Solution: To find the maximum value of $ f(t) = t^2 - \frac{t^4}{4} $, take the derivative $ f'(t) = 2t - t^3 $. Set $ f'(t) = 0 $: $ 2t - t^3 = 0 \Rightarrow t(2 - t^2) = 0 $. Critical points at $ t = 0 $ and $ t = \pm\sqrt{2} $. Evaluate $ f(\sqrt{2}) = (\sqrt{2})^2 - \frac{(\sqrt{2})^4}{4} = 2 - \frac{4}{4} = 1 $. The maximum value is $ \boxed{1} $.

Sun, 03 May 2026 09:00:17 +0000

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