Question

Find the maximum and minimum values of f(x,y)=x+2y on the disk x^2+y^2 less or equal to one.

Answer 1

find double derivative.. and equate them to zero!

Answer 2

oh wait.. there is more to it than i thought.. i just read half the question.. i dunno.. lets wait for others to answer.. sorryr :D

Answer 3

I have found the f_1(x,y) = 1 and f_2(x,y)=2 how can these be equated to zero?

Answer 4

It's okay :) I hope I get help, this one is hard

Answer 5

I think that means the extrema will be in the boundary

Answer 6

TuringTest, yes that is true. I am aware of that part and I suppose that x=cost and y=sint......but I have no Idea what to do next

Answer 7

\[x^2+y^2\le1\]\[x=\sqrt{1-y^2}\]plug that into f(x,y)=x+2y and call this new function g(y)\[g(y)=\sqrt{1-y^2}+y^2\]find the extrema of this function and check them against the value of f(x,y) at the endpoints

Answer 8

So I have to put g(y) = 0 ?

Answer 9

typo above

Answer 10

\[g(y)=\sqrt{1-y^2}+2y\]we need g'(y)=0 just like calc 1

Answer 11

I hope you're not expecting pretty numbers because I'm not getting them

Answer 12

The facit says sqrt(5) and -sqrt(5)

Answer 13

I get +-sqrt(3)/2

Answer 14

sqrt5 is the final answer?

Answer 15

+/-

Answer 16

Yes it is

Answer 17

oh wait I made a silly mistake on my paper
I think we'll get it, hold on...

Answer 18

I attached a file with the solution, but I dont quite understand it

Answer 19

I dont get it from the part where they start with the : f (x, y) = f (cos t, sin t) = cos t + 2 sin t = g(t)

Answer 20

\[f(x,y)=x+2y\]on the disk\[x^2+y^2\le1\]at the boundary\[x=\pm\sqrt{1-y^2}\]writing the funciton in terms of y we get\[g(y)=(1-y^2)^{1/2}+2y\]finding its extrema\[g'(y)=-y(1-y^2)^{-1/2}+2=0\]\[4(1-y^2)=y^2\]\[y=\pm\frac{2\sqrt5}5\]we also need the corresponding x values, which we can find from our boundary expression for x above\[x=\pm\frac{\sqrt5}5\]This is going to give the right answer, but obviously does not involve parameterization
shall I continue, or...?
there are also a lot of things we failed to check like the endpoints of g(y)

Answer 21

Im not quite forllowing....when you write "writing the funciton in terms of y we get" I dont know what is done....

Answer 22

I just plugged in the expression for x at the boundary into f(x,y) so that it's all in terms of y

Answer 23

\[f(x,y)=x+2y\]\[x=\sqrt{1-y^2}\]plug the expression for x into f and rename it g(y)

Answer 24

Did you see the file I attached? It seems like they solved it differently which confuses me

Answer 25

it's actually the exact same thing
actually theirs is a bit better in my opinion

Answer 26

Thank you very much for this and for your time. I appreciate it :-) This is my first time here and I already have a very good impression.

Answer 27

instead of minimizing it in terms of y they did it in terms of a parameter t by using\[x=\cos t\]\[y=\sin t\]so f(x,y) becomes\[g(t)=x(t)+2y(t)=\cos t+2\sin t\]when we solve g'(t)=0 and check what x and y turn out to be we get the exact same as I gave above

I'm glad you were able to get help
welcome to open study, hope to see you around :D

Answer 28

:D