Absolute extrema - Conceptual

KindofSlow

Junior Member
Joined
Mar 5, 2010
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90
Hello,
Problem is find abs min & abs max of f(x,y)=2(x^2)-(y^2)+6y on disk x^2+y^2<or=16
Step 1 - Finding internal critical point (0,3) and f(0,3)=9 is easy - did this part with no problem.

Then must find critical point on boundary.
x^2=16-y^2 so g(y)=-3(y^2)+6y+32

I understand the mechanics but I don't understand conceptually what g(y) is.
I thought it would be the intersection of f(x,y) and outside of the disk (looking down would be a circle and side view would follow contours of f(x,y).
But g(y) is obviously parabolic so I thought wrong.
g(y) does not equal x, g(y) does not equal z, and g(y) does not equal f(x,y)
Can you give me any insight regarding either g(y) the curve and/or g(y) for a specific y?
Hope this makes sense.
Thank you
 
Given the relationship, x^2 + y^2 = 16, g(y) IS f(x,y). Just for that cyllinder. Inside or outside the cyllinder, it is not so.

g(0) = 32

x^2 = 16 - 0^2 = 16, x = 4 or x = -4

f(4,0) = 2(16) = 32
f(-4,0) = 2(16) = 32
 
I think I am slowly getting all the moving parts straight in my head.

Thank you TK
 
Observe symmetries as well. Substitute x with -x and see what changes? Symmetries!

Complete the square on the y parts and discover more symmetries.

Humorous note -- Don't forget your algebra! There's a reason why we played with all that stuff back in high school. :D
 
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