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 \( 5 / 3,2^{\wedge} 3,5+4 \) ). \( x= \) \( y= \) \( z= \)

We begin by finding the critical point by setting the partial derivatives equal to zero. 1. The partial derivative with respect to \( x \) is: \[ f_x(x,y)=\frac{\partial}{\partial x}\left(2x^2-5x-y+3y^2\right)=4x-5 \] Set it equal to zero: \[ 4x-5=0 \quad \Longrightarrow \quad x=\frac{5}{4} \] 2. The partial derivative with respect to \( y \) is: \[ f_y(x,y)=\frac{\partial}{\partial y}\left(2x^2-5x-y+3y^2\right)=-1+6y \] Set it equal to zero: \[ -1+6y=0 \quad \Longrightarrow \quad y=\frac{1}{6} \] Since the graph of the function is represented by \( z=f(x,y) \), we substitute the values of \( x \) and \( y \) to find \( z \): \[ z=f\left(\frac{5}{4},\frac{1}{6}\right)=2\left(\frac{5}{4}\right)^2-5\left(\frac{5}{4}\right)-\frac{1}{6}+3\left(\frac{1}{6}\right)^2 \] Compute each term: - Compute \( \left(\frac{5}{4}\right)^2 \): \[ \left(\frac{5}{4}\right)^2=\frac{25}{16} \] So, \[ 2\left(\frac{25}{16}\right)=\frac{50}{16}=\frac{25}{8} \] - Compute: \[ 5\left(\frac{5}{4}\right)=\frac{25}{4} \] - Compute: \[ 3\left(\frac{1}{6}\right)^2=3\left(\frac{1}{36}\right)=\frac{3}{36}=\frac{1}{12} \] Now substitute back: \[ z=\frac{25}{8}-\frac{25}{4}-\frac{1}{6}+\frac{1}{12} \] To combine these, express each term with a common denominator. The least common denominator (LCD) for \( 8, 4, 6, 12 \) is \( 24 \): - Convert \(\frac{25}{8}\): \[ \frac{25}{8}=\frac{75}{24} \] - Convert \(\frac{25}{4}\): \[ \frac{25}{4}=\frac{150}{24} \] - Convert \(\frac{1}{6}\): \[ \frac{1}{6}=\frac{4}{24} \] - Convert \(\frac{1}{12}\): \[ \frac{1}{12}=\frac{2}{24} \] Thus, \[ z=\frac{75}{24}-\frac{150}{24}-\frac{4}{24}+\frac{2}{24}=\frac{75-150-4+2}{24}=\frac{-77}{24} \] The critical point is then: \[ x=\frac{5}{4}, \quad y=\frac{1}{6}, \quad z=\frac{-77}{24} \]