Find the maximum and minimum values of f(x,y)=xy on the ellipse 6x^2+y^2=3.?

Question

Find the maximum and minimum values of f(x,y)=xy on the ellipse 6x^2+y^2=3.

Answer 1

Using lagrangian multipliers
F(x,y,k) = xy +k(6x^2+y^2-3)
Fx=y+k*12x=0
Fy= x+2ky=0
y=-12kx and x-24k^2x=0
x=0 and 1-24k^2=0
x=0 gives y=0 which is not on the ellipse
so k=+-sqrt(24)/24
for k=sqrt(24)/24 y=-sqrt(24)/2x
and
6x^2+24/4 x^2=3 so x=+-1/2 and y = -+sqrt(24)/4
for k=-sqrt(24)/24 we get x=+-1/2 and y=+-sqrt(24)/4

so(1/2,-sqrt(24)/4) ,(-1/2 ,+sqrt(24)/4) ,(1/2,+sqrt(24)/4) and
(-1/2,-sqrt(24)/4)

Answer 2

use the method that you were taught
i think they aren called lagrangian constraints, smt like that