A geometry problem by Iftekher Ahmed

98 is the MINIMUM value of the areas of those two squares. Even if we restricted the sides to integer values, the maximum is 49^2+1=2402. Otherwise, the maximum can be infinite.

We have $xy=49\Rightarrow y=\frac{49}{x}$ . We wish to maximize $x^2+y^2 = x^2 + \frac{49^2}{x^2}$ . Taking the derivative and setting it equal to zero yields $\begin{aligned} 2x - \frac{2\times 49^2}{x^3} &= 0 \\ 2x &= \frac{2\times 49^2}{x^3} \\ x^4 &= 49^2 \\ x &= 7 && \color{#3D99F6}\text{Since }x>0 \end{aligned}$ This is indeed a minimum (take the second derivative at $x=7$ : $2+\frac{6\times 49^2}{7^4}>0$ ). Thus, $x=y=7$ and the area of the two squares is $7^2 + 7^2 = \boxed{98}$ .