Absolute max and min values Problem 1 (Multivariable Calculus)
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- Опубліковано 7 жов 2024
- 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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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!
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
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Thank you so much! This makes a lot more sense!
You’re welcome! I’m glad it helped.
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Thanks! Sending some love back to you. Good luck on your exams!
Nice!
Thanks, fellow mathematician!
Thank you sir
You’re welcome, fellow mathematician!
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
Great video. This helped a lor
THANK YOU SO MUCH
You’re welcome, fellow mathematician!
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?🙏
thanks boss
You’re welcome, fellow mathematician!
-4
why do we split the boundary as top and bottom and not left and right?
You could but you would need to solve for x instead of y
Do Lagrange multipliers work for this?
Yes, but only for checking for maximum/minimum values on the boundary of the region. The constraint would be x^2 + y^2 = 16
@@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?
That’s right!
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?
yes it is possible. you always have to compute the endpoints of the boundary.
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
Calculus is basically a fixed procedure. When familiar enough with computations you can remove some steps
Very complicated