits when x approaches some value and the result is infinity, and horizontal asymptotes are related to limits when x approaches infinity and the result is some value.

Horizontal asymptotes are found on slide 6 of today's slide show. To find the asymptotes we lower the our highest given degree of a polynomial to 0 (the power on x). We multiply the numerator and denominator by the reciprocal of the given polynomial, as found in slide 6. After doing this, we should be left with the coefficients of those polynomials we reduced. All the other values should be over "x". When infinity is substituted into these x-values, then those values won't exist, and your just left with the coefficients, which will be your horizontal asymptote.

Doing these steps are important because if you substitute infinity into the x-values of the original equation, then most likely the result will be infinity divided by infinity, which does not equal 1! This is called an indeterminate form, other indeterminate forms are 0/infinity, infinity/o and o/o, in university if you decide to take calculus, you will further learn about these indeterminate forms. If you get a indeterminate value of 2/0, when trying to find the horizontal asymptote, then there is no horizontal asymptote, but instead might be a slanted asymptote. Also, if trying to find the horizontal asymptote, and the value is 0/1 then the horizontal asymptote will be the x-axis, regardless.

That pretty much sums up our class. All fun stuff aside. Hopefully you guys watched that video like 20 times like me.The homework was exercise 5.3, all the odd questions including 6 and 20. The next scribe will be Kristina, because I guess she's first on the list.