To find the maximum distance from the origin to the object on the time interval $[4, 15]$, we need to maximize the function:

$d(t) = \frac{100}{5 + 4 \sin(t)}$

Steps:

1. Find the derivative $d'(t)$: We'll use the quotient rule to differentiate $d(t)$. The quotient rule is:

   $\frac{d}{dt} \left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}$

   Here, $f(t) = 100$ (a constant) and $g(t) = 5 + 4 \sin(t)$. So:

   • $f'(t) = 0$
   • $g'(t) = 4 \cos(t)$

   Applying the quotient rule:

   $d'(t) = \frac{0 \cdot (5 + 4 \sin(t)) - 100 \cdot 4 \cos(t)}{(5 + 4 \sin(t))^2}$

   Simplifying:

   $d'(t) = \frac{-400 \cos(t)}{(5 + 4 \sin(t))^2}$

2. Set $d'(t) = 0$ to find critical points:

   The derivative is zero when the numerator is zero:

   $-400 \cos(t) = 0$

   This simplifies to:

   $\cos(t) = 0$

   The values of $t$ where $\cos(t) = 0$ are:

   $t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$

   In the interval $[4, 15]$, the relevant values of $t$ are approximately:

   $t = \frac{3\pi}{2} \approx 4.71 \quad \text{and} \quad t = \frac{7\pi}{2} \approx 10.99$

3. Check the endpoints and critical points:

   We now evaluate $d(t)$ at the critical points and the endpoints $t = 4$ and $t = 15$:

   • At $t = 4$:

     $d(4) = \frac{100}{5 + 4 \sin(4)} \approx \frac{100}{5 + 4( -0.7568)} = \frac{100}{5 - 3.0272} = \frac{100}{1.9728} \approx 50.68$

   • At $t = 15$:

     $d(15) = \frac{100}{5 + 4 \sin(15)} \approx \frac{100}{5 + 4(0.6503)} = \frac{100}{5 + 2.6012} = \frac{100}{7.6012} \approx 13.16$

   • At $t = \frac{3\pi}{2} \approx 4.71$:

     $d\left(\frac{3\pi}{2}\right) = \frac{100}{5 + 4 \sin\left(\frac{3\pi}{2}\right)} = \frac{100}{5 + 4(-1)} = \frac{100}{1} = 100$

   • At $t = \frac{7\pi}{2} \approx 10.99$:

     $d\left(\frac{7\pi}{2}\right) = \frac{100}{5 + 4 \sin\left(\frac{7\pi}{2}\right)} = \frac{100}{5 + 4(1)} = \frac{100}{9} \approx 11.11$

4. Conclusion:

   The maximum value of $d(t)$ in the interval $[4, 15]$ is at $t = \frac{3\pi}{2} \approx 4.71$, where the distance is:

   $d\left(\frac{3\pi}{2}\right) = 100$

Thus, the maximum distance from the origin to the object is 100 units.

Would you like further details on any step? Here are 5 related questions to explore:

1. How would the result change if the interval were extended beyond [4, 15]?
2. Can you explain why $\cos(t) = 0$ gives critical points?
3. What is the significance of the derivative in finding maximum values?
4. How does the function $\sin(t)$ influence the behavior of $d(t)$?
5. What role do endpoints play in optimization problems?

Tip: Always check both the critical points and the endpoints when looking for absolute maximum or minimum values on a closed interval.