find all relative extrema and saddle points of the function. use the second partials test where applicable…

find all relative extrema and saddle points of the function. use the second partials test where applicable. (if an answer does not exist, enter dne.)\n$z = x ^ { 2 } + x y + \\frac { 1 } { 2 } y ^ { 2 } - 6 x + y$\nrelative minimum\n$( x , y , z ) = ( quad )$\nrelative maximum\n$( x , y , z ) = ( quad )$\nsaddle point\n$( x , y , z ) = ( quad )$
Answer
Explanation:
Step1: Find first - order partial derivatives
The function is (z = x^{2}+xy+\frac{1}{2}y^{2}-6x + y). The first - order partial derivative with respect to (x) is: (z_{x}=\frac{\partial z}{\partial x}=2x + y-6) The first - order partial derivative with respect to (y) is: (z_{y}=\frac{\partial z}{\partial y}=x + y + 1)
Step2: Find critical points
Set (z_{x}=0) and (z_{y}=0). We have the system of equations: (\begin{cases}2x + y-6=0\x + y+1 = 0\end{cases}) Subtract the second equation from the first equation: ((2x + y-6)-(x + y + 1)=0) (2x + y-6-x - y-1=0) (x-7 = 0), so (x = 7) Substitute (x = 7) into (x + y+1 = 0), we get (7 + y+1=0), so (y=-8) The critical point is ((7,-8))
Step3: Find second - order partial derivatives
(z_{xx}=\frac{\partial^{2}z}{\partial x^{2}}=2) (z_{xy}=\frac{\partial^{2}z}{\partial x\partial y}=1) (z_{yy}=\frac{\partial^{2}z}{\partial y^{2}}=1)
Step4: Apply the Second Partials Test
The discriminant (D=z_{xx}z_{yy}-(z_{xy})^{2}) Substitute (z_{xx}=2), (z_{xy}=1), (z_{yy}=1) into the formula: (D=(2\times1)-1^{2}=2 - 1=1) Since (z_{xx}=2>0) and (D = 1>0)
Answer:
relative minimum ((x,y,z)=(7,-8,z(7,-8))) First, find (z(7,-8)): (z(7,-8)=7^{2}+7\times(-8)+\frac{1}{2}\times(-8)^{2}-6\times7+( - 8)) (=49-56 + 32-42-8) (=49+32-(56 + 42+8)) (=81 - 106=-25) relative minimum ((x,y,z)=(7,-8,-25)) saddle point ((x,y,z)=\text{DNE}) relative maximum ((x,y,z)=\text{DNE})