Use the method of Lagrange multipliers to find the maximum and minimum values of the function f(x;y) = xyz subject to the constraint x^2 +
2y^2 + 3z^2 = 6.

Question

Use the method of Lagrange multipliers to find the maximum and minimum values of the function f(x;y) = xyz subject to the constraint x^2 +
2y^2 + 3z^2 = 6.

anonymous · Answer

yz=2xλ
xz=4yλ
xy=6zλ
x^2 + 2y^2 + 3z^2 = 6.

yz/2x=λ
xz/4y=λ
xy/6z=λ

yz/2x=xz/4y
2y^2=x^2
xz/4y=xy/6z
3z^2=2y^2

2y^2+2y^2+2y^2=6
6y^2=6
y=+/-1

y=1: z=+/-sqrt(2/3); x=+/-sqrt(2)
y=-1:z=+/-sqrt(2/3); x=+/-sqrt(2)

critical points:
(1,sqrt(2/3),sqrt(2))--->a
(-1,-sqrt(2/3),-sqrt(2)--->b

f(a)=sqrt(2/3)*sqrt(2)..............maximum
f(b)=-sqrt(2/3)*sqrt(2)............minimum

i apologize if i made any mistakes i haven't done these since last term