
A2 Find the maximum value of $z+x$, if $(x, y, z, t)$ satisfies the conditions:

$$
\left\{\begin{array}{l}
x^{2}+y^{2}=4 \\
z^{2}+t^{2}=9 \\
x t+y z \geq 6
\end{array}\right.
$$


Solution I: From the conditions we have

$$
36=\left(x^{2}+y^{2}\right)\left(z^{2}+t^{2}\right)=(x t+y z)^{2}+(x z-y t)^{2} \geq 36+(x z-y t)^{2}
$$

and this implies $x z-y t=0$.

Now it is clear that

$$
x^{2}+z^{2}+y^{2}+t^{2}=(x+z)^{2}+(y-t)^{2}=13
$$

and the maximum value of $z+x$ is $\sqrt{13}$. It is achieved for $x=\frac{4}{\sqrt{13}}, y=t=\frac{6}{\sqrt{13}}$ and $z=\frac{9}{\sqrt{13}}$.
