1. **State the problem:** We want to find the maximum value of $ (2x + 3y)^2 $ given the constraint $ x^2 + y^2 = 1 $ where $x$ and $y$ are real numbers. 2. **Recall the constraint and expression:** The constraint $ x^2 + y^2 = 1 $ represents a circle of radius 1 centered at the origin. 3. **Use the Cauchy-Schwarz inequality:** For vectors $\mathbf{u} = (2,3)$ and $\mathbf{v} = (x,y)$, the inequality states: $$ |\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\| \|\mathbf{v}\| $$ where $\mathbf{u} \cdot \mathbf{v} = 2x + 3y$. 4. **Calculate the norms:** $$ \|\mathbf{u}\| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} $$ $$ \|\mathbf{v}\| = \sqrt{x^2 + y^2} = \sqrt{1} = 1 $$ 5. **Apply the inequality:** $$ |2x + 3y| \leq \sqrt{13} \times 1 = \sqrt{13} $$ 6. **Square both sides to find the maximum of the square:** $$ (2x + 3y)^2 \leq (\sqrt{13})^2 = 13 $$ 7. **Conclusion:** The maximum value of $ (2x + 3y)^2 $ subject to $ x^2 + y^2 = 1 $ is $\boxed{13}$. This maximum is achieved when $ (x,y) $ is in the direction of the vector $ (2,3) $ normalized, i.e., $$ x = \frac{2}{\sqrt{13}}, \quad y = \frac{3}{\sqrt{13}} $$