Question:
Find the maximum and minimum of the function $f(x,y,x)=3x^2 + 2xz + 5y^2 +3z^2$ on the sphere $x^2 + y^2 + z^2 = 25$
Attempt:
This seems that it should be solvable using the Lagrange multiplier method so my attempt is focused on that.
$f(x,y,x)=3x^2 + 2xz + 5y^2 +3z^2$
$g(x,y,z)=x^2+y^2+z^2-25$
$\nabla f = \lambda \nabla g$
$\nabla f = (6x+2z,10y,2x+6z)$
$\nabla g = (2x, 2y, 2z) $
therefore we can make a system of equations:
$6x+2z=\lambda (2x)$ [1]
$10y=\lambda (2y)$ [2]
$2x+6z= \lambda (2z)$ [3]
$x^2 + y^2 + z^2 = 25$ [4]
From equation [2] it is clear to see that $\lambda = 5$
Subbing $\lambda$ into eqn [1] and [3] yeilds
$6x + 2z = 10x$ [1]
$2x + 6z = 10z$ [3]
It is also clear to see that $x=z$
This however creates a contradiction because if $x$ and $z$ are equal $\lambda$ cannot be equal to $5$ it must be equal to $4$ as shown below
From eqn [1] subbing x=z yeilds
$6x + 2x = \lambda (2x)$
$\lambda = 4$
I am unsure what I am doing wrong that is resulting in this system of equations to be unsolvable I know for a fact this question has a solution and I don't see why using Lagrange multipliers wouldn't work.
Any help would be appreciated Thank you.
Best Answer
Your mistakes:
From equation [2] you get $ \lambda=5$ or $y=0$ !
With $\lambda=5$ we get $x=z$ and then $x=z=0$, hence $y= \pm 5$.
You have also to investigate the case $y=0$ (see 1.).