Find the values of \( x, y \) and \( z \) that correspond to the critical point of the function: \( f(x, y)=2 x^{2}-5 x-1 y+3 y^{2} \) Enter your answer as a number (like \( 5,-3,2.2 \) ) or as a calculation (like \( 5 / 3,2^{\wedge} 3,5+4 \) ). \( x= \) \( y= \) \( z= \)

To find the critical points of the function \( f(x, y) = 2x^{2} - 5x - 1y + 3y^{2} \), we first compute the partial derivatives with respect to \( x \) and \( y \), and then set them to zero: 1. Partial derivative with respect to \( x \): \[ f_x = \frac{\partial f}{\partial x} = 4x - 5 \] 2. Partial derivative with respect to \( y \): \[ f_y = \frac{\partial f}{\partial y} = -1 + 6y \] Next, we set both partial derivatives to zero to find the critical points: \[ 4x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{4} \] \[ -1 + 6y = 0 \quad \Rightarrow \quad 6y = 1 \quad \Rightarrow \quad y = \frac{1}{6} \] Finally, we can now find \( z \) by evaluating \( f \) at the critical values of \( x \) and \( y \): \[ f\left(\frac{5}{4}, \frac{1}{6}\right) = 2 \left(\frac{5}{4}\right)^{2} - 5 \left(\frac{5}{4}\right) - 1 \left(\frac{1}{6}\right) + 3 \left(\frac{1}{6}\right)^{2} \] Calculating \( z \): \[ = 2 \left(\frac{25}{16}\right) - \frac{25}{4} - \frac{1}{6} + 3 \left(\frac{1}{36}\right) \] \[ = \frac{50}{16} - \frac{100}{16} - \frac{8}{48} + \frac{3}{36} \] \[ = \frac{50 - 100 - \frac{4}{24} + \frac{8}{48}}{16} = \text{(After combining)} \frac{-8}{16} = -\frac{1}{2} \] Thus, we have: \( x = \frac{5}{4} \) \( y = \frac{1}{6} \) \( z = -\frac{1}{2} \) Final answer: \( x= \frac{5}{4} \) \( y= \frac{1}{6} \) \( z= -\frac{1}{2} \)