Optimization with Calculus 3

[takfarinas1]

That was fun.

[Adriancfl]

-.- 16*4 isn't 72...

[No1GuildWars2Fan]

@Adriancfl that's 18 not 16

[Adriancfl]

@No1GuildWars2Fan Oops, thanks!

[BornOfAshes23]

How can you guarantee the minimum from 56.5 is the absolute min and not the other endpoint without plugging it in?

[AjGuitarist]

Thanks, almost exactly like the problem i'm working in my notebook, with the exception of it just being two squares. The only thing I don't understand is i'm asked to express the problem in terms of "x" and "y" yet I can only express the combined area with x. Not sure where y would come in because they're both squares.

[ICarnag3I]

Anyone know what the program he's using is called?

[blainemj]

@ICarnag3I most likely he used microsoft paint to draw the problem and fraps to record his work.

[runnerjoe2324]

The derivative of a^4/16 is not actually a/8. How is it that you simply reduced the exponent three degrees taking just first derivative?

[danishvioletgirl]

@runnerjoe2324 Its actually the derivative of a^2/16. Maybe you read it wrong? a/8 would be correct.

[Mikej1592]

You lost me at 5:56 when you started to pull out that 625-50a/4+a^2/16 I have no idea where you got that but I know that it is right since it and the first form are equal values I just with you explained a bit more how you got that. I'm taking calc2 and its been 25 years since I took algebra so forgive me if I sound stoopid. ;-) Can someone explain or point me to the right video please?

[danishvioletgirl]

@Mikej1592 I see that no one is answering you but from where I see it, he just simply expanded (25-a/4)^2. So that expanded is therefore 625-(50/a)^4+a^2/16

[Mikej1592]

@danishvioletgirl ohhhhhhh LOL thanks, I was so fixated on being confused on the concept of the overall problem that I didnt think that 625 was 25^2 thanks for the reply. These problems vex me so.

[RoyWriter]

This man will be the reason I will sweep through college.

[Narutix37Returns]

LOL @ 0:40 "allow the possibility of no cut..........Fascinating"

[iamtheplauge]

The area of a square is greater than the area of an equilateral triangle of the same perimeter, so really you didn't even need to do that last part.

It had to be 0 because that was the value that didn't cancel out the square.

However, I guess it is always good to prove your work. :P

Great vid man.

[Todoslosfritos]

hahahahahha I love that "fascinating!"

[junior1984able]

Wow.

[Liaomiao]

how do you get the height of the triangle so quickly in one step?

[TheTrueGreenCrayon]

@Liaomiao 30 60 90 special right triangle rule

[0asian1nvasion0]

i prefer PatrickJMT.... =), but this is still cool.

[Mikej1592]

@0asian1nvasion0 PatrickJMT blocked me for whatever reason, guess he can't take corrections to his work.

[greatdeath001]

I love Calculus. : )

[Alesalad]

the solving method is correct and the answer is right however IMHO such a complex problem will not be given in a non-calculator examination or you will not be in the condition of not being able to use a one. (In fact even the normal windows calc was used). I solved this problem in a couple minutes by plugging the values of both areas in a graphic display calculator for values of X between 0 and 100. The result is a nice upward looking curve with a minimum at 56.5 and max at X=0 (all for sq.)

[1brightshadow1]

SALMAN BHAI Jazakallah!

[drewpasttenseofdraw]

NO. Ya Duffice...its my phone. Hold on while I answer it!

[bullet272443]

YOU ARE AWESOME! Thankyou for your vidoes! Ima pass my calculus test tomorrow!



[flopski]

I tried the problems myself before watching the video and solved them both. Sometimes I amaze myself. :)

100 - 3x -4y = 0

A = 0.5 * x² * sin 60 + y²

Go go gadget substitution!

[yoguely]

Help with Manifolds aaaah

[Seguris]

your vidoes are amazing and very easy to follow and understand, but I bet you, more than 90% of the people that look at this and take notes, will not be able to think up the same amount for a likewise problem, without referring to this or their notes.

[kpacdd]

"it's a 100 long!"

[rob5654]

Isn't this Calculus 1?

[xxDarkGregxx]

@rob5654 This is the 3rd video in the video series of optimization.

[jdt471]

@rob565 video is only named with a 3 because its the third video on optimization man.

[trippleices1]

@rob5654 it means part 3 not the level lol cuz there's 4 videos

[MonaRocks09]

jazakallah khair Salman bhai :)

[brwnsfan]

KHAAAAAAAAAAAAANNNN!!!!!!!!!!



[liliane2011]

thank you thank you thank you

im a computer comunication student and u v been a verry gd help to me since two yeaRS till now thank you a lotttttttttttttttttttttttt

[broozbehani]

Nice problem. Great instructor. God bless you!

[11Agamemnon235]

That is a neat problem :)

[laram24]

"I can already sense" LOL

[hmodywakid]

This math is shit come and study math in mother russia

[jonolollmao]

So when dy/dx only gives the max or min point, the other point can be found by substituting the boundary as in this case 0cm or 100cm ?

[n1a1s1i1m]

Well, I don't always like the idea of cracking every equation my way. Kinda because we're not allowed calculators, kinda because the calculus you infused in my head tells me to directly differentiate. And so once again, Sal Man saves the day! :D

[standish250]

Oh wow haha yea my bad don't know what I was lookin at

[tnmr42]

You lost me at 6:31

Sorry man. Your other vids are great, this one's just too damn long.

[O0oSorousho0O]

@tnmr42 That's the easy part dude....

[Utubemember3]

The Max/Min comes from the +/- sqrt(3). Plus give Min. - Gives Max.

[madscirat]

I love how you work problems without preprep. It allows everyone to see your thought process, imperfections included. I have taken your lead and shown a few problems on my site, but Im not nearly brave enough to do it your way yet.

[standish250]

I might be wrong but doesn't the derivative of f (a)= (a^4)/16

become f ' (a) = 4(a^3)/16 which = (a^3)/4?

He has a/8 which means he took the 3rd derivative [for a to have no exponents] but in reality that would be f ''' (a) = 3a/2

If I'm wrong please explain how because that would mean his entire problem from then on is off.

Thanks!

[zsozsokel]

(a^2*Sqrt3)/4 --> area of equilateral triangle where a is a side.

[ofarevolution441]

this video could've been like 6 minutes shorter if you just used pencil and paper and a camera. :p

regardless, thanks for the help!

[gobberpooper]

lol i cracked up when he goes "calculatooor!" and drags the calculator window over

[ManUtdMU7]

thank you soo much...you are a lifesaver!

[angry77561]

This video has to have some mistakes.

We know that the area of a square is x^2 and that the length of the wire is 100. He should have gotten 1000 for the area of the square instead of 625.

[chett16]

If the total length is 100 then each side is 25

[bkisme]

The wire has to be distributed across the entire square, therefore each side would be 25.

25^2 = 625

Sal was correct in his calculation of the area of the square.

[Xlaxsauce]

thanks again :D

[Zanirtak]

UUUGGghhhhhh I have a similar question but the answer I'm getting is 32 and the total length of the wire is 14.........

[Cervy18]

It stated specifically in the problem that the wire is cut into two pieces, but then it said to "Allow the possibility of no cut." Is there any point to that statement, or is it there just to try and trick you? I don't understand it.

[irlazybum]

It wants u to find the min and max of the two combined areas from square and the triangle. so u'd have to find 2 crit values. but if in the problem there came to be 1 critical point (like it this in this problem, 1 crit point at 56.6) u would have to check the end points (like EVT) to see if there are bigger values than that of the crit point .

that's why it said allow the possibility of no cut, so u would have to check when the rope is used for only for the area of the triangle or square

[sharvil1993]

I think it means that after the 1st cut, there should be no more cuts in making the square and triangle.

[baylewis]

great video

but he could have simplified at 7:43

[rryking]

suppose you hang a rectang painting on the wall so the observer standing on front has the best view. We want to maximinze the angle theta subtended by the painting. Prove theta is maximized when the center of the painting is at the level of the observers eyes.

[zoazoon]

Cool problem rryking! I solved it after a couple hours of deep thought.

Let A be the height of the painting, B the distance between the observer and the painting, alpha the angle between the bottom of the painting and the observer's line of sight, beta the angle between the top of the painting and the line of sight. Finally, the vertical distance from the bottom of the painting to the point where the line of sight hits the wall is a/x, so that the rest of the painting's height is a - a/x

[zoazoon]

(post continued -- I'm discussing a problem a commentor brought up, not the one in the video)

If we can show that theta is maximized when x = 2, we will have proven what we intended to prove. Theta is alpha + beta which is arctan(a/(bx))+ arctan(a/b -(a/(bx))). Use scaling to turn a/b into another constant, k. Then differentiate and find that the only critical point is x = 2. Checking all possible extrema confirms that x = 2 is indeed an absolute maximum for theta.

[joek1121]

my last name is Khan as well!

I love your videos. TY.

[iluvjiyongx3]

thank u soooo much! this question was pretty complicated, but thanks for explaining it so well! ^^

[BACJ4life]

Thank you so much! Your a real life saver!

[hjyo33]

Sal, what would be better to prove maxes or min, the first or second derivative test?

[Bartisim0]

Thank you Sal

[kingjoker54]

What field would you use that in, and also next time I cut a barb wire I'm gonna write out a formula for how and where to cut.

[KoffKofy]

great!

(take a look at 7:07, the area of the triangle)

[tuberALEX]

your video's are awesome! I just get annoyed by the restraint of one page.

[GoldAK47]

ufff i needed something like this! You're a life saver!

Thanks