sir i want to say a big THANKS to you, you don't even waste our time when you write questions on board. Sometimes looks like it don't affect anything but no, it really saves a lot of time of ours !
Example 1: Absolute max value =2 at point (1/2,1/2) Absolute Min value= -32 at point (1,0) Example 2: Max temp = 9/4 occurs at points (-1/2,(sqrt(3))/2) and (-1/2,(-sqrt(3))/2) Min Temp = -1/4 occurs at point (1/2,0)
@@thorrr7613 hw-1-absolute min at (0,0) is 1 and absolute max at (0,4) and (4,4) is 17 hw 2-absolute minima at (5,-5/2) is -18.5 and absolute maxima at(0,-3) is 11
sir i have a doubt , in case 2 ( using boundary points) , we don't include boundary points as critical point's , but when ur listing all the points you included the boundary points at last for the square boundary question , why?
Good question.. It only tell us about local extremas reason being, when we prove this theorem involving hessian, we use Taylor's theorem and Taylor's theorem is local in nature ( not global)..
Sir, in first example at (1,0) value is -32 so it is absolute minima so it should be critical point. But when I used equation in online calculator it only gives (0.5,0.5) as maxima and (0,0) as saddle point. Can you please explain?
@@atharvabhomle3191 no..it's not the interior point.. At the end we take 1.Critical points then 2. interior points from the segments and 3. the corner points Then value of function at these points
sir i want to say a big THANKS to you, you don't even waste our time when you write questions on board. Sometimes looks like it don't affect anything but no, it really saves a lot of time of ours !
Thank you for such acute observation. 😊😊 I am happy that it helps you people so much 🙏😊
Thank you so much for providing such valuable knowledge. It really helped me to fully understand the course of MVC. Thank you so muchh!
Welcome..
Very happy to hear that 😊
Example 1:
Absolute max value =2 at point (1/2,1/2)
Absolute Min value= -32 at point (1,0)
Example 2:
Max temp = 9/4 occurs at points (-1/2,(sqrt(3))/2) and (-1/2,(-sqrt(3))/2)
Min Temp = -1/4 occurs at point (1/2,0)
Thank you for your efforts in posting the answer..
hw1- absolute minima 0 at (0,0)
absolute maxima 13 at (0,4) and (2,4)
hw2-absolute minima -10 at (4,-2)
absolute maxima 71.75 at (-9/2,-3)
Thanks for posting your answer..
in hw 1 how to find local maxima ? values of x & y are coming out complex.
hw-1-absolute min at (0,0) is 1 and absolute max at (0,4) and (4,4) is 17
hw 2-absolute minima at (4,-2) is -10 and absolute maxima at(0,-3) is 9
@@thorrr7613 hw-1-absolute min at (0,0) is 1 and absolute max at (0,4) and (4,4) is 17
hw 2-absolute minima at (5,-5/2) is -18.5 and absolute maxima at(0,-3) is 11
sir i have a doubt , in case 2 ( using boundary points) , we don't include boundary points as critical point's , but when ur listing all the points you included the boundary points at last for the square boundary question , why?
V nice👍👏
Thanks 😊
Sir in 2nd homework problem min value occur at -9.25 in two points (5,-2.5) and(4.5,-3).which point is considered as local minima
If that's what the answer is coming then local minima is occurring at 2 points.
One can have more than one points of local minima
Why didnt we use Hessian matrix to check if the points we got are abs max or abs min or saddle pt?
Good question..
It only tell us about local extremas reason being, when we prove this theorem involving hessian, we use Taylor's theorem and Taylor's theorem is local in nature ( not global)..
@@DrMathaholic Ooh ohkk...Thank you sir👍🏻
Sir, in first example at (1,0) value is -32 so it is absolute minima so it should be critical point. But when I used equation in online calculator it only gives (0.5,0.5) as maxima and (0,0) as saddle point. Can you please explain?
At 6:09 , We get (0.5,0.5) as critical point.
So at the end you take this point along with the corner points and interior points on the segments..
@@DrMathaholic So is (1,0) critical point?
@@atharvabhomle3191 no..it's not the interior point..
At the end we take
1.Critical points then
2. interior points from the segments and
3. the corner points
Then value of function at these points
@@DrMathaholic Sir but online calculator gives (0,0) as critical point too on wolframalpha
@@atharvabhomle3191 on wolfram did you give the region is square???