Let's break down the problem and solve it step by step: **Given:** - Position of Boat A at time \( t \): \( A(t) = (5t^2 + 4t + 3) \) - Position of Boat B at time \( t \): \( B(t) = (t^2 + 6t^2 + 4) \) - Units: miles (distance), hours (time) The question asks: **When are the boats closest to each other?** (You don't need to find the minimum distance, just the time.) --- ### Step 1: Clarify the positions It seems \( B(t) = (t^2 + 6t^2 + 4) \) has a typo. Let’s assume it means \( B(t) = (t^2 + 6t + 4) \). If not, please clarify! --- ### Step 2: Distance between the boats The distance between A and B at time \( t \) is: \[ D(t) = |A(t) - B(t)| \] Substitute the given expressions: \[ D(t) = |(5t^2 + 4t + 3) - (t^2 + 6t + 4)| \] \[ = |5t^2 + 4t + 3 - t^2 - 6t - 4| \] \[ = |(5t^2 - t^2) + (4t - 6t) + (3 - 4)| \] \[ = |4t^2 - 2t - 1| \] --- ### Step 3: Find the time when they are closest Since the distance is a quadratic function, its minimum occurs at the vertex. Let \( d(t) = 4t^2 - 2t - 1 \). The vertex of \( at^2 + bt + c \) is at: \[ t = -\frac{b}{2a} \] Here, \( a = 4, b = -2 \): \[ t = -\frac{-2}{2 \cdot 4} = \frac{2}{8} = \frac{1}{4} \] --- ### Step 4: Check the sign Since the quadratic opens upwards (positive \( 4t^2 \)), the minimum occurs at \( t = \frac{1}{4} \). --- ### **Final Answer:** \[ \boxed{\text{The boats are closest to each other at time } t = \frac{1}{4} \text{ hours.}} \]

The closest approach of the boats occurs at the vertex of the quadratic distance function, which is at \( t = \frac{1}{4} \) hours.