Sunday, December 31, 2006

Happy New Year!

Happy New Year! I hope you all had a great holiday. Seniors, I hope all those essays were finished. Juniors, I hope you all enjoyed not having to write them (this year!). I spent the last week having a great time up in Oregon with my family, and if you thought it was cold here…

You will notice when you log on to our blogs to creat a post that we have switched over to the “new Blogger.” It’s supposed to make things easier and run smoother (we’ll see!) but it will require each of you to switch over as well (and create a Google account) before you can edit or create new posts. Nothing serious, but I wanted to give you a heads up before your turn came around and you panicked!

See you all on the 8th!

Tuesday, December 19, 2006

Wednesday's Quiz Topics

Here’s a list of topics that will be covered on this Wednesday’s Quiz.

Quiz – Sections 5.3-4
Fundamental Theorem of Calculus, Part I (Sec. 5.3, #9,13,17)
Fundamental Theorem of Calculus, Part II (Sec. 5.3, #31,41)
Determine general indefinite integrals (+C!) (Sec. 5.4, #9)
Evaluate definite integrals (Sec. 5.4, #19,25,29)
Explain the meaning of a definite integral expression. (Sec. 5.4, #47,48)
Displacement vs. Total Distance Traveled (Sec. 5.4, #55)

Oh, by the way, no calculators on this one...

I’ll be in early Wednesday and I’ll check in tonight online. See you in class!

"Success is a peace of mind which is a direct result of... knowing that you did your best to become the best you are capable of becoming."
-John Wooden

Monday, December 18, 2006

5.4 Indefinite Integrals and The Net Change Theorem

Hey, it's Zak. I'm filling in for Sophie because she's all bogged down with other stuff. So she'll do 5.5 over break.

Anyway here's 5.4:

An Indefinate Integral is basically just like a definite integral except for the fact that it has no boundaries. A definite integral is limited by the values a and b whereas an indefinite integral is infinite, meaning the boundaries of a and b do not exist. Therefore, the notation for writing an indefinite integral is as follows:


Notice how the notation does not have any values for a or b inputed above or below the integral sign. That's how an indefinate integral should be written.

Important: There is a big distinction between definite and indefinite integrals. A definite integral is a number whereas an indefinite integral is a function or family of functions. Also, since indefinite integrals are functions, the "+ C" cannot be disregarded when writing out the function. For example:

When finding the antiderivative of the function:

We get:

Whereas with boundaries, we'd get an exact number and not incorporate the "+ C"

Net Change Theorem: The integral of a rate of change is the net change. In other words, the value of the integral is an object's displacement from the starting point, not the total distance traveled. For example, if I walk 10 feet and come back 9 feet, my displacement is 1 foot but my total distance is 19 feet.

Sample Problem:
Given the equationfind the displacement and total distance an object traveled in the interval

For this we can use our calculators:

Step 1: Enter the equation into Y1
Step 2: Go to the home screen and press (math) then (9) to get "fnInt" on the screen
Step 3: Enter after "fnInt"
Step 4: Press (Enter) and you should get an answer around 1.1715. This is your displacement.

To find the total distance:

Step 5: Go back to the equations screen and insert "abs" in front of the equation. This will prevent the graph from ever dipping below the x-axis, thus resulting in no negative distance and therefore representing the total distance and not the displacement.

Step 6: Follow steps 2, 3, and 4 again but this time you should get an answer around 14.828. This is the total distance traveled.

That's about it, hope that helped.

In the Humorous Spirit of the Holidays:
Mr. Church and Virginia may very well believe in Santa Claus, but some scientists have done some mathematical studies into the physics behind Santa Claus in order to prove that Santa Claus cannot physically exist. Needless to say, their results are very convincing (and humorous). So here they are:

A Scientific Inquiry into Santa Claus:

1) First of all, no known species of reindeer can fly. BUT there are 300,000 species of living organisms yet to be classified, and while most of these are insects and germs, this does not COMPLETELY rule out flying reindeer which only Santa has ever seen.

2) Secondly, there are 2 billion children (persons under 18) in the world. BUT since Santa doesn't appear to handle the Muslim, Hindu, Jewish, & Buddhist children, that reduces the workload to 15% of the total - which is 378 million according to Population Reference Bureau. So at an average (census) rate of 3.5 children per household, that's 91.8 million homes. And of course, one must presume that there's at least one good child in each.

3) Additionally, Santa has 31 hours of Christmas to work with due to the different time zones and the rotation of the earth, assuming he travels east to west (which seems logical). This works out to 822.6 visits/second which means that for each Christian household with good children, Santa has .001 second to park, hop out of the sleigh, jump down the chimney, deliver the presents, eat his milk and cookies, go back up the chimney, get back into the sleigh and move on to the next house. Assuming that each of these 91.8 million stops are evenly distributed around the earth (which, of course, we know to be false but for the purposes of our calculations we will accept), we are now talking about .78 miles/household, a total trip of 75.5 million miles. So Santa's sleigh must be moving at 650 miles/second, 3,000 times the speed of sound. For purposes of comparison, the fastest man-made vehicle on earth, the Ulysses space probe, moves at a poky 27.4 miles/second. A conventional reindeer can run, tops, 15 miles/hour.

4) Moreover, the payload on the sleigh adds another interesting element. Assuming that each child gets nothing more than a medium-sized lego set (2 lb.), the sleigh is carrying 321,300 tons, not counting Santa, who is invariably described as overweight. On land, conventional reindeer can pull no more than 300 lb. Even granting that "flying reindeer" (see #1) could pull 10 TIMES the normal amount, we cannot do the job with 8, or even 9 reindeer. Instead, we need 214,200. This therefore increases the payload - not counting the weight of the sleigh - to 353,430 tons which is four times the weight of the ocean-liner Queen Elizabeth.

5) 353,000 tons travelling at 650 miles/second creates enormous air resistance which will heat up the reindeer up in the same fashion as a spacecraft reentering the earth's atmosphere. Therefore the lead pair of reindeer will absorb 14.3 QUINTILLION joules of energy. Per second. Each. In short, they will burst into flame almost instantaneously, exposing the reindeer behind them thus vaporizing the entire reindeer team within .00426 of a second and all the while creating deafening sonic booms in their wake. Meanwhile, Santa will be subjected to centrifugal forces 17,500.06 times greater than gravity. Therefore, a slim 250-lb Santa would be pinned to the back of his sleigh by 4,315,015 lb. of force.

So if Santa ever DID deliver presents on Christmas Eve, there's no way he could've survived.

Sorry Virginia, sometimes facts are facts, but hopefully you never find out.

Sophie, you're up next, and you have all break to do it...

Mr. Jones is not Mr. French

I have proven that Mr. Jones does indeed reside in Scotland.

His website is, correct? Some of you have insinuated that Mr. French registered a UK hostname just to fool us. So, I decided to find the IP address of this website. The IP address is This particular IP is assigned to the UK, and presumably Mr. Jones is from Scotland. Just to make even surer, I used the IP to find the host for his website.

As you can see, this host is located in the UK. If you want further proof, I can show you how I did this in class tomorrow. If you want even further proof, I will ping the host myself. There is no way that Mr. French could have done this himself. It defies the laws of computer science. I think it's safe to say that this issue is now resolved. See you all tomorrow!


Sunday, December 17, 2006


Because I have way too much time on my hands, I was checking to see if the YouTube was working, and stumbled across Mr. French's comment on Mr. Jones blog. Thought I'd share with you all!! Hahaha I love our class.


(Mr. French writing on Mr. Jones' blog)

Mr. French
Dec 18th, 2006 at 3:27 am

Hello Mr. Jones! It’s Mr. French from across the sea with the crazy calculus class. I just thought you’d like to know you’ve become something of a cult figure with my students. In fact, you made it into some of their calculus/math holiday songs this year! Here’s a link to one of their videos. The girl on the right is Sophie…
Thank you for your comments!
Mr. French

Our Calculus Song on YouTube!

Hey everyone!!! I think that I have just successfully uploaded our Caluculus Christmas song to YouTube. If you are wondering, it IS Kristin Heintz laughing in the backround because she was too "Embarassed" to be in the video. Oh well... It's pretty much the tightest song ever! Enjoy! (Here is the link below, but if you click on the title, it will take you there too!)


Thursday, December 14, 2006

5.3: The Fundamental Theorem of Calculus

Hey guys, this is a brief update on the Fundamental Theorem of Calculus. We learned the first part of it in the previous section, which is:

If f is continuous on [a,b], then the function g defined by

is continuous on [a,b] and differentiable on (a,b) and g'(x) = f(x).

The Fundamental Theorem of Calculus Part II is as follows:

where F is the antiderivative of f so that F' = f.

Let's try a problem now, shall we?

So substitute 0 for a and (pi/2) in for b.

so the area of the graph is pi^2 +5.

sophie, i believe you are next up.

Wednesday, December 13, 2006

Thursday's Quiz Topics

Here’s a list of topics that will be covered on this Thursday’s Quiz.

Quiz – Sections 5.1-2
Estimate distance traveled from a velocity graph. (5.1, #15)
Express a Riemann sum as a definite integral. (5.2, #17,19)
Evaluate an integral in terms of areas – show your work! (5.2 #37)
Sketch a graph and estimate the area under the curve using RRAM, LRAM or MRAM (5.1, #3)
Evaluate definite integrals based on a graph (5.2, #33)

I’ll be in early Thursday, available after school this afternoon until 4, and I’ll check in tonight online. See you in class!

In the spirit of the holidays:
Yes, Virginia, There is a Santa Claus
By Francis P. Church, first published in The New York Sun in 1897. [See The People’s Almanac, pp. 1358–9.]
We take pleasure in answering thus prominently the communication below, expressing at the same time our great gratification that its faithful author is numbered among the friends of The Sun:

Dear Editor—
I am 8 years old. Some of my little friends say there is no Santa Claus. Papa says, “If you see it in The Sun, it’s so.” Please tell me the truth, is there a Santa Claus?
Virginia O’Hanlon

Virginia, your little friends are wrong. They have been affected by the skepticism of a skeptical age. They do not believe except they see. They think that nothing can be which is not comprehensible by their little minds. All minds, Virginia, whether they be men’s or children’s, are little. In this great universe of ours, man is a mere insect, an ant, in his intellect as compared with the boundless world about him, as measured by the intelligence capable of grasping the whole of truth and knowledge.

Yes, Virginia, there is a Santa Claus. He exists as certainly as love and generosity and devotion exist, and you know that they abound and give to your life its highest beauty and joy. Alas! how dreary would be the world if there were no Santa Claus! It would be as dreary as if there were no Virginias. There would be no childlike faith then, no poetry, no romance to make tolerable this existence. We should have no enjoyment, except in sense and sight. The external light with which childhood fills the world would be extinguished.

Not believe in Santa Claus! You might as well not believe in fairies. You might get your papa to hire men to watch in all the chimneys on Christmas eve to catch Santa Claus, but even if you did not see Santa Claus coming down, what would that prove? Nobody sees Santa Claus, but that is no sign that there is no Santa Claus. The most real things in the world are those that neither children nor men can see. Did you ever see fairies dancing on the lawn? Of course not, but that’s no proof that they are not there. Nobody can conceive or imagine all the wonders there are unseen and unseeable in the world.

You tear apart the baby’s rattle and see what makes the noise inside, but there is a veil covering the unseen world which not the strongest man, nor even the united strength of all the strongest men that ever lived could tear apart. Only faith, poetry, love, romance, can push aside that curtain and view and picture the supernal beauty and glory beyond. Is it all real? Ah, Virginia, in all this world there is nothing else real and abiding.

No Santa Claus! Thank God! he lives and lives forever. A thousand years from now, Virginia, nay 10 times 10,000 years from now, he will continue to make glad the heart of childhood.

About the Exchange
Francis P. Church’s editorial, “Yes Virginia, There is a Santa Claus” was an immediate sensation, and went on to became one of the most famous editorials ever written. It first appeared in the The New York Sun in 1897, almost a hundred years ago, and was reprinted annually until 1949 when the paper went out of business.
Thirty-six years after her letter was printed, Virginia O’Hanlon recalled the events that prompted her letter:
“Quite naturally I believed in Santa Claus, for he had never disappointed me. But when less fortunate little boys and girls said there wasn’t any Santa Claus, I was filled with doubts. I asked my father, and he was a little evasive on the subject.
“It was a habit in our family that whenever any doubts came up as to how to pronounce a word or some question of historical fact was in doubt, we wrote to the Question and Answer column in The Sun. Father would always say, ‘If you see it in the The Sun, it’s so,’ and that settled the matter.
“ ‘Well, I’m just going to write The Sun and find out the real truth,’ I said to father.
“He said, ‘Go ahead, Virginia. I’m sure The Sun will give you the right answer, as it always does.’ ”
And so Virginia sat down and wrote her parents’ favorite newspaper.
Her letter found its way into the hands of a veteran editor, Francis P. Church. Son of a Baptist minister, Church had covered the Civil War for The New York Times and had worked on the The New York Sun for 20 years, more recently as an anonymous editorial writer. Church, a sardonic man, had for his personal motto, “Endeavour to clear your mind of cant.” When controversal subjects had to be tackled on the editorial page, especially those dealing with theology, the assignments were usually given to Church.
Now, he had in his hands a little girl’s letter on a most controversial matter, and he was burdened with the responsibility of answering it.
“Is there a Santa Claus?” the childish scrawl in the letter asked. At once, Church knew that there was no avoiding the question. He must answer, and he must answer truthfully. And so he turned to his desk, and he began his reply which was to become one of the most memorable editorials in newspaper history.
Church married shortly after the editorial appeared. He died in April, 1906, leaving no children.
Virginia O’Hanlon went on to graduate from Hunter College with a Bachelor of Arts degree at age 21. The following year she received her Master’s from Columbia, and in 1912 she began teaching in the New York City school system, later becoming a principal. After 47 years, she retired as an educator. Throughout her life she received a steady stream of mail about her Santa Claus letter, and to each reply she attached an attractive printed copy of the Church editorial. Virginia O’Hanlon Douglas died on May 13, 1971, at the age of 81, in a nursing home in Valatie, N.Y.

5.2 The Definite Integral

Oh my god. This is a huge disaster. My whole blog is teeny tiny! I think if you click on it, and then press the little magnifying glass with the plus in it, it gets bigger. God I hope so.

IN OTHER NEWS...Here is a cool link to a UT website. You can basically plug in the function, your interval, whether you want it to use LRAM, RRAM, or midpoints, the number of rectangles you want, and it will give you both the area of the rectangles as well as the integral value. I'm a big fan.

I generally hate anything that glorifies seniors as somehow superior to other students. I also love the Onion. So combine those two and you get this.

Alison, you are next!

Friday, December 08, 2006

5.1: Areas and Distances

Here we go with Section 5.1, also known as finding the area on confusing graphs.

The first thing we talked about in class was how to find areas of different geometrical shapes. Of course, we can easily find the area of a triangle or a circle. But do we find the area of this graph?

Well let's take a look at a more specific example: estimate the area under the parabola y=x^2 from 2 to 4.

Well, we can't use a big triangle to find the area of this parabola because then the hypotenuse side would be curved. And sry Kate, we can't just not do the problem. But we can use rectangles! So let's divide the graph into four different strips (S1, S2, S3 and S4) by drawing vertical lines at x=2.5, x=3 and x=3.5.

Then if we make each of the strips rectangles, they will each have a base of 0.5. The parabola will look like this:

Now we can use these rectangles to estimate the area of the parabola. We know that the width of each rectangle is 0.5. The height is the y-value at each x coordinate. Therefore the heights are (2.5)^2, (3)^2, (3.5)^2 and (4)^2 respectively.
Let's use L4 to represent the sum of the areas of these rectangles.
L4 = (0.5)(6.25) + (0.5)(9) + (0.5)(12.25) + (0.5)(16)
L4 = 21.75

*Note that we don't include 2 as a height, because we can't find the y-value since it is not on the curve.*

However, part of the rectangles are drawn outside of the parabola. Therefore, the exact area of this parabola must be less than 21.75. If we make the rectangles smaller, we can find what the area has to be greater than.

Here the heights are (2)^2, (2.5)^2, (3)^2 and (3.5)^2 respectively.
Let's use R4 to represent the sum of the area of these rectangles.
R4 = (0.5)(4) + (0.5)(6.25) + (0.5)(9) + (0.5)(12.25)
R4 = 15.75
*Be careful NOT to inculude (4)^2 in the heights because the point (4, 16) is outside the rectangle.*

Therefore the area is greater than 15.75, but less than 21.75. If use the same parabola but generalize the interval, we can show that the sum of the areas of the rectangles approach 1/3.

The graph would be:

Each rectangle has a width of 1/n and the heights are the values of f(x)=x^2 at the points 1/n, 2/n, 3/n... The heights are (1/n)^2, (2/n)^2...

R4 = (1/n)(1/n)^2 + (1/n)(2/n)^2 + ......+ (1/n)(n/n)^2 R4 = (1/n)(1/n^2)(1^2 + 2^2 + 3^2+......+ n^2) R4 = (1/n^3)(1^2 + 2^2 + 3^2+......+ n^2)

We can use the formula for the sum of the squares of the first n positive integers:
1^2 + 2^2 +.....+n^2 = n(n+1)(2n+1)/6

After we substitute this expression for limit of Rn, we get:

If you think really really hard, you'll realize that if we keep on decreasing the width of the rectangles, the heights will eventually match the curve. Take a look:

The width of each rectangle in the interval [a,b] is b-a. The formula is:

Defintion: The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles:

Alright. Now that we know the formulas, let's try a problem. Find the area under the graph of f(x) = sin x between x = 0 and x = 1.

1) Find width
2) Estimate the area by taking the sample points to be midpoints & use 4 subintervals.

1) Since a=0 and a=1, the width is:

x1 = 1/n, x2 = 2/n, x3=3/n, xi=1i/n. The sum is:

2) n= 4 because we have 4 approximating rectangles. Therefore the width is:

Subintervals: [0, .25], [.25, .5], [.5, .75], [.75,1]

Midpoints are:

Finally we can calculate the sum:

The distance problem. Remember this equation: distance = velocity x time??? of course

But what if the velocity is hard to find? Well then we can find the area of the function! We simply make a table with our x-value as time and our y-value as velocity. (just like on pg. 376)

Now i bet you guys got lost back at the colorful graphs. Well no worries because this site is awesome. There are like 9 pages of explanations on how to find the area of any curve. GO CHECK IT OUT! This site is good too:

Kristin H! I believe you are up next!

Now for some Chistmas PUNS!: Which elf sings "Love me tender?" Santa's little Elvis!

Oh and this is just funny:

Wednesday, December 06, 2006

Solution to Question #1, Quiz 4.5-7

Greetings all!

I've put together a solution to the first question on Quiz 4.5-7 and posted it to our class website. Before we spend time going over it online or in tutoring, take a look at my solution...

See you in the morning!

And remember, math doesn't have to be boring:

Thursday's Test Topics

Here’s a list of topics that will be covered on this Thursday’s Chapter 4 Test.

Chapter 4 Test Topics
Identify absolute extrema (4.1, #29,41,49(
Mean Value Theorem (4.2, #9,11, bonus - #34)
Limits – Indeterminate form (4.4. #15,21,27)
Maximize Profit/Revenue/Cost (4.8, #9,11,13)
Antiderivatives (#4.10, #9,17,19)
Marginal Profit/Revenue/Cost (4.8, #9,11,13; 4.10, #35,37)
Interpretation/meaning of derivative concepts (Sections 4.1.,,4.10)
Optimization (4.7, #7,11,15)
More antiderivatives(4.10, #35,37)
More antiderivatives (position/velocity/acceleration) (4.10, #61)
Interpretation of the graph of f ‘ (4.3, #31)

That’s it! I’ll be in early on Thursday, and available after school on Wednesday. See you in class!

"I shall be telling this with a sigh
Somewhere ages and ages hence:
Two roads diverged in a wood, and I --
I took the one less traveled by,
And that has made all the difference."

- Robert Frost
The Road Not Taken

Tuesday, December 05, 2006

4.10: Antiderivatives

Hey guys!! So, I'm back to talk about the last section of Chapter 4: ANTIDERIVATIVES!! YAY! ok so let's go...

The book definition of an antiderivative:
A function F is called an antiderivative of f on interval I if

Now, let's take that definition and apply it to this next situation:

if this is my derivative equation, what is the original f(x) equation?
YES!! it would be, its general form,

Now, you might be wanting to ask me, "Gianna, what does the antiderivative have to do with this? And, furthermore, what the heck does 'general form' mean?" To those questions, I would respond by saying that firstly, the antiderivate is the original f(x) equation so that answers your first question. As for the second one, an antiderivative's general form is, simply put, the generic one that contains the + C, which needs to become your new best friend by the way. Furthermore, the general form( general antiderivative) can be found easily. LOOK:

Now that we have this extra tool to help us, let's do some examples.

Ok, FANTASMIC!! Now that we are very comfortable with, or should be at least, the concept of the antiderivative, let's apply it to the real world questions about acceleration, velocity and position equations.

Ok so now that we have a full knowledge of the antiderivative, we can call ourselves COMPLETE PEOPLE!! YAY!! So, Kristin, good luck and here's a comic to help you get going:

Monday, December 04, 2006

4.8: Applications to Business and Economics

If any of you used to think that calculus does not have to do with the real world you were mistaken!!! This lesson talks about using calculus concepts and applying them to economics, business, and money- things that can relate to our every day lives!

There are the different functions that were represented in this lesson- cost, marginal cost, average cost, demand (or price), revenue, marginal revenue, profit, and marginal profit. I know all of these sound really similar, like, what's the difference between revenue and marginal revenue? Well, the word marginal should make you think of a derivative. Because the revenue function is represented as R, the marginal revenue function is R'(x) (or the derivative of R). The same thing applies to the rest of the equations as well- ie. if profit is P marginal profit is P'(x) (or the derivative of P).

So here are the actual equations for the money functions:

Cost: C(x)
Marginal cost: C'(x)
Average cost: C(x)/ (x)
Demand: p(x) --- note that this p is lower case!
Revenue: R= xp(x)
Marginal revenue: R'(x)
Profit: P(x) = R(x) - C(x)
Marginal profit: P'(x)

Now that we know all the equations we can start to understand them!

Important things to remember:

-If the average cost is a minimum (if you want to minimize the cost of an item), the marginal cost must equal the average cost, so C'(x) = C(x) / x

-When you're talking about the marginal cost, you're talking about how much $ you are making by selling or producing items.

-The average cost is the total cost in terms of units.

-If the profit is a maximum (you want to maximize the profit), the marginal cost must equal the marginal revenue, so R'=C' and R'-C'=0.

A problem that might be asked:

The cost of producing x amount of ribbon is C(x)= 2000 + 10x + .005x^2.

A. Find the cost (C(x)) the average cost, (c(x)) and the marginal cost (C'(x))of producing 50, 500, and 5000 ribbons?

B. When will the average cost of the ribbons be the lowest?

First you have to find the equations for the cost, av. cost, and marginal cost.

You know the equation for the cost is C(x)=2000 + 10x + .02x^2.
The equation for the average cost is C(x) / x, so when you divide the cost equation by x you get: 2000/x + 10 + .005x. So c(x)= 2000/x + 10 + .005x.
The marginal cost is just the derivative, so C'(x)= 10 + .01x.

Now, because there are so many equations and you have to find different values, plug in 50, 500, and 5000 into each equation. You can even make table to make it easier.

So now we have part A answered! (The answers would be the numbers in the table)

On to part B.

The average cost of the ribbons will be the lowest when the marginal cost= average cost! So this means the equation we found for c(x) = C'(x).

2000/x + 10 + .005x = 10 + .01x

Subtract 10 from both sides.

2000/x + .005x = .01x

Subtract .005x from .01x.

2000/x = .005x

Multiply x to both sides.

.005x^2 = 2000

x= 632.45

Now, plug 632.45 back into the c(x) equation.

2000/ 632.45 + 10 + .005(632.45)

You get 16.32.

So the minimum average cost per ribbon is $16.32.

Now we're done with the question!


This is another question you'll be asked- its like a word problem.

There's a Joshua Radin concert coming up next week at the Troubador. If the ticket price is $12, 2,000 tickets will be sold. But if the price is dropped to $10, 5,000 tickets will be sold. How many tickets should be sold to maximize the revenue?

Find the slope of (2000,12) and (5000,10) and put this into slope intercpet form.

(The slope is 2 / -3000 which simplifies to 1 / -1500)

p- p1 = m(x- x1)

p - 12 = - 1 / 1500 (x - 2000)


p-12= -1/1500x + 4 / 3

Add 12 to both sides.

p = -1/1500x + 40/3

Now that you have your demand function you have to find the revenue function.

The revenue function (R) = xp(x). So multiply the equation we just found by x. You get:

R= -1/1500x^2 + 40/3x

Find the marginal revenue function (take the derivative)

R' = - 1/750 x + 40/3

The maximum revenue is when the marginal revenue is equal to zero.

So, 0= R'= -1/750x + 40/3

-40/3= -1/750x

x= 10000

Now plug 1000 back into the equation that we have for p.

p = -1/1500(10000) + 40/3

p= 20

So to maximize ticket sales the price of the tickets must be $20!

(I can't believe that with a problem I made up the numbers for I got whole numbers for my answer! Yay!)

Here's an amazing link that explains marginal cost. It even talks to you and it really interactive, theres information on the site on fixed and variable costs & I don't think we're learning that, so just skip down to the bottom to the part on marginal cost.

This site is good too:

I know it's a little hard to read so i'll just tell you what it says:
"What matters most in your pricing decision, average or marginal cost?"
And the guy in the car is saying: "Officer, my trip to Chicago only took 2 and 1/2 hours which equates to 60 miles per hour!" But the officer is saying he went 90 miles per hour. You can see this page to see how the speed the car is going relates to the marginal cost/revenue etc.

Gianna you're next!!!