Use Lagrange multiplier to find extrema of $f$ subject to the stated constraints.
$f(x,y,z,t)=xyzt$
$x-z=2$ and $y^2+t=4$
My attempt:
Let $g(x,z)=x-z-2=0$ and $h(y,t)=y^2+t-4=0$
$\nabla f=\lambda \nabla g+\mu \nabla h$
$(yzt)i+(xzt)j+(xyt)k+(xyz)l=\lambda (i-k)+\mu (2y j + l)$
Therefore,
$xyzt=\lambda x$
$xyzt=2\mu y^2$
$xyzt=-\lambda z$
$xyzt=\mu t$
$x-z-2=0$
$y^2+t-4=0$
Now, we wnat to solve this system, and by add the first four equations, we have
$\frac{7}{2} xyzt=\lambda (x-z)+\mu (y^2+t)$
$=2 \lambda +4\mu=2(\lambda +2\mu)$
$xyzt=\frac{4}{7} (\lambda+\mu)$
Then,
$\lambda x= \frac{4}{7} (\lambda+\mu)$
$\mu y^2= \frac{2}{7} (\lambda+\mu)$
$\lambda z= \frac{-4}{7} (\lambda+\mu)$
$\mu t=\frac{4}{7} (\lambda+\mu)$
Now what can I do? Can I divide these equations by $\lambda$ or $\mu$ to find $x,y,z$ and $t$? But I don’t know what the value of $\lambda$ and $\mu$ it may be will zero!
Please help :) thanks.
The stationary points can be found as follows:
We have the following first order conditions:
$$yzt = \lambda \tag{1}$$
$$xzt=2\mu y\tag{2}$$
$$xyt=-\lambda\tag{3}$$
$$xyz=\mu \tag{4}$$
$$x-z=2\tag{5}$$
$$y^2+t=4 \tag{6}$$
From $(1)$ and $(3)$,
$$yzt=-xyt \iff (x+z)yt=0$$ From $(2)$ and $(4)$, $$xzt=2xy^2z \iff xz(t-2y^2)=0$$
We consider the case where the objective function is non-zero.
$$x+z=0$$ and $$x-z=2$$
Then we have $(x,z)=(1,-1)$.
Also if $$t-2y^2=0$$
and $$y^2+t=4$$
Then we have $$2y^2=4-y^2 \iff y^2=\frac{4}{3}$$
$$(t,y)=\left( \frac83, \pm\frac{2}{\sqrt3}\right)$$
Remark: We still have to examine if the function can be unbounded.
If we let $x=2, z=1$, $t=4-y^2$, then the objective function becomes $2y(4-y^2)$, it is indeed unbounded.