find max and min of the following function: $f(x, y, z) = yz+xz$ on a given set

Question

I need to find max and min (if they exist) of the following function: $f(x, y, z) = yz+xz$

On the set $C = \{(x, y, z) ∈ R^3: y^2+z^2 = 1, xz =3\}$

I have checked that the Lagrange theorem assumptions hold, so that I can find the critical points solving the following system

$z-\lambda_2 z=0$

$z-\lambda_1 2y=0$

$y+x-\lambda_1 2z=0$

$y^2+z^2-1=0=0$

$xz-3=0$

Then I found the critical points $x_1=(3/\sqrt2,1/\sqrt2,1/\sqrt2)$ and $x_2=(-3/\sqrt2,1/\sqrt2,1/\sqrt2)$. However the lagrange theorem only gives the necessary condition, how can I actually check that they are global max/min without using the bordered hessian method?

Answer 1

$$f(x,y,z) = yz+3\leq {y^2+z^2\over 2}+3 ={1\over 2} + 3=3,5$$

with equality iff $y=z = \pm\sqrt{1/2}$ and $x = 3/z$.

Answer 2

This is like the previous question you asked a day ago. You can actually parametrize $C$ by $(3/\sin\theta,\cos\theta,\sin\theta)$, as $\theta$ varies from $0$ to $2\pi$ (not including $0,\pi,2\pi$). Note that evaluating $f$ at such a point gives you $g(\theta) = \sin\theta\cos\theta + 3$, so even though $C$ is unbounded, you have a very nice continuous function on the closed interval $[0,2\pi]$.

Answer 3

Reduce the problem with the elimination of $z$,

$$f(x,y) = 3+3\frac yx,\>\>\>\>\>y^2+\frac 9{x^2} =1$$

Note $\left(y \pm \frac 3x\right)^2 =1\pm 6\frac yx \ge 0$, which leads to $-\frac16 \le \frac yx \le \frac16$. Thus,

$$\frac52 \le f(x,y) \le \frac72$$

Answer 4

$f(x,y,z)=yz+3$;

1) $2f(x,y,z)=2yz+6+(z^2+y^2-1);$

$2f(x,y,z)=(z+y)^2+5$;

$f_{\min}(x,y,z)= 2.5 $;

2)$2f(x,y,z)=2yz +6 -(z^2+y^2-1)$;

$2f(x,y,z)=-(z-y)^2 +7$;

$f_{\max}(x,y,z)=3.5$;