Answer with Step-by-step explanation:
We are given that
[tex]f(x,y)=4x^2+6y^2[/tex]
Let g(x,y)=[tex]x^2+y^2=1[/tex]
We have to find the extreme values of the given function
[tex]\nabla f(x,y)<8x,12y>[/tex]
[tex]\nabla g(x,y)=<2x,2y>[/tex]
Using Lagrange multipliers
[tex]\nabla f(x,y)=\lambda \nabla g(x,y)[/tex]
[tex]f_x=\lambda g_x[/tex]
[tex]8x=\lambda 2x[/tex]
Possible value x=0 or [tex]\lambda=4[/tex]
If x=0 then substitute the value in g(x,y)
Then, we get [tex]y=\pm 1[/tex]
[tex]f_y=\lambda g_y[/tex]
[tex]12y=\lambda 2y[/tex]
If [tex]\lambda=4[/tex] and substitute in the equation
Then , we get possible value of y=0
When y=0 substitute in g(x,y) then we get
[tex]x=\pm 1[/tex]
Hence, function has possible extreme values at points (0,1),(0,-1), (1,0) and (-1,0).
[tex]f(0,1)=6[/tex]
[tex]f(0,-1)=6[/tex]
[tex]f(1,0)=4[/tex]
[tex]f(-1,0)=4[/tex]
Therefore, the maximum value of f on the circle[tex]x^2+y^2=1[/tex] is [tex]f(0,\pm1)=6[/tex] and minimum value of [tex]f(\pm1,0)=4[/tex]
