Find local minimizers for x21 x22 subject to x21 2x1x2 x2

Find local minimizers for x^2_1 + x^2_2 subject to x^2_1 + 2x_1x_2 + x^2_2 = 1, x^2_1 - x^2_2 lessthanorequalto 0.

Let x₁=x and x₂=y.

Let f:=x²+y²

g:= x²+2xy+y²-1

h:=x²-y²

The equations are: (f_x , f_y)=K(g_x, g_y)+L(h_x ,h_y) , g=0, and h=0.

2x=K(2x+2y)+L(2x) -----(1)

2y=K(2x+2y)+L(-2y) -----(2)

x²+2xy+y²-1=0 ------(3)

x²-y²=0 -------(4)

From (3) x²+2xy+y²-1=0 i.e. (x+y)² =1 i.e (x+y)=+1 or x+y=-1.------(5)

From (4) x²-y²=0 implies (x-y)(x+y)=0 implies x-y=0....(Since from (5) x+y is non zero).So, (x-y)=0---(6)

Now, subtracting eq. (2) from eq.( 1) we get,

2(x-y)=0+2L(x+y).

(Since x-y=0) we get 2L(x+y)=0

which implies L=0.(since x+y is non zero). So, L=0-----(7)

Now putting L in(1) and (2) we get,

2x=2K(x+y) ----(8)

2y=2K(x+y) -------(9)

Now adding (8) and (9) we get

2(x+y)=4K(x+y) (since (x+y) is nonzero)

2=4K which implies K=1/2. K=1/2 ----------(10)

case (i)... (x+y)=1

Substitute K in (8) and (9) we get x=1/2 and y=1/2.

case (ii)... (x+y)=-1

Substitute K in (8) and (9) we get x=-1/2 and y=-1/2.

These two points satisfy the conditions x²+2xy+y² =1 and x²-y² <=0.

The value of f at (1/2,1/2) is 1/2 and at (-1/2,-1/2) is 1/2.

Hence both points (1/2,1/2) and (-1/2, -1/2) are minimizers of f and the minimium value is 1/2.