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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[magaplex]

I always thought finding the critical points of a circular region was the hardest thing ever, but this video has given me such a clear understanding of it, and for that I thank you!

[Mustafa]

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[Aaronmtambo-mz]

Thank you so much Sir, I have been struggling to find critical points and absolute maximum and minimum in circular region now, I have something to jot down. Once again thank you

[eulzzzz]

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?🙏

[gretchenhe]

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

[muffindestroyeritsmumblintime.]

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

[wryanihad]

Another way to find min

f(x, y)=(x+2)²+(y-3)²-13

Since the braced always positive then minf(x, y)=-13

[QuocAnhHandsome]

-4 <= x <= 4, so x can be -4 or 4.
Therefor sqrt(16 - x^2) can be 0. How about this case ?

[brightknight]

Do Lagrange multipliers work for this?

[SweetBonanza]

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?