A graphing calculator is recommended. When a foreign object that is lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward, causing an increase in pressure in the lungs. At the same time, the trachea contracts, causing the expelled air to move faster and increasing the pressure on the foreign object. According to a mathematical model of coughing, the velocity \( v \) (in \( \mathrm{cm} / \mathrm{s} \) ) of the airstream through an average-sized person's trachea is related to the radius \( r \) of the trachea (in cm) by the function \[ v(r)=3.1(1-r) r^{2} \quad \frac{1}{2} \leq r \leq 1 . \] Determine the value of \( r \) for which \( v \) is a maximum. (Round your answers to two decimal places.) \( r=\square \mathrm{cm} \)

To find the value of \( r \) that maximizes the velocity \( v(r) \), we first need to derive the function and find its critical points. Given the function: \[ v(r) = 3.1(1 - r)r^2 \] We'll differentiate \( v(r) \) with respect to \( r \) and then set the derivative equal to zero to find critical points. The derivative is found using the product rule: \[ v'(r) = 3.1 \left( (1 - r) \cdot 2r + r^2 \cdot (-1) \right) \] This simplifies to: \[ v'(r) = 3.1 \left( 2r(1 - r) - r^2 \right) \] \[ v'(r) = 3.1 \left( 2r - 2r^2 - r^2 \right) \] \[ v'(r) = 3.1 \left( 2r - 3r^2 \right) \] Setting the derivative to zero to find critical points: \[ 2r - 3r^2 = 0 \] Factoring out \( r \): \[ r(2 - 3r) = 0 \] Thus, the critical points are: \[ r = 0 \quad \text{or} \quad r = \frac{2}{3} \] Now, since we are constrained by \( \frac{1}{2} \leq r \leq 1 \), we only consider \( r = \frac{2}{3} \). Next, we confirm it's a maximum by evaluating the second derivative or testing points in the original function. However, since \( r = 0 \) is outside our interval, we know the maximum must occur at \( r = \frac{2}{3} \). Calculating \( r \): \[ r = \frac{2}{3} \approx 0.67 \] Thus, rounding to two decimal places, we find: \( r = 0.67 \, \mathrm{cm} \)