Given the equation $3x^2 - 4x + k = 0$ with real roots. The value of $k$ for which the product of the roots of the equation is a maximum is:
$\textbf{(A)}\ \frac{16}{9}\qquad \textbf{(B)}\ \frac{16}{3}\qquad \textbf{(C)}\ \frac{4}{9}\qquad \textbf{(D)}\ \frac{4}{3}\qquad \textbf{(E)}\ -\frac{4}{3}$
By [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas), the product of the roots is $\frac{k}{3}$.  This value increases as $k$ increases.
Also, the quadratic’s roots are real, then the [discriminant](https://artofproblemsolving.com/wiki/index.php/Discriminant) is greater than or equal to zero, so
\[16-12k \ge 0\]
\[-12k \ge -16\]
\[k \le \frac{4}{3}\]
Thus, the $k$ value that maximizes the product of the roots is $\frac{4}{3}$, which is answer choice $\boxed{\textbf{(D)}}$.