The equations that must be solved for maximum or minimum values of a differentiable function w=f(x, y,z) subject to two constraints g(x, y,z)=0 and h(x, y,z)=0, where g and h are also differentiable, are gradientf=lambdagradientg+mugradien th, g(x, y,z)=0, and h(x, y,z)=0, where lambda and mu (the lagrange multipliers) are real numbers. use this result to find the maximum and minimum values of f(x, y,z)=xsquared+ysquared+zsquared on the intersection between the cone zsquared=4xsquared+4ysquared and the plane 2x+4z=2.

The Lagrangian is

with partial derivatives (set equal to 0)

Case 1: If , then

Then

So we have two critical points, and

Case 2: If , then in the first equation we get

and from the third equation,

Then

but there are no real solutions for , so this case yields no additional critical points.

So at the two critical points we've found, we get extreme values of

(min)

and

(max)

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d

$6.42

step-by-step explanation: