For any $2$ squares, the sum of whose area is $1$, a rectangle of area $A$ exists into which the squares can be placed without overlapping of interior points (Assume that the square are to be placed with their sides parallel to the sides of the rectangle). The least value of A can be expressed as $\frac{1+\sqrt\alpha}{\beta}$ where $\alpha$ and $\beta$ are natural numbers. Find $(\alpha+\beta)$.
For the least area, the squares should be stacked to the sides of the rectangle (i.e. without a gap between a side of the rectangle and that of squares). So, if the length of the side of one square is $x$ units and then the length of the rectangle is $x+y$ units. And along the width of the rectangle, another square is placed (sides of two squares overlapping each other) then the length of the side of this square is $\sqrt{1-x^2}$ (because sum of area of two squares is $1$). So, the width of the rectangle is $x+\sqrt{1-x^2}$.
To minimize area, we need to take derivative of $A=(x+y)(x+\sqrt{1-x^2})$
Not able to eliminate $y$.