We are to find the range of the values of a real number $t$ such that the maximum value of the cubic function $$f(x) = \frac{1}{3}x^3 - \frac{t+2}{2}x^2 + 2tx + \frac{2}{3}$$ over the interval $x \leqq 4$ is greater than 6.
First of all, since the derivative of $f(x)$ is $$f'(x) = (x - \mathbf{A})(x - t),$$ we consider the problem by dividing the range of the values of $t$ as follows:
(i) When $t > \mathbf{A}$, $f(x)$ has a local maximum at $x = \mathbf{A}$ and a local minimum at $x = t$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(\mathbf{A}) > 6$.
(ii) When $t = \mathbf{A}$, the maximum value of $f(x)$ over the interval $x \leqq 4$ is $f(\mathbf{C}) = \mathbf{D}$, and hence the condition is not satisfied.
(iii) When $t < \mathbf{A}$, $f(x)$ has a local maximum at $x = t$ and a local minimum at $x = \mathbf{A}$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(t) > 6$.
Here, we note $$f(t) - 6 = -\frac{1}{6}(t + \mathbf{E})(t - \mathbf{EF})^2.$$
From the above, the range of the values of $t$ is to be determined.