A function \( f \) and a point \( P \) are given. Let \( \theta \) correspond to the direction of the directional derivative. Complete parts (a) through (e). \( f(x, y)=16-3 x^{2}-2 y^{2}, P(2,3) \) a. Find the gradient and evaluate it at \( P \). The gradient at \( P \) is \( \langle-12,-12) \). b. Find the angles \( \theta \) (with respect to the positive \( x \)-axis) associated with the directions of maximum increase, maximum decrease, and zero change. What angle(s) is/are associated with the direction of maximum increase? \( \frac{5 \pi}{4} \) (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.) What angle(s) is/are associated with the direction of maximum decrease? \( \square \) (Type any angles in radians between 0 and \( 2 \pi \). Type an exact answer, using \( \pi \) as needed. Use a comma to separate answers as needed.)

To find the angles associated with maximum increase and maximum decrease for the function \( f(x, y) = 16 - 3x^2 - 2y^2 \) evaluated at point \( P(2,3) \), we start with the gradient, which is given as \( \langle -12, -12 \rangle \). The direction of maximum increase occurs along the direction of the gradient, so we need to find the angle corresponding to this vector. The angle \( \theta \) can be found using: \[ \tan(\theta) = \frac{y}{x} = \frac{-12}{-12} = 1. \] Thus, \( \theta = \frac{5\pi}{4} \) corresponds to the maximum increase direction (which is the direction opposite to \( \frac{3\pi}{4} \) in quadrant III). For maximum decrease, this occurs in the opposite direction of the gradient, which corresponds to rotating by \( \pi \) radians to obtain the angle for maximum decrease: \[ \theta = \frac{5\pi}{4} + \pi = \frac{9\pi}{4} \equiv \frac{\pi}{4} \quad(\text{mod } 2\pi). \] Thus, the angle associated with the direction of maximum decrease is \( \frac{\pi}{4} \). So the answers are: Maximum Increase: \( \frac{5\pi}{4} \) Maximum Decrease: \( \frac{\pi}{4} \)