Question

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+mugradient​h, ​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.

Answer 1

The Lagrangian is

$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(4x^2+4y^2-z^2)+\mu(2x+4z-2)$

with partial derivatives (set equal to 0)

$L_x=2x+8\lambda x+2\mu=0\implies x(1+4\lambda)+\mu=0$

$L_y=2y+8\lambda y=0\implies y(1+4\lambda)=0$

$L_z=2z-2\lambda z+4\mu=0\implies z(1-\lambda)+2\mu=0$

$L_\lambda=4x^2+4y^2-z^2=0$

$L_\mu=2x+4z-2=0\implies x+2z=1$

Case 1: If $y=0$, then

$4x^2-z^2=0\implies4x^2=z^2\implies2|x|=|z|$

Then

$x+2z=1\implies x=1-2z\implies2|1-2z|=|z|\implies z=\dfrac25\text{ or }z=\dfrac23$

$\implies x=\dfrac15\text{ or }x=-\dfrac13$

So we have two critical points, $\left(\dfrac15,0,\dfrac25\right)$ and $\left(-\dfrac13,0,\dfrac23\right)$

Case 2: If $\lambda=-\dfrac14$, then in the first equation we get

$x(1+4\lambda)+\mu=\mu=0$

and from the third equation,

$z(1-\lambda)+2\mu=\dfrac54z=0\implies z=0$

Then

$x+2z=1\implies x=1$

$4x^2+4y^2-z^2=0\implies1+y^2=0$

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

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

$f\left(\dfrac15,0,\dfrac25\right)=\dfrac15$ (min)

and

$f\left(-\dfrac13,0,\dfrac23\right)=\dfrac59$ (max)