Find all two-digit numbers such that the sum of the digits constituting the number is not less than $7$; the sum of the squares of the digits is not greater than $30$; the number consisting of the same digits written in the reverse order is not larger than half the given number.

Let $n(a,b)$ be two digit number

Given that, $a+b\ge 7\to \left(1\right)$

$a^2+b^2\le 30\to \left(2\right)$

$n(b,a)\le \frac{n(a,b)}{2}$

$10b+a\le \frac{10a+b}{2}$

$19b\le 8a$

$\Rightarrow 8a\ge 19b\to \left(3\right)$

$8a^2\ge 19ab$

$91)\Rightarrow (a+b{)}^2\ge 49$

$a^2+b^2+2ab\ge 49$

$30+2ab\ge 49$

$2ab\ge 19$

$\frac{16a^2}{19}=19(\because \frac{8a^2}{19}\ge ab)$

$a\ge \frac{19}{4}\Rightarrow a\ge 4-25$

Minimum value of $a=5$

$30\ge a^2+b^2$

$30\ge \frac{{19}^2}{4^2}+b^2(\because a\ge \frac{19}{4})$

$b=\le \frac{\sqrt{119}}{4}$

Maximum value of $b=2$

If $b=0,a\ge 7$ but $a^2+b^2>30$

$\therefore b\ne 0$

If $b=1,a\ge 6$ but $a^2+b^2>30$

$\therefore b\ne 1$

$\therefore a=5,b=2$

Only one number is possible

$\therefore 52$ is required number.