Question

Find the extreme values of subject to constraints and .

f(x, y, z) y+ z = 2

Answer 1

given,

f(x, y, z) y+ z

let,

g(ar, y, ) + y - 2 CN

h(x, y, z x y 1

by Lagrange multipliers,

Fx, y, 2, A, u f(x,y,z) - Ag(x, y, 2)-uh(a, y, z)

partially differentiating with respect to the variables and equating to zero,

1,1,1 >

2.x, 2y, 0 > Vg(r, y, z)

Vh(x, y, < 1,1,0>

by Lagrange multipliers,

2Ar

2Ay + u

solving,

there fore, x=y

substituting in g(x,y,z)

$x=\pm 1=y$

clearly $\lambda =0$,

$\mu =1$

so the points are,

$(x,y,z)=(\pm 1,\pm 1,1)$

one extreme value is,

f(x, y, =f(1,1,1)

$f(x,y,z)=3$

and other extreme value is

f(,y, =f(-1, -1, 1)

$f(x,y,z)=-1$

hence the answer.