Absolute max and min values Problem 1 (Multivariable Calculus)

This problem goes over how to find the absolute maximum and absolute minimum values of a function of two variables on a closed, bounded region. It's very similar to how this is done in calculus 1, where you check the values of the function at the critical points and endpoints of the interval. Now, the boundary of a region is a curve.
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[@magaplex6476]

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[@BlackTshirtMathProfessor]

Thank you! As a former physics student myself, this makes me really happy to see that you found some of my videos useful. I do still actively reply to comments but I haven’t been able to make time for recording new videos after some things changed in my personal life. I’m still planning to get back to it at some point.

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[@zizobob1082]

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[@BlackTshirtMathProfessor]

+0.0000000127 bonus points

[@halbarad7932]

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[@BlackTshirtMathProfessor]

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[@gretchenhe6726]

Thank you so much! This makes a lot more sense!

[@BlackTshirtMathProfessor]

You’re welcome! I’m glad it helped.

[@SweetBonanza2001]

love you man

[@BlackTshirtMathProfessor]

Thanks! Sending some love back to you. Good luck on your exams!

[@azizkash286]

Thank you sir

[@BlackTshirtMathProfessor]

You’re welcome, fellow mathematician!

[@eulzzzz7682]

Sir how did you check without inserting the boundary values on to the function is that possible to find the max and min value just by seeing the boundary value max and min?🙏

[@Jarrodjohn2007]

Nice!

[@BlackTshirtMathProfessor]

Thanks, fellow mathematician!

[@philippelevesque4801]

THANK YOU SO MUCH

[@BlackTshirtMathProfessor]

You’re welcome, fellow mathematician!

[@Bedoroski]

Great video. This helped a lor

[@muffindestroyeritsmumblintime.]

why do we split the boundary as top and bottom and not left and right?

[@BlackTshirtMathProfessor]

You could but you would need to solve for x instead of y

[@QuocAnhHandsome]

-4

[@brightknight0623]

Do Lagrange multipliers work for this?

[@BlackTshirtMathProfessor]

Yes, but only for checking for maximum/minimum values on the boundary of the region. The constraint would be x^2 + y^2 = 16

[@brightknight0623]

⁠@@BlackTshirtMathProfessorso the process would be to use Lagrange multipliers to find the candidates on the boundary, and then find other critical points where it could occur?

[@BlackTshirtMathProfessor]

That’s right!

[@rakibulhaque3444]

thanks boss

[@BlackTshirtMathProfessor]

You’re welcome, fellow mathematician!

[@SweetBonanza2001]

Why did you have to calculate the end points (4,0) & (-4, 0) for both curve functions. Is it possible for a point to have different values?

[@punnawitrinnasak2576]

yes it is possible. you always have to compute the endpoints of the boundary.

[@umar5834]

In this case I don’t think he needed to recalculate those values since they are shared by both curves. But you should definitely calculate them initially and since they share those points they should have the same function value

[@Bedoroski]

Calculus is basically a fixed procedure. When familiar enough with computations you can remove some steps

[@rogersssali9722]

Very complicated