2024 AMC 10B Problem 13

$1183 = 7 \cdot 13^2,$ so $\sqrt{1183} = 13\sqrt{7}$

Then $\sqrt{x} = a\sqrt{7}$ and $\sqrt{y} = b\sqrt{7}$ with $a + b = 13;$ minimize $x + y = 7(a^2 + b^2)$

Since $1183 = 7 \cdot 13^2,$ we have $\sqrt{1183} = 13\sqrt7.$ Squaring the given equation shows that $\sqrt{xy}$ is rational. Thus $x$ and $y$ have the same squarefree part: write $x=da^2$ and $y=db^2,$ where $d$ is squarefree and $a,b$ are positive integers. Then $(a+b)\sqrt d=13\sqrt7,$ so $d=7$ and $a+b=13.$ Therefore $x+y=7(a^2+b^2),$ which is smallest when $a$ and $b$ are as close as possible. Taking $a=6$ and $b=7$ gives $x+y=7(36+49)=595.$ Thus, B is the correct answer.