If xy = 10, then minimum value of $12x^2+13y^2$ is equal to:

$40\sqrt{39}$

$xy=10⇒y=\frac{10}{x}⇒\frac{dy}{dx}=\frac{-10}{x^2}$

$f=12x^2+13y^2$

$\frac{df}{dx}=24x+\frac{26×10}{x}×(\frac{-10}{x^2})=0$

$⇒24x=\frac{26×10×10}{x^3}$

$x^4=\frac{5^2×13}{3}⇒x^2=5×\sqrt{\frac{13}{3}}$ and $y^2=\frac{100}{x^2}=20\sqrt{\frac{3}{13}}$

$f=60\sqrt{\frac{13}{3}}+13×20\sqrt{\frac{3}{13}}$

$⇒f=20\sqrt{39}+20\sqrt{39}=40\sqrt{39}$

To check it is minimum value at x = 1  $f = 12 +1300>40\sqrt{39}$