Multivariable Optimization with Boundaries

Рет қаралды 44,528

4 жыл бұрын

Suppose we want to find the maximums and minimums of a function. Previously in our Calc III playlist we saw how to do this with the second derivative test. But what if we demand that are optimizing on a closed, bounded region? That is, a region that has a boundary. Similar to first year calculus, the idea is that the maximums and minimums either occur in the interior of the region (found by analyzing critical points with the second derivative test), OR the extrema occur along the boundary. In this video we will learn how to do this process along the boundary.
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Пікірлер: 51

@devashishshah9021 3 жыл бұрын

Please keep creating such great playlists..... very helpful while all of us study from home

@DrTrefor 3 жыл бұрын

Thank you I will!

@coltonlemasters 9 ай бұрын

I'm a TA for a multivariable calculus class and your videos are the best refreshers! Thank you!!

@konstantinakalimeri3650 4 жыл бұрын

I really appreciate your videos. I have recently discovered your channel and it has helped me a lot. I’m going to take my calculus III and differential geometry exams in a few days. Keep up the good work!! Greetings from Greece 🇬🇷

@konstantinakalimeri3650 4 жыл бұрын

Trefor Bazett Thank you!!

@youssefdirani 2 жыл бұрын

Smooth smooth smooth. Just fantastic

@zheka47 2 жыл бұрын

You are awesome! Very clear explanations

@CartwheelPig 3 жыл бұрын

Thank you so much for these videos, they're a huge help!!!!

@DrTrefor 3 жыл бұрын

So glad they are helping!

@stevehof 4 жыл бұрын

Amazing, as always

@aashsyed1277 2 жыл бұрын

awesome super great explanation , great content!

@kartikshrivastava582 4 жыл бұрын

Nicely explained !!

@kennethkau2332 2 жыл бұрын

You are the goddamn best Dr. Bazett.

@avinavkashyap8802 2 жыл бұрын

Nicely explained

@joaomattos9271 Жыл бұрын

Excellent!

@ankuryuvraj6117 2 жыл бұрын

Thank you from India 🇮🇳🙏❤. I cleared my doubt.

@sandeshkumarchandrakar1622 3 жыл бұрын

good work keep going ahead

@declanallan885 2 жыл бұрын

awesome video dude, honestly, clearer explanation then my uni lecturers [Newcastle University Aus; The quality of education at this institution is horrible though admittedly, we have the WORST teachers, they basically don't have teaching skills and are so out of touch with reality/industry it's a joke!] Anyhow, thank you youtube teacher who is better then some of the guys i have paid thousands to over the years to!

@giorgostarnaras5658 2 жыл бұрын

Nice there's actually a third method that's good for more complicated functions/boundaries.

@avinavkashyap8802 2 жыл бұрын

Really appreciated

@HosRo4161 Жыл бұрын

Thank you!

@richlovepherodin2050 Жыл бұрын

You earned a sub.

@sergiolucas38 2 жыл бұрын

excellent video, very important subject :)

@DrTrefor 2 жыл бұрын

Glad it was helpful!

@lenag3329 3 жыл бұрын

thank u!

@AbjSir 8 ай бұрын

Thanks sir

@avinavkashyap8802 2 жыл бұрын

Ty very much

@bilalnaseem7830 4 жыл бұрын

Hi how do you make these 3d graphs?

@markburton6182 Жыл бұрын

Finally, a professor with diction!

@sureshkeerthi9820 Жыл бұрын

Thank you

@hubenbu 2 жыл бұрын

Great visualization! imao, the solution presented is on the closed region x^2 +y^2 = 1, which is a hollow cylinder, not on the solid cylinder x^2 +y^2

@yatooosenpaaii9771 2 жыл бұрын

Omg thank you u saved me lotssss>< My professor sucks TT

@avinavkashyap8802 2 жыл бұрын

Plzzz make videos on probability and statistics also

@lucieneyvrard5414 3 жыл бұрын

Amazing content as usual ! Very clear and unbelievably helpful ! However anyone could help me to get this clearer ? I understand until the equation sin(theta)= -1/2 After that where does this triangle come from ?! And how did he solve it after that ? Cheers Pardon my french

@anubhavbhura13 3 жыл бұрын

The triangle he draws is from sin(theta)=1/2, not -1/2. If you know basic high school trigonometry you can easily solve sin(theta)=1/2. To solve for sin(theta)=-1/2, you need to know how sine acts in all 4 quadrants, using the definition of radians and unit circle. The graph he draws for sin comes from these 4 quadrant sine values. you can either see from that graph, or you can use rules for different quadrants, such as sin(theta)=-1/2 corresponds to sin(pi - theta)=1/2. this gives pi - theta = pi/6, hence theta = 7pi/6. These are the rules of inverse trigonometric functions, the fact that different trigonometric functions may be negative or positive in different quadrants requires that we have several tools to bring these angles back to our first quadrant (from which you may infer that values theta only goes from 0 to pi/2). Ill leave the other value of 11pi/6 up to you to calculate, because inverse trigonometry is very unpleasant to me!!

@KonstantinA.-bb8rt 2 ай бұрын

If I use the second method (i.e. take df/dy after substituting 1-y^2 for x^2), I will only find the Maxima with y = -1/2. The nothing points at x = 0 don't show up. Do they just fall out of the computation depending on which parametrization you choose and nothing points like these can just come up from to time? Thanks a lot for your videos, very concise, they allow me to understand what I just memorized before!

@atirmahmood7058 9 ай бұрын

If I choose substitution instead of parametrization then will I have to substitute once y in terms of x and solve it and then x in terms of y and solve this also please reply to this doubt of mine and if not both then why.

@sandrashappell6754 Жыл бұрын

What program are you using for your beautiful graphs

@Dollygehan012 2 жыл бұрын

sir, what is local maxima and local minima

@tillfriesewinkel5770 3 жыл бұрын

First of all, thanks a lot for this superb series! Not trying to be a smart ass, but I would like to understand: a) At ~14:30 you talked about global maximum/minimum. Was that a mispronounciation, did you mean global in terms of "global on R", or did I get something wrong? b) What about the case where f(x, y, ...) = c (some constant value)? I *guess* that then we also have the case of f_xx * f_yy - f_xy^2 = 0, and would like to know if that is true, and how that is function is considered inside a fixed region?

@SHAHHUSSAIN 4 жыл бұрын

Matchless

@christopherrosson2400 3 жыл бұрын

how does it clean up to 1+cos^2-sin at 8:08 ?

@anguskappa5637 3 жыл бұрын

using identities on the sin^2 term. 1-cos^2(theta)

@AAMMTECH 2 жыл бұрын

Basically, you separate the two cos^2(theta) and one of them you use it to simplify cos^2(theta) + sin^2(theta) = 1. So, cos^2(theta) + [cos^2(theta) + sin^2(theta)] - sin(theta) = cos^2(theta) + [1] - sin(theta)