Question

Find the extreme values of the function f(x, y) = 4x2 + 6y2 on the circle x2 + y2 = 1.

Answer 1

Answer with Step-by-step explanation:

We are given that

$f(x,y)=4x^2+6y^2$

Let g(x,y)=$x^2+y^2=1$

We have to find the extreme values  of the given function

$\nabla f(x,y)$

$\nabla g(x,y)=$

Using Lagrange multipliers

$\nabla f(x,y)=\lambda \nabla g(x,y)$

$f_x=\lambda g_x$

$8x=\lambda 2x$

Possible value x=0 or $\lambda=4$

If x=0 then substitute the value in g(x,y)

Then, we get $y=\pm 1$

$f_y=\lambda g_y$

$12y=\lambda 2y$

If $\lambda=4$ and substitute in the equation

Then , we  get possible value of y=0

When y=0 substitute in g(x,y) then we get

$x=\pm 1$

Hence, function has possible  extreme values at points (0,1),(0,-1), (1,0) and (-1,0).

$f(0,1)=6$

$f(0,-1)=6$

$f(1,0)=4$

$f(-1,0)=4$

Therefore, the maximum value of f  on the circle$x^2+y^2=1$  is $f(0,\pm1)=6$ and minimum value of $f(\pm1,0)=4$