The maximum value of f(x) = x√(1 - x) on the interval [0, 1] is

The correct answer is Option (1) → \(\displaystyle\frac{2\sqrt{3}}{9}\)

By the product rule, for 0 < x < 1,

    \( \displaystyle f'(x) = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} \)

    \( \displaystyle = \frac{2(1 - x) - x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}} \)

So f′(x) = 0 at \(x = \frac{2}{3}\). The values at this point and at the end points are

    \( f(0) = 0 \), \( f(1) = 0 \)

    \( \displaystyle f\left(\frac{2}{3}\right) = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9} \)

So the maximum value is \(\frac{2\sqrt{3}}{9} \approx 0.385\).

Why the other options are incorrect

• Option (2) \(\displaystyle\frac{\sqrt{2}}{4}\) — Incorrect. This is \(f\left(\frac{1}{2}\right)\), about 0.354, smaller than \(f\left(\frac{2}{3}\right)\).

• Option (3) \(\displaystyle\frac{1}{3}\) — Incorrect. It is about 0.333, below the value 0.385 that f reaches at \(x = \frac{2}{3}\).

• Option (4) \(\displaystyle\frac{2}{3}\) — Incorrect. This is the point of maximum, not the maximum value.