Find minimum value of sum of areas of squares 
I tried using some trigonometry but couldn't arrive at any helpful inequality
 A: Let $x$ be the side of square $P$ and $y$ be the side of square $Q$ then
$$\frac{x}{\tan(A)}+x+y+\frac{y}{\tan(B)}=|AB|.$$
that is
$$\frac{7x}{4}+\frac{7y}{3}=5\implies y=-\frac{3x}{4}+\frac{15}{7}$$
Now find the minimum of the quadratic function
$$f(x)=x^2+y^2=x^2+\left(-\frac{3x}{4}+\frac{15}{7}\right)^2$$
over $[0,|AC|\cos(A)]=[0,9/5]$.
Can you take it from here?
A: 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

A: like Robert Z suggested it is $$x+y+\frac{3}{4}x+\frac{4}{3}y=5$$ or $$\frac{7}{4}x+\frac{7}{3}y=5$$ from here we get
$$y=\frac{15}{7}-\frac{21}{28}x$$ plugging this in $$x^2+y^2$$ we have 
$$f(x)={\frac {25\,{x}^{2}}{16}}-{\frac {45\,x}{14}}+{\frac{225}{49}}$$
A: 
Split the side of length $5$ into $4$ parts and give them lengths $3w$,$x$,$y$ and $4z$ as indicated in the diagram above. By similar triangles $x=4w$ and $y= 3z$. Now consider the side of length $5$, we have $7(w+z)=5$. We wish to minimise $x^2+y^2$ subject to the aforementioned constraint; This is easily achieved by Lagrange multipliers
\begin{eqnarray*}
A= 16w^2+9z^2 + \lambda(7(w+z)-5)
\end{eqnarray*}
partially differentiate and do a little algebra and we have $w=\frac{9}{35},z=\frac{16}{35}$. So $x=\frac{36}{35}$ and $y=\frac{48}{35}$. I will leave you to figure out what your $r$ and $s$ values are.
