Given \(a^2 + b^2 = 1\), the maximum of \(ab\) is \(\frac{1}{2}\) when \(a = b = \frac{\sqrt{2}}{2}\). Then:
["### Given (a^2 + b^2 = 1), the Maximum of (ab) is (\frac{1}{2}) When (a = b = \frac{\sqrt{2}}{2})", "The equation (a^2 + b^2 = 1) describes a point ((a, b)) lying on the unit circle. When analyzing the product (ab) under this constraint, one discovers a captivating mathematical truth: the maximum value of (ab) is (\frac{1}{2}), achieved when (a = b = \frac{\sqrt{2}}{2}).", "#### Understanding the Constraint and Objective", "We are given the condition:\n[\na^2 + b^2 = 1\n]\nand we aim to maximize the expression:\n[\nab\n]\nwith (a) and (b) real numbers.", "This constraint limits ((a, b)) to the unit circle in the first quadrant (assuming (a, b \geq 0)), where both values range between (-1) and (1), but the product (ab) is non-negative only when (a) and (b) have the same sign. For maximum positive (ab), both should be non-negative.", "#### Using Algebra to Maximize (ab)", "From algebraic identities, recall that:\n[\n(a + b)^2 = a^2 + 2ab + b^2\n]\nUsing the constraint (a^2 + b^2 = 1), substitute:\n[\n(a + b)^2 = 1 + 2ab \quad \Rightarrow \quad 2ab = (a + b)^2 - 1\n]\nThus:\n[\nab = \frac{(a + b)^2 - 1}{2}\n]\nSince ((a + b)^2 \geq 0), the maximum occurs when ((a + b)^2) is maximized under (a^2 + b^2 = 1).", "By the Cauchy-Schwarz inequality, or by parametrization, the maximum (a + b) subject to (a^2 + b^2 = 1) is (\sqrt{2}), achieved when (a = b). Setting (a = b), substitute into the constraint:\n[\na^2 + a^2 = 1 \Rightarrow 2a^2 = 1 \Rightarrow a^2 = \frac{1}{2} \Rightarrow a = \frac{\sqrt{2}}{2}\n]\nSo:\n[\na = b = \frac{\sqrt{2}}{2}\n]\nNow compute:\n[\nab = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{4} = \frac{1}{2}\n]", "#### Verification via Trigonometric Substitution", "Alternatively, parametrize using trigonometric identities:\nLet (a = \cos\ heta), (b = \sin\ heta) (since (a^2 + b^2 = 1)) with (\ heta \in [0, \frac{\pi}{2}]). Then:\n[\nab = \cos\ heta \sin\ heta = \frac{1}{2} \sin 2\ heta\n]\nThe maximum value of (\sin 2\ heta) is (1), so:\n[\n\max(ab) = \frac{1}{2}\n]\nThis maximum occurs when (\sin 2\ heta = 1 \Rightarrow 2\ heta = \frac{\pi}{2} \Rightarrow \ heta = \frac{\pi}{4}), giving:\n[\na = \cos\frac{\pi}{4} = b = \frac{\sqrt{2}}{2}\n]", "#### Why Is (\frac{1}{2}) the Maximum?", "At (a = b = \frac{\sqrt{2}}{2}), the constraint (a^2 + b^2 = \frac{1}{2} + \frac{1}{2} = 1) holds perfectly. The symmetry of the circle ensures balanced values, and the product (ab) reaches its peak due to the quadratic nature of the expression and the symmetry imposed by equality in algebraic identities.", "Any deviation from (a = b) breaks the balance and results in a smaller product—this can be verified numerically or by optimizing using Lagrange multipliers.", "#### Conclusion", "The equation (a^2 + b^2 = 1) constrains ((a, b)) to the unit circle, and through algebraic or geometric analysis, the maximum of (ab) is clearly:\n[\n\max ab = \frac{1}{2}\n]\nachieved uniquely when:\n[\na = b = \frac{\sqrt{2}}{2}\n]\nThis elegant result showcases how symmetry and mathematical tools together reveal optimal configurations—essential in optimization, physics, and engineering applications.", "---", "Keywords: (a^2 + b^2 = 1), maximize (ab), maximum value (\frac{1}{2}), (a = b = \frac{\sqrt{2}}{2}), trigonometric substitution, geometry optimization.\nMeta Description: Explore the maximum of (ab) given (a^2 + b^2 = 1)—finding (\frac{1}{2}) at (a = b = \frac{\sqrt{2}}{2}) using algebra and trigonometry."]
