Find minimum value of sum of areas of squares 
I tried using some trigonometry but couldn't arrive at any helpful inequality

From trivial applications of similitude we get $y=\dfrac{15}{7}-\dfrac{3 x}{4}$
The sum of the area of the two squares is
$f(x)=x^2+\left(\dfrac{15}{7}-\dfrac{3 x}{4}\right)^2$
Derivative is
$f'(x)=2x+2\left(-\dfrac{3}{4}\right)\left(\dfrac{15}{7}-\dfrac{3 x}{4}\right)$
$f'(x)=0$ if $x=\dfrac{36}{35}$ 
This value is a minimum because $f'(x)<0$ for $x<\dfrac{36}{35}$ and $f'(x)>0$ for $x>\dfrac{36}{35}$
When $x=\dfrac{36}{35}$ we have $y=\dfrac{48}{35}$
The minimum value of the sum of the areas is $x^2+y^2=\left(\dfrac{36}{35}\right)^2+\left(\dfrac{48}{35}\right)^2=\dfrac{144}{49}$
$x+y=193$
Hope this helps

