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

Explanation:

Step1: Find the first - order partial derivatives

The first - order partial derivatives of (f(x,y)=-6x^{2}-5y^{2}+6x - 10y + 8) are: (f_{x}=\frac{\partial f}{\partial x}=-12x + 6) (f_{y}=\frac{\partial f}{\partial y}=-10y-10)

Step2: Find the critical points

Set (f_{x}=0) and (f_{y}=0) For (f_{x}=-12x + 6 = 0), we solve for (x): (-12x+6 = 0\Rightarrow12x = 6\Rightarrow x=\frac{1}{2}) For (f_{y}=-10y - 10=0), we solve for (y): (-10y-10 = 0\Rightarrow10y=-10\Rightarrow y=-1) So the critical point is ((\frac{1}{2},-1))

Step3: Find the second - order partial derivatives

(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=-12), (f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=-10), (f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=0)

Step4: Apply the Second Partials Test

The discriminant (D=f_{xx}f_{yy}-(f_{xy})^{2}) Substitute (f_{xx}=-12), (f_{yy}=-10), (f_{xy}=0) into the formula: (D=(-12)\times(-10)-0^{2}=120) Since (D = 120>0) and (f_{xx}=-12<0)

Step5: Calculate the (z) - value of the critical point

Substitute (x = \frac{1}{2}) and (y=-1) into (f(x,y)): (z=f(\frac{1}{2},-1)=-6\times(\frac{1}{2})^{2}-5\times(-1)^{2}+6\times\frac{1}{2}-10\times(-1)+8) (=-6\times\frac{1}{4}-5 + 3 + 10+8) (=-\frac{3}{2}-5 + 3 + 10+8) (=-\frac{3}{2}+16=\frac{-3 + 32}{2}=\frac{29}{2})

Answer:

relative minimum ((x,y,z)=(\frac{1}{2},-1,\frac{29}{2})) relative maximum: DNE saddle point: DNE