Given \(u = 2x(y - x)\) and \(x + 3y = 12\).

First, express \(y\) in terms of \(x\):

\(x + 3y = 12\)

\(3y = 12 - x\)

\(y = \frac{12 - x}{3}\)

Substitute \(y\) in the expression for \(u\):

\(u = 2x\left(\frac{12 - x}{3} - x\right)\)

\(u = 2x\left(\frac{12 - x - 3x}{3}\right)\)

\(u = 2x\left(\frac{12 - 4x}{3}\right)\)

\(u = \frac{2x(12 - 4x)}{3}\)

\(u = \frac{24x - 8x^2}{3}\)

\(u = 8x - \frac{8x^2}{3}\)

To find the stationary value, differentiate \(u\) with respect to \(x\):

\(\frac{du}{dx} = 8 - \frac{16x}{3}\)

Set \(\frac{du}{dx} = 0\):

\(8 - \frac{16x}{3} = 0\)

\(8 = \frac{16x}{3}\)

\(24 = 16x\)

\(x = \frac{3}{2}\)

Substitute \(x = \frac{3}{2}\) back into the equation for \(y\):

\(y = \frac{12 - \frac{3}{2}}{3}\)

\(y = \frac{21}{6}\)

\(y = \frac{7}{2}\)

Substitute \(x = \frac{3}{2}\) and \(y = \frac{7}{2}\) back into the expression for \(u\):

\(u = 2 \times \frac{3}{2} \times \left(\frac{7}{2} - \frac{3}{2}\right)\)

\(u = 3 \times 2\)

\(u = 6\)