Question

Find the absolute maximum and minimum of f(x; y) = 2x^2- y^2 + 6y on the disk x^ 2 + y^ 2 <= 16.

Answer 1

@wikiemol @RH

Answer 2

@mathgeek!

Answer 3

@iforgot

Answer 4

@dmezzullo

Answer 5

Hint:
it's a continuous function on closed domain, so it obtains it's máximum on the boundaries

Answer 6

so all i need to do is to find the max and min of f(x,Y)

Answer 7

you know, that the max, min are obtained at x^ 2 + y^ 2 = 16. Notice the equality

Answer 8

so you can express y as function of x, and viceversa

Answer 9

for example: \(y=\pm\sqrt{16-x^2}\)

Answer 10

plug this into the equation of f(x,y) and just find where the derivative \(f_x=0\) .

Answer 11

so \(f(x,y) = f(x,y(x))=2x^2 -19+x^2 \pm6\sqrt{16-x^2}\)

Answer 12

can i make a trig subsitution

Answer 13

sry, where there is 19 should be 16

Answer 14

ok thank

Answer 15

can i let x= 4cos(u)

Answer 16

yes, that's can be too

Answer 17

\(32cos^2u-16sin^2u +24sinu\)

Answer 18

huh?

Answer 19

i have \[12\cos ^{2}u +24\sin u-16\]

Answer 20

\[12\sin ^{2}u-24\sin u-4\]

Answer 21

\(2(16cos^2u)-16sin^2u+24sinu=32cos^2u-16sin^2u+24sinu\)

Answer 22

which can be writen as:
\(32-32sin^2u-16sin^2u+24sinu=-48sin^2u+24sinu+32\)

Answer 23

i cant see what you did to get this line 32cos2u−16sin2u+24sinu

Answer 24

just substitute x=4cosu, y=4sinu
into the f(x; y) = 2x^2- y^2 + 6y

Answer 25

ok

Answer 26

now find the derivative of this ,set it to 0 and find values of u that make that happen

Answer 27

i got pi/2 and \[\sin ^{-1}(1/4)\]

Answer 28

thats it