To find the maximum of \(f(y)\) on \( (0,1] \), take the derivative:

February 22, 2026

$To find the maximum of \(f(y)\) on \( (0,1] \), take the derivative:$

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Using the identity \((\sin \theta + \cos \theta)^2 = 1 + \sin 2\theta\), and \(\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta\), we rewrite it as:
Let \(y = \sin 2\theta\), where \(y \in (0, 1]\). Define:
f(y) = \frac{4(1 + y)}{y^2}.
f'(y) = 4 \cdot \frac{(1)(y^2) - (1 + y)(2y)}{y^4} = 4 \cdot \frac{y^2 - 2y(1 + y)}{y^4} = 4 \cdot \frac{y^2 - 2y - 2y^2}{y^4} = 4 \cdot \frac{-y^2 - 2y}{y^4} = -4 \cdot \
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