1. Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point.

\[ f(x,y)=x^4-2x^2+y^3-27y-15\\ f_x(x,y)=4x^3-4x\\ 4x^3-4x=0\\ 4x(x+1)(x-1)=0\\ x=\{-1,0,1\}\\ f_y(x,y)=3y^2-27\\ 3y^2-27=0\\ y^2=9\\ y=\pm3\\ f(-1,-3)=(-1)^4-2(-1)^2+(-3)^3-27(-3)-15=38\\ f(-1,3)=(-1)^4-2(-1)^2+(3)^3-27(3)-15=-70\\ f(0,-3)=(0)^4-2(0)^2+(-3)^3-27(-3)-15=39\\ f(0,3)=(0)^4-2(0)^2+(3)^3-27(3)-15=-69\\ f(1,-3)=(1)^4-2(1)^2+(-3)^3-27(-3)-15=38\\ f(1,3)=(1)^4-2(1)^2+(3)^3-27(3)-15=-70\\ \]

(-1)^4-2*(-1)^2+(-3)^3-27*(-3)-15
## [1] 38
(-1)^4-2*(-1)^2+(3)^3-27*(3)-15
## [1] -70
(0)^4-2*(0)^2+(-3)^3-27*(-3)-15
## [1] 39
(0)^4-2*(0)^2+(3)^3-27*(3)-15
## [1] -69
(1)^4-2*(1)^2+(-3)^3-27*(-3)-15
## [1] 38
(1)^4-2*(1)^2+(3)^3-27*(3)-15
## [1] -70

Apply the second derivative test:

\[ f_x(x,y)=4x^3-4x\\ f_{xx}(x,y)=12x^2-4\\ f_{xy}(x,y)=0\\ f_y(x,y)=3y^2-27\\ f_{yy}(x,y)=6y\\ D=(12x^2-4)(6y)-0^2\\ D=(12x^2-4)(6y)\\ D=(12(-1)^2-4)(6(-3))\\ D=-144 \]

\((-1,-3,38)\) is a saddle point.

\[ D=(12(-1)^2-4)(6(3))\\ D=144 \] \((-1,3,-70)\) is a relative minimum.

\[ D=(12(0)^2-4)(6(-3))\\ D=72 \] \((0,-3,39)\) is a relative maximum.

\[ D=(12(0)^2-4)(6(3))\\ D=-72 \] \((0,3,-69)\) is a saddle point.

\[ D=(12(1)^2-4)(6(-3))\\ D=-144 \] \((1,-3,38)\) is a saddle point.

\[ D=(12(1)^2-4)(6(3))\\ D=144 \] \((1,3,-70)\) is a relative minimum.