It is obvious that $\max\left\{ \frac{1}{x}, \frac{x}{y}, y \right\} = y$, and $\min\left\{ \frac{1}{x}, \frac{x}{y}, y \right\} = \begin{cases} \frac{1}{x}, & \text{if } y < x^2 \\ \frac{x}{y}, & \text{if } y \geq x^2 \end{cases}$.

1. When $y < x^2$, we have $t = \frac{y}{x}$. The feasible region is defined by $\begin{cases} 1 \leq x \leq y \\ y < x + 1 \\ y < x^2 \end{cases}$. Since the parabola $y = x^2$ intersects the lines $y = x$ and $y = x + 1$ within the first quadrant at the points $(1, 1)$ and $\left( \frac{1 + \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2} \right)$ respectively, we have $1 < t < \frac{1 + \sqrt{5}}{2}$.

2. When $y \geq x^2$, $t = x$. The feasible region is defined by $\begin{cases} 1 \leq x \leq y \\ y < x + 1 \\ y \geq x^2 \end{cases}$. Since the parabola $y = x^2$ intersects the lines $y = x$ and $y = x + 1$ within the first quadrant at the points $(1, 1)$ and $\left( \frac{1 + \sqrt{5}}{2}, \frac{3 + \sqrt{5}}{2} \right)$ respectively, we have $1 \leq t < \frac{1 + \sqrt{5}}{2}$.

In summary, the range of $t$ is $\boxed{[1, \frac{1 + \sqrt{5}}{2})}$.

This problem examines the range of $t$, involves knowledge of parabolas, and requires defining a new feasible region and finding the coordinates of the intersection points within the first quadrant as the key steps.