$x$,$y$ and $z$ are real numbers.
We have two equations which are given by the following:
$x+y+z=5$
$xy+yz+zx=3$
By using the two equations above, find the maximum value of $z$.
How can we solve that question by using derivatives or without using
derivatives? Can we find an answer without using Lagrange multiplier? Because I
have to tell the solution to my student in a high school:)
You can formalize this by using the target function $f(x,y,z)=z$ and equality constraints $g(x,y,z) = x+y+z-5=0$ and $h(x,y,z) = xy+yz+zx-3=0$. You can use the method of Lagrange-multiplier. Namely solving the (unconstraint) optimization-problem $$\mathcal{L}(x,y,z,\lambda_1, \lambda_2) = f(x,y,z) + \lambda_1 \cdot g(x,y,z) + \lambda_2\cdot h(x,y,z) \\ =z + \lambda_1 \cdot (x+y+z-5) + \lambda_2 \cdot (xy +yz+zx-3)$$ You know have two more variables but are unconstrained. Solve this by finding all fives partial derivatives to zero and you find your critical point.
The first equation is that of a plane, and the second one that of a quadric. Their intersection must be a conic, and because of the symmetry $x\leftrightarrow y\leftrightarrow z$ it must be a circle, centered at $C=\left({5\over3},{5\over3},{5\over3}\right)$.
The point of that circle with the largest $z$ must be of the form $P=\left({5\over3}-d,{5\over3}-d,{5\over3}+2d\right)$: plugging that into the second equation gives $d={4\over3}$ and consequently $z={13\over3}$.
Notice that,
$$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=25$$
So,
$$x^2+y^2+z^2+6=25$$
$$z^2=19-x^2-y^2=19-r^2$$
From the second equation we have,
$$xy+z(x+y)=3$$
From the first,
$$z=5-x-y$$
So,
$$xy+(5-x-y)(x+y)=3$$
This is an ellipse, and when we switch $x$ with $y$ and $y$ with $x$ we end up with the same equation so it is symmetric about $x=y$. Hence the minimum of $r^2$ must occur when $x=y=\frac{1}{3}$ (or $=3$ which doesn't work), since no two points on an ellipse can both be the closest to a point outside an ellipse. That implies $z=\sqrt{19-2(\frac{1}{3})^2}=\frac{13}{3}$ is maximum.
Substituting $y=5-x-z$ from the first equation into the second gives:
$$ x^2+(z-5)x + z^2 - 5z + 3 =0 $$
Considering this as a quadratic in $x\,$, for it to have real roots, its discriminant must be non-negative:
$$ 0 \le \Delta = (z-5)^2 - 4(z^2-5z+3) = -3z^2+10z+13=-(z+1)(3z-13) $$
Therefore $z \in \left[-1, \cfrac{13}{3}\right]\,$, so the maximum value of $z$ is $\cfrac{13}{3}\,$.