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 } - 7 x + y )\nrelative minimum ( ( x , y , z ) = )\nrelative maximum ( ( x , y , z ) = )\nsaddle point ( ( x , y , z ) = )

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 } - 7 x + y )\nrelative minimum ( ( x , y , z ) = )\nrelative maximum ( ( x , y , z ) = )\nsaddle point ( ( x , y , z ) = )

Answer

Explanation:

Step1: Find first - order partial derivatives

The function is (z = x^{2}+xy+\frac{1}{2}y^{2}-7x + y). The first - order partial derivatives: (z_{x}=\frac{\partial z}{\partial x}=2x + y-7) (z_{y}=\frac{\partial z}{\partial y}=x + y + 1)

Step2: Solve the system of equations (z_{x}=0) and (z_{y}=0)

Set (z_{x}=0): (2x + y-7 = 0), so (y=7 - 2x) Substitute (y = 7-2x) into (z_{y}=0): (x+(7 - 2x)+1=0) (x + 7-2x + 1=0) (-x+8 = 0), then (x = 8) Substitute (x = 8) into (y=7 - 2x), we get (y=7-2\times8=-9) The critical point is ((8,-9))

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: Use 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)\times(1)-(1)^{2}=2 - 1=1) Since (z_{xx}=2>0) and (D = 1>0)

Answer:

relative minimum ((x,y,z)=(8,-9,-36)); relative maximum: DNE; saddle point: DNE