Evaluate $|z|$'s maximum and minimum value under condition:$ \begin{cases} x^2+9y^2-2z^2=0 \\ x+3y+3z=5\\ \end{cases} $

Question

$$ \begin{cases} x^2+9y^2-2z^2=0,(1) \\ x+3y+3z=5, (2)\\ \end{cases} $$

I try to solve maximum and minimum value of $z^2$, maybe Lagrange Multiplier is useful but I can't formulate the equation properly.

From $(1)$ and $(2)$ I can get $$g(x,y)=7x^2+20x+63y^2-12xy+60y=0$$ So my Lagrange multiplier is $$F(x,y,\lambda)=\frac{x^2+9y^2}{2}+\lambda g(x,y)$$ But I can't solve the following equation $$ \begin{cases} F'_x=0 \\ F'_y=0\\ F'_{\lambda}=0\\ \end{cases} $$ So I think there must be something wrong with my method. Any ideas are welcome.

Answer 1

By Cauchy-Schwarz, we have $$4z^2=(1+1)(x^2+9y^2) \ge (x+3y)^2 = (5-3z)^2$$ giving $$(z-1)(z-5) \le 0$$ which implies $$1 \le z \le 5$$ $$1 \le z^2 \le 25$$

The equality holds at $(x,y,z)=(1,\frac{1}{3},1)$ and $(x,y,z)=(-5,-\frac{5}{3},5)$

Answer 2

One thing we need to realize is $z^2$ is maximized when $h(x,y,z)=z$ is either maximized or minimized.

By taking the partial derivatives with respect to $x,y,z$, you have

$$2x\lambda+u=0\implies x=-{u\over2\lambda}$$ $$18y\lambda+3u=0\implies y=-{u\over6\lambda}$$ $$-4z\lambda+3u+v=0\implies z={3u+v\over4\lambda}$$

Now sub in to $x+3y+3z=5$ and $x^2+9y^2-2z^2=0$, let ${u\over\lambda}=a$ and ${v\over\lambda}=b$ you get

$$-{1\over2}a-{1\over2}a+{9\over4}a+{3\over4}b=5\implies 5a+3b=20$$

$${1\over4}a^2+{1\over4}a^2-{1\over8}(3a+b)^2=0\implies 4a^2=(3a+b)^2$$

(1) $2a=3a+b$, we have $a=10,b=-10\implies z=5$

(2) $2a=-3a-b$, we have $a=-2,b=10\implies z=1$

Hence $z$ is maximized at $5$ and minimized at $1$ and $z^2$ is maximized at $25$.

Answer 3

Hint: the linear system $$\eqalign{ \lambda(- 12 y + 2 x + 20) = F_x' &= 0 \cr \lambda(126 y - 12 x + 60) + 9 =F_y'&= 0\cr }$$ has a unique solution for $12\lambda^2 + 16\lambda + 1\ne 0$ (why?).

What happens when $12\lambda^2 + 16\lambda + 1 = 0$?